Learn4Less Logo

Navigation

Multiple Integrals

3.1 Double Integrals

29 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

For each of the following, evaluate the given double integral without using iteration. Instead, interpret the integral as, for example, an area or a volume.

  1. ∫−13∫−41dy dx\dst\int_{-1}^3\int_{-4}^1 \dee{y}\,\dee{x}

  2. ∫02∫04−y2dx dy\dst\int_0^2\int_0^{\sqrt{4-y^2}} \dee{x}\,\dee{y}

  3. ∫−33∫09−y29−x2−y2 dx dy\dst\int_{-3}^3\int_0^{\sqrt{9-y^2}}\sqrt{9-x^2-y^2}\ \dee{x}\,\dee{y}

Answer

(a) 2020 (b) π\pi (c) 9π9\pi

Full solution

(a) The given double integral ∫−13∫−41dy dx=∬Rdx dy\int_{-1}^3\int_{-4}^1 \dee{y}\,\dee{x}=\dblInt_R \dee{x}\,\dee{y} where

R={ (x,y) ∣ −1≤x≤3, −4≤y≤1 }\begin{equation*} R=\Set{(x,y)}{ -1\le x\le 3,\ -4\le y\le1} \end{equation*}

and so the integral is the area of a rectangle with sides of lengths 44 and 55. Thus ∫−13∫−41dy dx=4×5=20\int_{-1}^3\int_{-4}^1 \dee{y}\,\dee{x}=4\times 5=20.

(b) The given double integral ∫02∫04−y2dx dy=∬Rdx dy\dst\int_0^2\int_0^{\sqrt{4-y^2}} \dee{x}\,\dee{y} =\dst\dblInt_R \dee{x}\,\dee{y} where

R={ (x,y) ∣ 0≤y≤2, 0≤x≤4−y2 }={ (x,y) ∣ x≥0, y≥0, y≤2, x2+y2≤4 }\begin{align*} R&=\Set{(x,y)}{ 0\le y\le 2,\ 0\le x\le\sqrt{4-y^2}} \\ &=\Set{(x,y)}{ x\ge 0,\ y\ge 0,\ y\le 2,\ x^2+y^2\le 4} \end{align*}

So RR is the first quadrant part of the circular disk of radius 22 centred on (0,0)(0,0). The area of the full disk is π 22=4π\pi\,2^2=4\pi. The given integral is one quarter of that, which is π\pi.

(c) The given double integral ∫−33∫09−y29−x2−y2 dx dy=∬Rz(x,y) dx dy\dst\int_{-3}^3\int_0^{\sqrt{9-y^2}}\sqrt{9-x^2-y^2}\ \dee{x}\,\dee{y} =\dst\dblInt_{\cR} z(x,y)\ \dee{x}\,\dee{y} where z(x,y)=9−x2−y2z(x,y)=\sqrt{9-x^2-y^2} and

R={ (x,y) ∣ −3≤y≤3, 0≤x≤9−y2 }={ (x,y) ∣ x≥0, −3≤y≤3,  x2+y2≤9 }\begin{align*} \cR&=\Set{(x,y)}{ -3\le y\le 3,\ 0\le x\le\sqrt{9-y^2}} \\ &=\Set{(x,y)}{ x\ge 0,\ -3\le y\le 3,\ \ x^2+y^2\le 9} \end{align*}

So R\cR is the right half of the circular disk of radius 33 centred on (0,0)(0,0). By Equation (3.1.9) in the CLP-3 text, the given integral is the volume of the solid

V={ (x,y,z) ∣ (x,y)∈R, 0≤z≤9−x2−y2 }={ (x,y,z) ∣ (x,y)∈R, z≥0, x2+y2+z2≤9 }\begin{align*} \cV&=\Big\{\,(x,y,z)\,\Big|\,(x,y)\in\cR,\ 0\le z\le \sqrt{9-x^2-y^2}\,\Big\}\\ &=\Big\{\,(x,y,z)\,\Big|\,(x,y)\in\cR,\ z\ge 0,\ x^2+y^2+z^2\le 9\,\Big\} \end{align*}

Thus V\cV is the one quarter of the spherical ball of radius 33 and centre (0,0,0)(0,0,0) with x≥0x\ge 0 and z≥0z\ge 0. So

∫−33∫09−y29−x2−y2 dx dy=14(43π33)=9π\begin{equation*} \int_{-3}^3\int_0^{\sqrt{9-y^2}}\sqrt{9-x^2-y^2}\ \dee{x}\,\dee{y} =\frac{1}{4}\Big(\frac{4}{3}\pi 3^3\Big) =9\pi \end{equation*}
Q2Stage 1

Let f(x,y)=12x2y3f(x,y)= 12 x^2y^3. Evaluate

  1. ∫03f(x,y) dx\dst\int_0^3 f(x,y)\,\dee{x}

  2. ∫02f(x,y) dy\dst\int_0^2 f(x,y)\,\dee{y}

  3. ∫02∫03f(x,y) dx dy\dst\int_0^2\int_0^3 f(x,y)\,\dee{x}\,\dee{y}

  4. ∫03∫02f(x,y) dy dx\dst\int_0^3\int_0^2 f(x,y)\,\dee{y}\,\dee{x}

  5. ∫03∫02f(x,y) dx dy\dst\int_0^3\int_0^2 f(x,y)\,\dee{x}\,\dee{y}

Hint

Be careful to match each integration variable with its own limits. Remember that the integral with respect to xx treats yy as a constant and the integral with respect to yy treats xx as a constant.

Answer

(a) 108y3108 y^3 (b) 48x248x^2 (c), (d) 432432 (e) 648648

Full solution

(a) The integral with respect to xx treats yy as a constant. So

∫03f(x,y) dx=∫0312x2y3 dx=[4x3y3]x=0x=3=108y3\begin{align*} \int_0^3 f(x,y)\,\dee{x} &= \int_0^3 12 x^2y^3\,\dee{x} = \Big[4x^3y^3\Big]_{x=0}^{x=3} = 108 y^3 \end{align*}

(b) The integral with respect to yy treats xx as a constant. So

∫02f(x,y) dy=∫0212x2y3 dy=[3x2y4]y=0y=2=48x2\begin{align*} \int_0^2 f(x,y)\,\dee{y} &= \int_0^2 12 x^2y^3\,\dee{y} = \Big[3x^2y^4\Big]_{y=0}^{y=2} = 48x^2 \end{align*}

(c) By part (a)

∫02∫03f(x,y) dx dy=∫02[∫03f(x,y) dx]dy=∫02108y3 dy=[27y4]y=0y=2=27×16=432\begin{align*} \int_0^2\int_0^3 f(x,y)\,\dee{x}\,\dee{y} &=\int_0^2\left[\int_0^3 f(x,y)\,\dee{x}\right]\dee{y} =\int_0^2 108y^3\,\dee{y} =\Big[27y^4\Big]_{y=0}^{y=2} \\ &=27\times 16 = 432 \end{align*}

(d) By part (b)

∫03∫02f(x,y) dy dx=∫03[∫02f(x,y) dy]dx=∫0348x2 dy=[16x3]x=0x=3=16×27=432\begin{align*} \int_0^3\int_0^2 f(x,y)\,\dee{y}\,\dee{x} &=\int_0^3\left[\int_0^2 f(x,y)\,\dee{y}\right]\dee{x} =\int_0^3 48x^2\,\dee{y} =\Big[16x^3\Big]_{x=0}^{x=3} \\ &=16\times 27 = 432 \end{align*}

(e) This time

∫03∫02f(x,y) dx dy=∫03[∫0212x2y3 dx]dy=∫03[4x3y3]02dy=∫0332y3dy=[8y4]03=8×81=648\begin{align*} \int_0^3\int_0^2 f(x,y)\,\dee{x}\,\dee{y} &=\int_0^3\left[\int_0^2 12 x^2y^3\,\dee{x}\right]\dee{y} =\int_0^3\left[4 x^3y^3\right]_0^2\dee{y} =\int_0^3 32 y^3\dee{y} \\ & =\Big[8y^4\Big]_0^3 =8\times 81 = 648 \end{align*}

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Questions 3 through 8 provide practice with limits of integration for double integrals in Cartesian coordinates.

Q3Stage 2

For each of the following, evaluate the given double integral using iteration.

  1. ∬R(x2+y2) dx dy\displaystyle\dblInt_R (x^2+y^2)\,\dee{x}\,\dee{y} where RR is the rectangle 0≤x≤a, 0≤y≤b0\le x\le a,\ 0\le y\le b where a>0a>0 and b>0b>0.

  2. ∬T(x−3y) dx dy\displaystyle\dblInt_T (x-3y)\,\dee{x}\,\dee{y} where TT is the triangle with vertices (0,0), (a,0), (0,b)(0,0),\ (a,0),\ (0,b).

  3. ∬Rxy2 dx dy\displaystyle\dblInt_R xy^2\,\dee{x}\,\dee{y} where RR is the finite region in the first quadrant bounded by the curves y=x2y=x^2 and x=y2x=y^2.

  4. ∬Dxcos⁡y dx dy\displaystyle\dblInt_D x\cos y\,\dee{x}\,\dee{y} where DD is the finite region in the first quadrant bounded by the coordinate axes and the curve y=1−x2y=1-x^2.

  5. ∬Rxyey dx dy\displaystyle\dblInt_R {x\over y}e^y\,\dee{x}\,\dee{y} where RR is the region 0≤x≤1, x2≤y≤x0\le x\le 1,\ x^2\le y\le x.

  6. ∬Txy1+x4 dx dy\displaystyle\dblInt_T {xy\over 1+x^4}\,\dee{x}\,\dee{y} where TT is the triangle with vertices (0,0), (0,1), (1,1)(0,0),\ (0,1),\ (1,1).

Answer

(a) 13(a3b+ab3)\frac{1}{3}\big(a^3b+ab^3\big) (b) a2b6−ab22\frac{a^2b}{6}-\frac{ab^2}{2} (c) 356\frac{3}{56} (d) 12(1−cos⁡1)\frac{1}{2}(1-\cos 1) (e) 12(e−2)\frac{1}{2}(e-2)

(f) 14(π4−12ln⁡2)\frac{1}{4}\left(\frac{\pi}{4}-\frac{1}{2}\ln 2\right)

Full solution

The following figures show the domains of integration for the integrals in this problem.

(a)

Figure from prob_s3.1, line 220

Figure from prob_s3.1, line 220

(b)

Figure from prob_s3.1, line 220

Figure from prob_s3.1, line 220

(c)

Figure from prob_s3.1, line 220

Figure from prob_s3.1, line 220

(d)

Figure from prob_s3.1, line 220

Figure from prob_s3.1, line 220

(e)

Figure from prob_s3.1, line 220

Figure from prob_s3.1, line 220

(f)

Figure from prob_s3.1, line 220

Figure from prob_s3.1, line 220

∬R(x2+y2) dx dy=∫0adx∫0bdy (x2+y2)=∫0adx (x2b+13b3)=13(a3b+ab3)[0]∬T(x−3y) dx dy=∫0adx∫0b(1−xa)dy (x−3y)=∫0adx [bx(1−xa)−32b2(1−xa)2]=[b2x2−b3ax3+a2b2(1−xa)3]0a=a2b2−a2b3−ab22=a2b6−ab22[0]∬Rxy2 dx dy=∫01dx∫x2xdy xy2=13∫01dx x(x3/2−x6)=13(27−18)=356[0]∬Dxcos⁡y dx dy=∫01dx∫01−x2dy xcos⁡y=∫01dx xsin⁡(1−x2)=12[cos⁡(1−x2)]01=12(1−cos⁡1)[0]∬Rxyey dx dy=∫01dy∫yydx xyey=∫01dy y−y22yey=12∫01dy (1−y)ey=12[−yey+2ey]01=12(e−2)[0]∬Txy1+x4 dx dy=∫01dx∫x1dy xy1+x4=12∫01dx x(1−x2)1+x4=14∫01dt 1−t1+t2 where t=x2=14[arctan⁡t−12ln⁡(1+t2)]01=14(π4−12ln⁡2)\begin{align*} \dblInt_R (x^2+y^2)\,\dee{x}\,\dee{y} &=\int_0^a \dee{x}\int_0^b\dee{y}\ (x^2+y^2) =\int_0^a \dee{x}\,\left(x^2b+\frac{1}{3}b^3\right) \tag{a} \\ &=\frac{1}{3}\big(a^3b+ab^3\big) \notag [0]\\ {} \dblInt_T (x-3y)\,\dee{x}\,\dee{y} &=\int_0^a \dee{x}\int_0^{b(1-{x\over a})}\dee{y}\,(x-3y) \tag{b} \\ &=\int_0^a \dee{x}\,\left[bx\left(1-\frac{x}{a}\right) -\frac{3}{2}b^2\left(1-\frac{x}{a}\right)^2\right] \notag \\ &=\left[\frac{b}{2}x^2-\frac{b}{3a}x^3 +\frac{a}{2}b^2\left(1-\frac{x}{a}\right)^3\right]_0^a \notag \\ &=\frac{a^2b}{2}-\frac{a^2b}{3}-\frac{ab^2}{2} =\frac{a^2b}{6}-\frac{ab^2}{2} \notag [0]\\ {} \dblInt_R xy^2\,\dee{x}\,\dee{y} &=\int_0^1 \dee{x}\int_{x^2}^{\sqrt{x}}\dee{y}\ xy^2 =\frac{1}{3}\int_0^1 \dee{x}\ x\big(x^{3/2}-x^6 \big) =\frac{1}{3}\left(\frac{2}{7}-\frac{1}{8}\right) \tag{c} \\ &=\frac{3}{56} \notag [0]\\ \dblInt_D x\cos y\,\dee{x}\,\dee{y} &=\int_0^1 \dee{x}\int_{0}^{1-x^2}\dee{y}\ x\cos y =\int_0^1 \dee{x}\ x\sin(1-x^2) \tag{d} \\ &=\frac{1}{2}\Big[\cos(1-x^2)\Big]_0^1 =\frac{1}{2}(1-\cos 1) \notag [0]\\ {} \dblInt_R \frac{x}{y}e^y\,\dee{x}\,\dee{y} &=\int_0^1 \dee{y}\int_{y}^{\sqrt{y}}\dee{x}\ \frac{x}{y}e^y =\int_0^1 \dee{y}\ \frac{y-y^2}{2y}e^y =\frac{1}{2}\int_0^1 \dee{y}\ (1-y)e^y \tag{e} \\ &=\frac{1}{2}\Big[-ye^y+2e^y\Big]_0^1 =\frac{1}{2}(e-2) \notag [0]\\ {} \dblInt_T \frac{xy}{1+x^4}\,\dee{x}\,\dee{y} &=\int_0^1 \dee{x}\int_{x}^{1}\dee{y}\ \frac{xy}{1+x^4} =\frac{1}{2}\int_0^1 \dee{x}\ \frac{x(1-x^2)}{1+x^4} \tag{f} \\ &=\frac{1}{4}\int_0^1 \dee{t}\ \frac{1-t}{1+t^2}\hbox{ where $t=x^2$} \notag \\ &=\frac{1}{4}\left[\arctan t-\frac{1}{2}\ln(1+t^2)\right]_0^1 =\frac{1}{4}\left(\frac{\pi}{4}-\frac{1}{2}\ln 2\right) \notag \end{align*}
Q4Stage 2

For each of the following integrals (i) sketch the region of integration, (ii) write an equivalent double integral with the order of integration reversed and (iii) evaluate both double integrals.

  1. ∫02dx∫1exdy\displaystyle\int_0^2\dee{x}\int_1^{e^x}\dee{y}

  2. ∫02dy∫−4−2y24−2y2dx y\displaystyle\int_0^{\sqrt{2}}\dee{y} \int_{-\sqrt{4-2y^2}}^{\sqrt{4-2y^2}}\dee{x}\ y

  3. ∫−21dx∫x2+4x3x+2dy\displaystyle\int_{-2}^1 \dee{x}\int_{x^2+4x}^{3x+2}\dee{y}

Answer

(a) e2−3e^2-3 (b) 83\frac{8}{3} (c) 92\frac{9}{2}

Full solution

The following figures show the domains of integration for the integrals in this problem.

(a)

Figure from prob_s3.1, line 320

Figure from prob_s3.1, line 320

(b)

Figure from prob_s3.1, line 320

Figure from prob_s3.1, line 320

(c)

Figure from prob_s3.1, line 320

Figure from prob_s3.1, line 320

∫02dx∫1exdy=∫02dx [ex−1]=[ex−x]02=e2−3∫1e2dy∫ln⁡y2dx=∫1e2dy [2−ln⁡y]=[2y−yln⁡y+y]1e2=e2−3[0]∫02dy∫−4−2y24−2y2dx y=∫02dy 2y4−2y2=−13[(4−2y2)3/2]02=83∫−22dx∫02−x22dy y=∫−22dx [1−x24]=2∫02dx [1−x24]=2[x−x312]02=83[0]∫−21dx∫x2+4x3x+2dy=∫−21dx [−x2−x+2]=[−x33−x22+2x]−21=92∫−45dy∫y−23−2+4+ydx=∫−45dy [ ⁣−43−y3+4+y]=[−4y3−y26+23(4+y)32]−45=92\begin{align*} \int_0^2 \dee{x}\int_1^{e^x} \dee{y} &=\int_0^2 \dee{x}\ \big[e^x-1\big] =\big[e^x-x\big]_0^2 =e^2-3 \tag{a} \\ \int_1^{e^2} \dee{y}\int_{\ln y}^2 \dee{x} &=\int_1^{e^2} \dee{y}\ \big[2-\ln y\big] =\big[2y-y\ln y+y\big]_1^{e^2} =e^2-3\notag [0]\\ {} \int_0^{\sqrt{2}}\dee{y}\int_{-\sqrt{4-2y^2}}^{\sqrt{4-2y^2}}\dee{x}\ y &=\int_0^{\sqrt{2}}\dee{y}\ 2y\sqrt{4-2y^2} =-\frac{1}{3}\Big[{(4-2y^2)}^{3/2}\Big]_0^{\sqrt{2}} =\frac{8}{3} \tag{b} \\ \int_{-2}^{2} \dee{x}\int_0^{\sqrt{2-{x^2\over 2}}} \dee{y}\ y &=\int_{-2}^{2} \dee{x}\ \left[1-\frac{x^2}{4}\right] =2\int_0^{2} \dee{x}\ \left[1-\frac{x^2}{4}\right] =2\left[x-\frac{x^3}{12}\right]_0^{2} =\frac{8}{3}\notag[0]\\ {} \int_{-2}^1 \dee{x}\int_{x^2+4x}^{3x+2}\dee{y} &=\int_{-2}^1 \dee{x}\ \big[-x^2-x+2\big] =\left[-\frac{x^3}{3}-\frac{x^2}{2}+2x\right]_{-2}^1 =\frac{9}{2} \tag{c} \\ \int_{-4}^{5} \dee{y}\int_{{y-2\over 3}}^{-2+\sqrt{4+y}} \dee{x} &=\int_{-4}^{5} \dee{y}\ \left[\!-\frac{4}{3}-\frac{y}{3}+\sqrt{4+y}\right] =\left[-\frac{4y}{3}-\frac{y^2}{6}+\frac{2}{3}{(4+y)}^{3\over 2}\right]_{-4}^{5} \notag \\ &=\frac{9}{2}\notag \end{align*}

In part (c), we used that the equation y=x2+4xy=x^2+4x is equivalent to y+4=(x+2)2y+4=(x+2)^2 and hence to x=−2±y+4x=-2\pm\sqrt{y+4}.

Q5Stage 2Past exam · M200 2006A

Combine the sum of the two iterated double integrals

∫y=0y=1∫x=0x=yf(x,y) dx dy+∫y=1y=2∫x=0x=2−yf(x,y) dx dy\begin{equation*} \int_{y=0}^{y=1}\int_{x=0}^{x=y} f(x,y)\ \dee{x}\,\dee{y} +\int_{y=1}^{y=2}\int_{x=0}^{x=2-y} f(x,y)\ \dee{x}\,\dee{y} \end{equation*}

into a single iterated double integral with the order of integration reversed.

Answer

∫x=0x=1∫y=xy=2−xf(x,y) dy dx\int_{x=0}^{x=1}\int_{y=x}^{y=2-x} f(x,y)\ \dee{y}\,\dee{x}

Full solution

In the given integrals

  • yy runs for 00 to 22, and

  • for each fixed yy between 00 and 11, xx runs from 00 to yy and

  • for each fixed yy between 11 and 22, xx runs from 00 to 2−y2-y

The figure on the left below contains a sketch of that region together with the generic horizontal slices that were used to set up the given integrals.

Figure from prob_s3.1, line 391

Figure from prob_s3.1, line 391

Figure from prob_s3.1, line 391

Figure from prob_s3.1, line 391

To reverse the order of integration, we switch to vertical, rather than horizontal, slices, as in the figure on the right above. Looking at that figure, we see that

  • xx runs for 00 to 11, and

  • for each fixed xx in that range, yy runs from xx to 2−x2-x.

So the desired integral is

∫x=0x=1∫y=xy=2−xf(x,y) dy dx\begin{align*} \int_{x=0}^{x=1}\int_{y=x}^{y=2-x} f(x,y)\ \dee{y}\,\dee{x} \end{align*}
Q6Stage 2Past exam · M200 2004A

Consider the integral

∫01∫x1ex/y dy dx\begin{equation*} \int_0^1\int_x^1 e^{x/y}\ \dee{y}\ \dee{x} \end{equation*}

  1. Sketch the domain of integration.

  2. Evaluate the integral by reversing the order of integration.

Answer

(a)

Figure from prob_s3.1, line 446

Figure from prob_s3.1, line 446

(b) 12(e−1)\half(e-1)

Full solution

(a) In the given integral

  • xx runs from 00 to 11 and

  • for each fixed xx between 00 and 11, yy runs from xx to 11

So the domain of integration is

D={ (x,y) ∣ 0≤x≤1, x≤y≤1 }\begin{equation*} D = \Set{(x,y)}{0\le x\le 1,\ x\le y\le 1} \end{equation*}

It is sketched in the figure on the left below.

Figure from prob_s3.1, line 446

Figure from prob_s3.1, line 446

Figure from prob_s3.1, line 456

Figure from prob_s3.1, line 456

(b) The given integral decomposed the domain of integration into vertical strips like the blue strip in the figure on the right above. To reverse the order of integration, we instead use horizontal strips. Looking at the pink strip in the figure on the right above, we see that this entails

  • having yy run from 00 to 11 and

  • for each fixed yy between 00 and 11, having xx run from 00 to yy

This gives

∫01dy∫0ydx ex/y=∫01dy [yex/y]0y=∫01dy y(e−1)=12(e−1)\begin{equation*} \int_0^1\dee{y}\int_0^y \dee{x}\ e^{x/y}=\int_0^1\dee{y}\ \Big[ye^{x/y}\Big]_0^y =\int_0^1\dee{y}\ y(e-1) =\frac{1}{2} (e-1) \end{equation*}
Q7Stage 2Past exam · M200 2006D

The integral II is defined as

I=∬Rf(x,y) dA=∫12∫1/yyf(x,y) dx dy+∫24∫y/2yf(x,y) dx dy\begin{equation*} I =\dblInt_R f(x,y)\ \dee{A} = \int_1^{\sqrt{2}} \int_{1/y}^{\sqrt{y}} f(x,y)\,\dee{x}\,\dee{y} +\int_{\sqrt{2}}^4 \int_{y/2}^{\sqrt{y}} f(x,y)\,\dee{x}\,\dee{y} \end{equation*}

  1. Sketch the region RR.

  2. Re–write the integral II by reversing the order of integration.

  3. Compute the integral II when f(x,y)=x/yf(x,y)= x/y.

Answer

(a) The region RR is the shaded region in the figure

Figure from prob_s3.1, line 520

Figure from prob_s3.1, line 520

(b) I=∫1/21∫1/x2xf(x,y) dy dx+∫12∫x22xf(x,y) dy dxI= \int_{1/\sqrt{2}}^1 \int_{1/x}^{2x} f(x,y)\,\dee{y}\,\dee{x} +\int_1^2 \int_{x^2}^{2x} f(x,y)\,\dee{y}\,\dee{x} (c) 12\frac{1}{2}

Full solution

(a) On RR

  • yy runs from 11 to 44 (from 11 to 2\sqrt{2} in the first integral and from 2\sqrt{2} to 44 in the second).

  • For each fixed yy between 11 and 2\sqrt{2}, xx runs from 1y\frac{1}{y} to y\sqrt{y} and

  • for each fixed yy between 2\sqrt{2} and 44, xx runs from y2\frac{y}{2} to y\sqrt{y}.

The figure on the left below is a sketch of RR, together with generic horizontal strips as were used in setting up the integral.

Figure from prob_s3.1, line 532

Figure from prob_s3.1, line 532

Figure from prob_s3.1, line 532

Figure from prob_s3.1, line 532

(b) To reverse the order of integration we use vertical strips as in the figure on the right above. Looking at that figure, we see that, on RR,

  • xx runs from 1/21/\sqrt{2} to 22.

  • For each fixed xx between 1/21/\sqrt{2} and 11, yy runs from 1x\frac{1}{x} to 2x2x and

  • for each fixed xx between 11 and 22, yy runs from x2x^2 to 2x2x.

So

I=∫1/21∫1/x2xf(x,y) dy dx+∫12∫x22xf(x,y) dy dx\begin{align*} I= \int_{1/\sqrt{2}}^1 \int_{1/x}^{2x} f(x,y)\,\dee{y}\,\dee{x} +\int_1^2 \int_{x^2}^{2x} f(x,y)\,\dee{y}\,\dee{x} \end{align*}

(c) When f(x,y)=xyf(x,y)=\frac{x}{y},

I=∫12∫1/yyxy dx dy+∫24∫y/2yxy dx dy=∫121y[y2−12y2] dy+∫241y[y2−y28] dy=[y2+14y2]12+[y2−y216]24=12+18−12−14+2−1−12+18=12\begin{align*} I &= \int_1^{\sqrt{2}} \int_{1/y}^{\sqrt{y}} \frac{x}{y}\,\dee{x}\,\dee{y} +\int_{\sqrt{2}}^4 \int_{y/2}^{\sqrt{y}} \frac{x}{y}\,\dee{x}\,\dee{y} \\ &=\int_1^{\sqrt{2}} \frac{1}{y}\left[\frac{y}{2}-\frac{1}{2y^2}\right] \,\dee{y} +\int_{\sqrt{2}}^4 \frac{1}{y}\left[\frac{y}{2}-\frac{y^2}{8}\right] \,\dee{y} \\ &=\left[\frac{y}{2}+\frac{1}{4y^2}\right]_1^{\sqrt{2}} +\left[\frac{y}{2}-\frac{y^2}{16}\right]_{\sqrt{2}}^4 =\frac{1}{\sqrt{2}} +\frac{1}{8}-\frac{1}{2}-\frac{1}{4} +2 -1 -\frac{1}{\sqrt{2}} +\frac{1}{8} \\ &=\frac{1}{2} \end{align*}
Q8Stage 2Past exam · M200 2007A

A region EE in the xyxy–plane has the property that for all continuous functions f

∬Ef(x,y) dA=∫x=−1x=3[∫y=x2y=2x+3f(x,y)dy]dx\begin{equation*} \dblInt_E f(x,y)\,\dee{A} = \int_{x=-1}^{x=3}\left[\int_{y=x^2}^{y=2x+3} f(x,y) \dee{y}\right] \dee{x} \end{equation*}

  1. Compute ∬Ex dA\dblInt_E x\,\dee{A}.

  2. Sketch the region E.

  3. Set up ∬Ex dA\dblInt_E x\,\dee{A} as an integral or sum of integrals in the opposite order.

Answer

(a) 323\frac{32}{3}

(b)

Figure from prob_s3.1, line 613

Figure from prob_s3.1, line 613

(c) I=∫01dy∫−yydx x+∫19dy∫(y−3)/2ydx xI = \int_0^1\dee{y} \int_{-\sqrt{y}}^{\sqrt{y}}\dee{x}\ x +\int_1^9\dee{y} \int_{(y-3)/2}^{\sqrt{y}}\dee{x}\ x

Full solution

(a) When f(x,y)=xf(x,y)=x,

∫x=−1x=3[∫y=x2y=2x+3xdy]dx=∫x=−1x=3[x(2x+3−x2)]dx=[2x33+3x22−x44]−13=18+272−814+23−32+14=18+12−20+23=323\begin{align*} \int_{x=-1}^{x=3}\left[\int_{y=x^2}^{y=2x+3} x \dee{y}\right] \dee{x} &= \int_{x=-1}^{x=3}\left[x(2x+3-x^2) \right] \dee{x} \\ &=\left[\frac{2x^3}{3}+\frac{3x^2}{2}-\frac{x^4}{4}\right]_{-1}^3 =18+\frac{27}{2}-\frac{81}{4} +\frac{2}{3}-\frac{3}{2}+\frac{1}{4} \\ &=18+12-20+\frac{2}{3} =\frac{32}{3} \end{align*}

(b) On the region EE

  • xx runs from −1-1 to 33 and

  • for each xx in that range, yy runs from x2x^2 to 2x+32x+3

Here are two sketches of EE, with the left one including a generic vertical strip as was used in setting up the given integral.

Figure from prob_s3.1, line 626

Figure from prob_s3.1, line 626

Figure from prob_s3.1, line 626

Figure from prob_s3.1, line 626

(c) To reverse the order of integration we use horizontal strips as in the figure on the right above. Looking at that figure, we see that, on the region EE,

  • yy runs from 00 to 99 and

  • for each yy between 00 and 11, xx runs from −y-\sqrt{y} to y\sqrt{y}

  • for each yy between 11 and 99, xx runs from (y−3)/2(y-3)/2 to y\sqrt{y}

So

I=∫01dy∫−yydx x+∫19dy∫(y−3)/2ydx x\begin{align*} I = \int_0^1\dee{y} \int_{-\sqrt{y}}^{\sqrt{y}}\dee{x}\ x +\int_1^9\dee{y} \int_{(y-3)/2}^{\sqrt{y}}\dee{x}\ x \end{align*}
Q9Stage 2Past exam · M200 2008A

Calculate the integral:

∬Dsin⁡(y2) dA\begin{equation*} \dblInt_D \sin(y^2)\ \dee{A} \end{equation*}

where DD is the region bounded by x+y=0x + y = 0, 2x−y=02x - y = 0, and y=4y = 4.

Hint

The antiderivative of the function sin⁡(y2)\sin(y^2) cannot be expressed in terms of familiar functions. So we do not want the inside integral to be over yy.

Answer

34[1−cos⁡(16)]\frac{3}{4}\big[1-\cos(16)\big]

Full solution

The antiderivative of the function sin⁡(y2)\sin(y^2) cannot be expressed in terms of familiar functions. So we do not want the inside integral to be over yy. So we'll use horizontal slices as in the figure

Figure from prob_s3.1, line 688

Figure from prob_s3.1, line 688

On the domain of integration

  • yy runs from 00 to 44, and

  • for each fixed yy in that range, xx runs from −y-y to y/2y/2

The given integral

∬Dsin⁡(y2) dA=∫04dy∫−yy/2dx sin⁡(y2)=∫04dy 32y sin⁡(y2)=[−34cos⁡(y2)]04=34[1−cos⁡(16)]\begin{align*} \dblInt_D \sin(y^2)\ \dee{A} &=\int_0^4\dee{y}\int_{-y}^{y/2}\dee{x}\ \sin(y^2) \\ &=\int_0^4 \dee{y}\ \frac{3}{2}y\,\sin(y^2) \\ &=\left[-\frac{3}{4}\cos(y^2)\right]_0^4 \\ &=\frac{3}{4}\big[1-\cos(16)\big] \end{align*}
Q10Stage 2Past exam · M200 2008D

Consider the integral

I=∫01∫y1sin⁡(πx2)x dx dy\begin{equation*} I = \int_0^1 \int_{\sqrt{y}}^1 \frac{\sin(\pi x^2)}{x}\ \dee{x}\,\dee{y} \end{equation*}

  1. Sketch the region of integration.

  2. Evaluate I.

Hint

The inside integral, ∫y1sin⁡(πx2)x dx\int_{\sqrt{y}}^1 \frac{\sin(\pi x^2)}{x}\ \dee{x}, in the given form of II looks really nasty. So try exchanging the order of integration.

Answer

(a)

Figure from prob_s3.1, line 737

Figure from prob_s3.1, line 737

(b) 1π\frac{1}{\pi}

Full solution

(a) On the domain of integration

  • yy runs from 00 to 11 and

  • for each fixed yy in that range, xx runs from y\sqrt{y} to 11.

The figure on the left below is a sketch of that domain, together with a generic horizontal strip as was used in setting up the integral.

Figure from prob_s3.1, line 746

Figure from prob_s3.1, line 746

Figure from prob_s3.1, line 746

Figure from prob_s3.1, line 746

(b) The inside integral, ∫y1sin⁡(πx2)x dx\int_{\sqrt{y}}^1 \frac{\sin(\pi x^2)}{x}\ \dee{x}, in the given form of II looks really nasty. So let's try exchanging the order of integration. Looking at the figure on the right above, we see that, on the domain of integration,

  • xx runs from 00 to 11 and

  • for each fixed xx in that range, yy runs from 00 to x2x^2.

So

I=∫01dx∫0x2dy sin⁡(πx2)x=∫01dx xsin⁡(πx2)=[−cos⁡(πx2)2π]01(Looks pretty rigged!)=1π\begin{align*} I &= \int_0^1\dee{x}\int_0^{x^2}\dee{y}\ \frac{\sin(\pi x^2)}{x} \\ &=\int_0^1\dee{x}\ x\sin(\pi x^2) \\ &=\left[-\frac{\cos(\pi x^2)}{2 \pi}\right]_0^1 \qquad\text{(Looks pretty rigged!)}\\ &=\frac{1}{\pi} \end{align*}
Q11Stage 2Past exam · M200 2009A

Let II be the double integral of the function f(x,y)=y2sin⁡xyf(x,y) = y^2 \sin xy over the triangle with vertices (0,0)(0, 0), (0,1)(0, 1) and (1,1)(1, 1) in the xyxy–plane.

  1. Write II as an iterated integral in two different ways.

  2. Evaluate II.

Answer

(a) I=∫01dx∫x1dy y2sin⁡xy=∫01dy∫0ydx y2sin⁡xyI=\int_0^1\dee{x}\int_x^1\dee{y}\ y^2\sin xy =\int_0^1\dee{y}\int_0^y\dee{x}\ y^2\sin xy

(b) 1−sin⁡12\frac{1-\sin 1}{2}

Full solution

(a) Let's call the triangle T\cT. Here are two sketches of T\cT, one including a generic vertical strip and one including a generic horizontal strip. Notice that the equation of the line through (0,0)(0,0) and (1,1)(1,1) is y=xy=x.

Figure from prob_s3.1, line 812

Figure from prob_s3.1, line 812

Figure from prob_s3.1, line 812

Figure from prob_s3.1, line 812

First, we'll set up the integral using vertical strips. Looking at the figure on the left above, we see that, on T\cT,

  • xx runs from 00 to 11 and

  • for each xx in that range, yy runs from xx to 11.

So the integral

I=∫01dx∫x1dy y2sin⁡xy\begin{align*} I=\int_0^1\dee{x}\int_x^1\dee{y}\ y^2\sin xy \end{align*}

Next, we'll set up the integral using horizontal strips. Looking at the figure on the right above, we see that, on T\cT,

  • yy runs from 00 to 11 and

  • for each yy in that range, xx runs from 00 to yy.

So the integral

I=∫01dy∫0ydx y2sin⁡xy\begin{align*} I=\int_0^1\dee{y}\int_0^y\dee{x}\ y^2\sin xy \end{align*}

(b) To evaluate the inside integral, ∫x1dy y2sin⁡xy\int_x^1\dee{y}\ y^2\sin xy, of the vertical strip version, will require two integration by parts to get rid of the y2y^2. So we'll use the horizontal strip version.

I=∫01dy∫0ydx y2sin⁡xy=∫01dy [−ycos⁡xy]0y=∫01dy [y−ycos⁡y2]=[y22−sin⁡y22]01(Look’s pretty rigged!)=1−sin⁡12\begin{align*} I&=\int_0^1\dee{y}\int_0^y\dee{x}\ y^2\sin xy \\ &=\int_0^1\dee{y}\ \Big[-y\cos xy\Big]_0^y \\ &=\int_0^1\dee{y}\ \big[y-y\cos y^2\big] \\ &=\left[\frac{y^2}{2}-\frac{\sin y^2}{2}\right]_0^1 \qquad\text{(Look's pretty rigged!)} \\ &= \frac{1-\sin 1}{2} \end{align*}
Q12Stage 2Past exam · M200 2009D

Find the volume (V)(V) of the solid bounded above by the surface

z=f(x,y)=e−x2,\begin{equation*} z = f (x,y) = e^{-x^2}, \end{equation*}

below by the plane z=0z = 0 and over the triangle in the xyxy–plane formed by the lines x=1x = 1, y=0y = 0 and y=xy = x.

Answer

1−e−12\frac{1-e^{-1}}{2}

Full solution

If we call the triangular base region T\cT, then the volume is

V=∬Tf(x,y) dA=∬Te−x2 dx dy\begin{align*} V=\dblInt_\cT f(x,y)\ \dee{A} = \dblInt_\cT e^{-x^2}\ \dee{x}\,\dee{y} \end{align*}

If we set up the integral using horizontal slices, so that the inside integral is the xx–integral, there will be a big problem — the integrand e−x2e^{-x^2} does not have an obvious anti–derivative. (In fact its antiderivative cannot be expressed in terms of familiar functions.) So let's try vertical slices as in the sketch

Figure from prob_s3.1, line 882

Figure from prob_s3.1, line 882

Looking at that sketch we see that

  • xx runs from 00 to 11, and

  • for each xx in that range, yy runs from 00 to xx.

So the integral is

V=∫01dx∫0xdy e−x2=∫01dx xe−x2=[−12e−x2]01=1−e−12\begin{align*} V&=\int_0^1\dee{x}\int_0^x\dee{y}\ e^{-x^2} \\ &=\int_0^1 \dee{x}\ xe^{-x^2} \\ &=\left[-\frac{1}{2}e^{-x^2}\right]_0^1 \\ &=\frac{1-e^{-1}}{2} \end{align*}
Q13Stage 2Past exam · M200 2009D

Consider the integral I=∫01∫y2−yyx dx dy\displaystyle I=\int_0^1 \int_y^{2-y}\frac{y}{x} \ \dee{x}\,\dee{y}.

  1. Sketch the region of integration.

  2. Interchange the order of integration.

  3. Evaluate II.

Answer

(a)

Figure from prob_s3.1, line 933

Figure from prob_s3.1, line 933

(b) I=∫01dx∫0xdy yx+∫12dx∫02−xdy yxI=\int_0^1\dee{x}\int_0^x\dee{y}\ \frac{y}{x} +\int_1^2\dee{x}\int_0^{2-x}\dee{y}\ \frac{y}{x} (c) 2ln⁡2−12\ln 2 -1

Full solution

(a) On the domain of integration

  • yy runs from 00 to 11 and

  • for each yy in that range xx runs from yy to 2−y2-y. So the left hand side of the domain is the line x=yx=y and the right hand side of the domain is x=2−yx=2-y.

The figure on the left below is a sketch of that domain, together with a generic horizontal strip as was used in setting up the integral.

Figure from prob_s3.1, line 944

Figure from prob_s3.1, line 944

Figure from prob_s3.1, line 944

Figure from prob_s3.1, line 944

(b) To reverse the order of integration we use vertical, rather than horizontal, strips. Looking at the figure on the right above, we see that, in the domain of integration

  • xx runs from 00 to 22 and

  • for each xx between 00 and 11, yy runs from 00 to xx, while

  • for each xx between 11 and 22, yy runs from 00 to 2−x2-x.

So the integral

I=∫01dx∫0xdy yx+∫12dx∫02−xdy yx\begin{align*} I=\int_0^1\dee{x}\int_0^x\dee{y}\ \frac{y}{x} +\int_1^2\dee{x}\int_0^{2-x}\dee{y}\ \frac{y}{x} \end{align*}

(c) Using the answer to part (b)

I=∫01dx∫0xdy yx+∫12dx∫02−xdy yx=12∫01dx x+12∫12dx (2−x)2x=14+12∫12dx (4x−4+x)=14+12[4ln⁡2−4+4−12]=2ln⁡2−1\begin{align*} I&=\int_0^1\dee{x}\int_0^x\dee{y}\ \frac{y}{x} +\int_1^2\dee{x}\int_0^{2-x}\dee{y}\ \frac{y}{x} \\ &=\frac{1}{2}\int_0^1\dee{x}\ x +\frac{1}{2}\int_1^2\dee{x}\ \frac{(2-x)^2}{x}\\ &=\frac{1}{4} +\frac{1}{2}\int_1^2\dee{x}\ \left(\frac{4}{x}-4+x\right) \\ &=\frac{1}{4} +\frac{1}{2}\left[4\ln 2 -4 + \frac{4-1}{2}\right] \\ &=2\ln 2 -1 \end{align*}
Q14Stage 2Past exam · M200 2010A

For the integral

I=∫01∫x11+y3 dy dx\begin{equation*} I = \int_0^1 \int_{\sqrt{x}}^1 \sqrt{1+y^3}\ \dee{y}\,\dee{x} \end{equation*}

  1. Sketch the region of integration.

  2. Evaluate II.

Hint

The inside integral, ∫x11+y3 dy\int_{\sqrt{x}}^1 \sqrt{1+y^3}\ \dee{y}, of the given integral looks pretty nasty. Try reversing the order of integration.

Answer

(a)

Figure from prob_s3.1, line 1011

Figure from prob_s3.1, line 1011

(b) 2(22−1)9\frac{2\big(2\sqrt{2}-1\big)}{9}

Full solution

(a) On the domain of integration,

  • xx runs from 00 to 11, and

  • for each fixed xx in that range, yy runs from x\sqrt{x} to 11. We may rewrite y=xy=\sqrt{x} as x=y2x=y^2, which is a rightward opening parabola.

Here are two sketches of the domain of integration, which we call DD. The left hand sketch also shows a vertical slice, as was used in setting up the integral.

Figure from prob_s3.1, line 1020

Figure from prob_s3.1, line 1020

Figure from prob_s3.1, line 1020

Figure from prob_s3.1, line 1020

(b) The inside integral, ∫x11+y3 dy\int_{\sqrt{x}}^1 \sqrt{1+y^3}\ \dee{y}, of the given integral looks pretty nasty. So let's reverse the order of integration, by using horizontal, rather than vertical, slices. Looking at the figure on the right above, we see that

  • yy runs from 00 to 11, and

  • for each fixed yy in that range xx runs from 00 to y2y^2.

So

I=∫01dy∫0y2dx 1+y3=∫01dy y21+y3=∫12du3 uwith u=1+y3, du=3y2 dy. Looks pretty rigged!=13[u3/23/2]12=2(22−1)9\begin{align*} I &= \int_0^1\dee{y} \int_0^{y^2}\dee{x}\ \sqrt{1+y^3} \\ &=\int_0^1\dee{y} \ y^2\sqrt{1+y^3} \\ &=\int_1^2\frac{\dee{u}}{3}\ \sqrt{u} \qquad\text{with }u=1+y^3,\ \dee{u}=3y^2\,\dee{y}\text{. Looks pretty rigged!} \\ &=\frac{1}{3} \left[\frac{u^{3/2}}{3/2}\right]_1^2 \\[0.05in] &=\frac{2\big(2\sqrt{2}-1\big)}{9} \end{align*}
Q15Stage 2Past exam · M200 2010D

  1. DD is the region bounded by the parabola y2=xy^2 = x and the line y=x−2y = x - 2. Sketch DD and evaluate JJ where

    J=∬D3y dA\begin{equation*} J = \dblInt_D 3y\ \dee{A} \end{equation*}

  2. Sketch the region of integration and then evaluate the integral II :

    I=∫04∫12x1ey3 dy dx\begin{equation*} I = \int_0^4 \int_{\frac{1}{2}\sqrt{x}}^1 e^{y^3}\ \dee{y}\,\dee{x} \end{equation*}
Hint

(b) The inside integral, ∫12x1ey3 dy\int_{\frac{1}{2}\sqrt{x}}^1 e^{y^3}\ \dee{y}, looks pretty nasty because ey3e^{y^3} does not have an obvious antiderivative. Try reversing the order of integration.

Answer

(a) J=274J=\frac{27}{4}

Figure from prob_s3.1, line 1085

Figure from prob_s3.1, line 1085

(b) I=43[e−1]I=\frac{4}{3}\big[e-1\big]

Figure from prob_s3.1, line 1085

Figure from prob_s3.1, line 1085

Full solution

(a) Observe that the parabola y2=xy^2=x and the line y=x−2y=x-2 meet when x=y+2x=y+2 and

y2=y+2  ⟺  y2−y−2=0  ⟺  (y−2)(y+1)=0\begin{equation*} y^2=y+2 \iff y^2-y-2=0 \iff (y-2)(y+1)=0 \end{equation*}

So the points of intersection of x=y2x=y^2 and y=x−2y=x-2 are (1,−1)(1,-1) and (4,2)(4,2). Here is a sketch of DD.

Figure from prob_s3.1, line 1100

Figure from prob_s3.1, line 1100

To evaluate JJ, we'll use horizontal slices as in the figure above. (If we were to use vertical slices we would have to split the integral in two, with 0≤x≤10\le x\le 1 in one part and 1≤x≤41\le x\le 4 in the other.) From the figure, we see that, on DD,

  • yy runs from −1-1 to 22 and

  • for each fixed yy in that range, xx runs from y2y^2 to y+2y+2.

Hence

J=∬D3y dA=∫−12dy∫y2y+2dx 3y=3∫−12dy y(y+2−y2)=3[y33+y2−y44]−12=3[83+4−4+13−1+14]=274\begin{align*} J &=\dblInt_D 3y\ \dee{A} =\int_{-1}^2\dee{y}\int_{y^2}^{y+2}\dee{x}\ 3y \\ &=3\int_{-1}^2\dee{y}\ y(y+2-y^2) \\ &=3\left[\frac{y^3}{3}+y^2-\frac{y^4}{4}\right]_{-1}^2 \\ &=3\left[\frac{8}{3}+4-4+\frac{1}{3}-1+\frac{1}{4}\right] \\ &=\frac{27}{4} \end{align*}

(b) On the domain of integration,

  • xx runs from 00 to 44 and

  • for each fixed xx in that range, yy runs from 12x\frac{1}{2}\sqrt{x} to 11.

The figure on the left below is a sketch of that domain, together with a generic vertical strip as was used in setting up the integral.

Figure from prob_s3.1, line 1100

Figure from prob_s3.1, line 1100

Figure from prob_s3.1, line 1100

Figure from prob_s3.1, line 1100

The inside integral, over yy, looks pretty nasty because ey3e^{y^3} does not have an obvious antiderivative. So let's reverse the order of integration. That is, let's use horizontal, rather than vertical, strips. From the figure on the right above, we see that, on the domain of integration

  • yy runs from 00 to 11 and

  • for each fixed yy in that range, xx runs from 00 to 4y24y^2.

So

I=∫01dy∫04y2dx ey3=∫01dy 4y2ey3=43∫01du euwith u=y3, du=3y2 dy(Looks rigged!)=43[e−1]\begin{align*} I &= \int_0^1\dee{y}\int_0^{4y^2}\dee{x}\ e^{y^3} \\ &= \int_0^1\dee{y}\ 4y^2 e^{y^3} \\ &= \frac{4}{3}\int_0^1\dee{u}\ e^u \qquad\text{with } u=y^3,\ \dee{u}=3y^2\,\dee{y}\qquad \text{(Looks rigged!)}\\ &=\frac{4}{3}\big[e-1\big] \end{align*}
Q16Stage 2Past exam · M200 2011D

Consider the iterated integral

∫−40∫−y2cos⁡(x3) dx dy\begin{equation*} \int_{-4}^0\int_{\sqrt{-y}}^2 \cos(x^3)\,\dee{x}\,\dee{y} \end{equation*}

  1. Draw the region of integration.

  2. Evaluate the integral.

Hint

(b) The inside integral, ∫−y2cos⁡(x3) dx\int_{\sqrt{-y}}^2 \cos(x^3)\,\dee{x} looks nasty. Try reversing the order of integration.

Answer

(a)

Figure from prob_s3.1, line 1192

Figure from prob_s3.1, line 1192

(b) sin⁡(8)3\frac{\sin(8)}{3}

Full solution

(a) On the domain of integration

  • yy runs from −4-4 to 00 and

  • for each yy in that range, xx runs from −y\sqrt{-y} (when y=−x2y=-x^2) to 22.

The figure on the left below provides a sketch of the domain of integration. It also shows the generic horizontal slice that was used to set up the given iterated integral.

Figure from prob_s3.1, line 1202

Figure from prob_s3.1, line 1202

Figure from prob_s3.1, line 1202

Figure from prob_s3.1, line 1202

(b) The inside integral, ∫−y2cos⁡(x3) dx\int_{\sqrt{-y}}^2 \cos(x^3)\,\dee{x} looks nasty. So let's reverse the order of integration and use vertical, rather than horizontal, slices. From the figure on the right above, on the domain of integration,

  • xx runs from 00 to 22 and

  • for each xx in that range, yy runs from −x2-x^2 to 00.

So the integral

∫−40∫−y2cos⁡(x3) dx dy=∫02dx∫−x20dy cos⁡(x3)=∫02dx x2 cos⁡(x3)=[sin⁡(x3)3]02=sin⁡(8)3\begin{align*} \int_{-4}^0\int_{\sqrt{-y}}^2 \cos(x^3)\,\dee{x}\,\dee{y} &=\int_0^2\dee{x}\int_{-x^2}^0\dee{y}\ \cos(x^3) \\ &=\int_0^2\dee{x}\ x^2\ \cos(x^3) =\left[\frac{\sin(x^3)}{3}\right]_0^2 \\ &=\frac{\sin(8)}{3} \end{align*}
Q17Stage 2Past exam · M200 2012A

  1. Combine the sum of the iterated integrals

    I=∫01∫−yyf(x,y) dx dy+∫14∫y−2yf(x,y) dx dy\begin{equation*} I = \int_0^1\int_{-\sqrt{y}}^{\sqrt{y}} f(x,y)\ \dee{x}\,\dee{y} + \int_1^4\int_{y-2}^{\sqrt{y}} f(x,y)\ \dee{x}\,\dee{y} \end{equation*}

    into a single iterated integral with the order of integration reversed.

  2. Evaluate II if f(x,y)=ex2−xf(x,y)=\frac{e^x}{2-x}.

Answer

(a) I=∫−12∫x2x+2f(x,y) dy dxI = \int_{-1}^2\int_{x^2}^{x+2} f(x,y)\ \dee{y}\,\dee{x} (b) 2e2+1e2e^2 + \frac{1}{e}

Full solution

(a) On the domain of integration

  • yy runs from 00 to 44 and

  • for each yy in the range 0≤y≤10\le y\le 1, xx runs from −y-\sqrt{y} to y\sqrt{y} and

  • for each yy in the range 1≤y≤41\le y\le 4, xx runs from y−2y-2 to y\sqrt{y}.

Both figures below provide sketches of the domain of integration.

Figure from prob_s3.1, line 1267

Figure from prob_s3.1, line 1267

Figure from prob_s3.1, line 1267

Figure from prob_s3.1, line 1267

To reverse the order of integration observe, from the figure on the right above that, on the domain of integration,

  • xx runs from −1-1 to 22 and

  • for each xx in that range, yy runs from x2x^2 to x+2x+2.

So the integral

I=∫−12∫x2x+2f(x,y) dy dx\begin{equation*} I = \int_{-1}^2\int_{x^2}^{x+2} f(x,y)\ \dee{y}\,\dee{x} \end{equation*}

(b) We'll use the integral with the order of integration reversed that we found in part (a). When f(x,y)=ex2−xf(x,y)=\frac{e^x}{2-x}

I=∫−12∫x2x+2ex2−x dy dx=∫−12(x+2−x2)ex2−x dx=−∫−12(x−2)(x+1)ex2−x dx=∫−12(x+1)ex dx=[xex]−12=2e2+1e\begin{align*} I &= \int_{-1}^2\int_{x^2}^{x+2} \frac{e^x}{2-x}\ \dee{y}\,\dee{x} \\ &= \int_{-1}^2 (x+2-x^2)\frac{e^x}{2-x}\ \dee{x} = -\int_{-1}^2 (x-2)(x+1)\frac{e^x}{2-x}\ \dee{x} \\ &= \int_{-1}^2 (x+1) e^x\ \dee{x} \\ &= \Big[xe^x\Big]_{-1}^2 \\ &= 2e^2 + \frac{1}{e} \end{align*}
Q18Stage 2Past exam · M200 2012D

Let

I=∫04∫y8−yf(x,y) dx dy\begin{equation*} I=\int_0^4\int_{\sqrt{y}}^{\sqrt{8-y}} f(x,y)\,\dee{x}\,\dee{y} \end{equation*}

  1. Sketch the domain of integration.

  2. Reverse the order of integration.

  3. Evaluate the integral for f(x,y)=1(1+y)2f(x,y)=\frac{1}{(1+y)^2}.

Hint

19−x2=16(1x+3−1x−3)\frac{1}{9-x^2} =\frac{1}{6}\left(\frac{1}{x+3}-\frac{1}{x-3}\right).

Answer

(a)

Figure from prob_s3.1, line 1333

Figure from prob_s3.1, line 1333

(b) ∫02∫0x2f(x,y) dy dx+∫28∫08−x2f(x,y) dy dx\int_0^2\int_0^{x^2} f(x,y)\,\dee{y}\,\dee{x} +\int_2^{\sqrt{8}}\int_0^{8-x^2} f(x,y)\,\dee{y}\,\dee{x} (c) 8−arctan⁡2−16[ln⁡3+83−8−ln⁡5]\sqrt{8}-\arctan 2 -\frac{1}{6}\left[\ln\frac{3+\sqrt{8}}{3-\sqrt{8}} -\ln 5\right]

Full solution

On the domain of integration

  • yy runs from 00 to 44. In inequalities, 0≤y≤40\le y\le 4.

  • For each fixed yy in that range, xx runs from y\sqrt{y} to 8−y\sqrt{8-y}. In inequalities, that is y≤x≤8−y\sqrt{y}\le x\le \sqrt{8-y}, or y≤x2≤8−yy\le x^2\le 8-y.

Here are two sketchs of the domain of integration.

Figure from prob_s3.1, line 1345

Figure from prob_s3.1, line 1345

Figure from prob_s3.1, line 1345

Figure from prob_s3.1, line 1345

(b) To reverse the order we observe, from the figure on the right above, that, on the domain of integration,

  • xx runs from 00 to 8\sqrt{8}. In inequalities, 0≤x≤80\le x\le \sqrt{8}.

  • For each fixed xx between 00 and 22, yy runs from 00 to x2x^2. In inequalities, that is 0≤y≤x20\le y\le x^2.

  • For each fixed xx between 22 and 8\sqrt{8}, yy runs from 00 to 8−x28-x^2. In inequalities, that is 0≤y≤8−x20\le y\le 8-x^2.

So the integral is

∫02∫0x2f(x,y) dy dx+∫28∫08−x2f(x,y) dy dx\begin{equation*} \int_0^2\int_0^{x^2} f(x,y)\,\dee{y}\,\dee{x} +\int_2^{\sqrt{8}}\int_0^{8-x^2} f(x,y)\,\dee{y}\,\dee{x} \end{equation*}

(c) We'll use the form of part (b).

∫02∫0x21(1+y)2 dy dx+∫28∫08−x21(1+y)2 dy dx=−∫02[11+y]0x2 dx−∫28[11+y]08−x2 dx=∫02[1−11+x2] dx+∫28[1−19−x2] dx=8−arctan⁡x∣02−16∫28[13+x+13−x] dx=8−arctan⁡2−16[ln⁡(3+x)−ln⁡(3−x)]28=8−arctan⁡2−16[ln⁡3+83−8−ln⁡5]\begin{align*} &\int_0^2\int_0^{x^2} \frac{1}{(1+y)^2}\,\dee{y}\,\dee{x} +\int_2^{\sqrt{8}}\int_0^{8-x^2} \frac{1}{(1+y)^2}\,\dee{y}\,\dee{x} \\ &=-\int_0^2 \left[\frac{1}{1+y}\right]_0^{x^2}\,\dee{x} -\int_2^{\sqrt{8}} \left[\frac{1}{1+y}\right]_0^{8-x^2}\,\dee{x} \\ &=\int_0^2 \left[1-\frac{1}{1+x^2}\right]\,\dee{x} +\int_2^{\sqrt{8}} \left[1-\frac{1}{9-x^2}\right]\,\dee{x} \\ &=\sqrt{8}-\arctan x\bigg|_0^2 -\frac{1}{6}\int_2^{\sqrt{8}} \left[\frac{1}{3+x}+\frac{1}{3-x}\right]\,\dee{x} \\ &=\sqrt{8}-\arctan 2 -\frac{1}{6}\Big[\ln(3+x)-\ln(3-x)\Big]_2^{\sqrt{8}} \\ &=\sqrt{8}-\arctan 2 -\frac{1}{6}\left[\ln\frac{3+\sqrt{8}}{3-\sqrt{8}} -\ln 5\right] \end{align*}
Q19Stage 2Past exam · M200 2013D

Evaluate

∫−10∫−22xey2 dy dx\begin{equation*} \int_{-1}^0 \int_{-2}^{2x} e^{y^2}\ \dee{y}\,\dee{x} \end{equation*}
Hint

The antiderivative of the function e−y2e^{-y^2} cannot be expressed in terms of elementary functions. So the inside integral ∫−22xey2 dy\int_{-2}^{2x} e^{y^2}\ \dee{y} cannot be evaluated using standard calculus 2 techniques. Try reversing the order of integration.

Answer

14[e4−1]\frac{1}{4}\big[e^4-1\big]

Full solution

The antiderivative of the function e−y2e^{-y^2} cannot be expressed in terms of elementary functions. So the inside integral ∫−22xey2 dy\int_{-2}^{2x} e^{y^2}\ \dee{y} cannot be evaluated using standard calculus 2 techniques. The trick for dealing with this integral is to reverse the order of integration. On the domain of integration

  • xx runs from −1-1 to 00. In inequalities, −1≤x≤0-1\le x\le 0.

  • For each fixed xx in that range, yy runs from −2-2 to 2x2x. In inequalities, −2≤y≤2x-2\le y\le 2x.

The domain of integration, namely

{ (x,y) ∣ −1≤x≤0, −2≤y≤2x }\begin{equation*} \Set{(x,y)}{-1\le x\le 0,\ -2\le y\le 2x} \end{equation*}

is sketched in the figure on the left below.

Figure from prob_s3.1, line 1416

Figure from prob_s3.1, line 1416

Figure from prob_s3.1, line 1416

Figure from prob_s3.1, line 1416

Looking at the figure on the right above, we see that we can also express the domain of integration as

{ (x,y) ∣ −2≤y≤0, y/2≤x≤0 }\begin{equation*} \Set{(x,y)}{-2\le y\le 0,\ y/2\le x\le 0} \end{equation*}

So the integral

∫−10∫−22xey2 dy dx=∫−20∫y/20ey2 dx dy=−12∫−20yey2 dy=−12[12ey2]−20=14[e4−1]\begin{align*} \int_{-1}^0 \int_{-2}^{2x} e^{y^2}\ \dee{y}\,\dee{x} &=\int_{-2}^0 \int_{y/2}^{0} e^{y^2}\ \dee{x}\,\dee{y} \\ &=-\frac{1}{2}\int_{-2}^0 y e^{y^2}\ \dee{y} \\ &=-\frac{1}{2}\left[\frac{1}{2}e^{y^2}\right]_{-2}^0 \\ &=\frac{1}{4}\big[e^4-1\big] \end{align*}
Q20Stage 2Past exam · M200 2014A

Let

I=∫02∫0xf(x,y) dy dx+∫26∫06−xf(x,y) dy dx\begin{equation*} I = \int_0^2 \int_0^x f(x,y)\ \dee{y}\,\dee{x} + \int_2^6 \int_0^{\sqrt{6-x}} f(x,y)\ \dee{y}\,\dee{x} \end{equation*}

Express II as an integral where we integrate first with respect to xx.

Answer

I=∫02∫y6−y2f(x,y) dx dyI = \int_0^2 \int_y^{6-y^2} f(x,y)\ \dee{x}\,\dee{y}

Full solution

We first have to get a picture of the domain of integration. The first integral has domain of integration

{ (x,y) ∣ 0≤x≤2, 0≤y≤x }\begin{equation*} \Set{(x,y)}{0\le x\le 2,\ 0\le y\le x} \end{equation*}

and the second integral has domain of integration

{ (x,y) ∣ 2≤x≤6, 0≤y≤6−x }\begin{equation*} \Set{(x,y)}{2\le x\le 6,\ 0\le y\le \sqrt{6-x}} \end{equation*}

Here is a sketch. The domain of integration for the first integral is the shaded triangular region to the left of x=2x=2 and the domain of integration for the second integral is the shaded region to the right of x=2x=2.

Figure from prob_s3.1, line 1473

Figure from prob_s3.1, line 1473

To exchange the order of integration, we use horizontal slices as in the figure below.

Figure from prob_s3.1, line 1473

Figure from prob_s3.1, line 1473

The bottom slice has y=0y=0 and the top slice has y=2y=2. On the slice at height yy, xx runs from yy to 6−y26-y^2. So

I=∫02∫y6−y2f(x,y) dx dy\begin{equation*} I = \int_0^2 \int_y^{6-y^2} f(x,y)\ \dee{x}\,\dee{y} \end{equation*}
Q21Stage 2Past exam · M200 2015D

Consider the domain DD above the xx–axis and below parabola y=1−x2y = 1-x^2 in the xyxy–plane.

  1. Sketch DD.

  2. Express

    ∬Df(x,y) dA\begin{equation*} \dblInt_D f(x,y)\ \dee{A} \end{equation*}

    as an iterated integral corresponding to the order dx dy\dee{x}\,\dee{y}. Then express this integral as an iterated integral corresponding to the order dy dx\dee{y}\,\dee{x}.

  3. Compute the integral in the case f(x,y)=ex−(x3/3)f(x,y) = e^{x-(x^3/3)}.

Answer

(a)

Figure from prob_s3.1, line 1530

Figure from prob_s3.1, line 1530

(b) ∫01∫−1−y1−yf(x,y) dx dy\int_0^1 \int_{-\sqrt{1-y}}^{\sqrt{1-y}} f(x,y)\ \dee{x}\,\dee{y}, ∫−11∫01−x2f(x,y) dy dx\int_{-1}^1 \int_0^{1-x^2} f(x,y)\ \dee{y}\,\dee{x}

(c) e2/3−e−2/3e^{2/3}-e^{-2/3}

Full solution

(a), (b) Looking at the figure on the left below, we see that we can write the domain

D={ (x,y) ∣ 0≤y≤1, −1−y≤x≤1−y }\begin{equation*} D = \Set{(x,y)}{0\le y\le 1,\ -\sqrt{1-y}\le x\le\sqrt{1-y}} \end{equation*}

So

∬Df(x,y) dA=∫01dy∫−1−y1−ydx f(x,y)=∫01∫−1−y1−yf(x,y) dx dy\begin{equation*} \dblInt_D f(x,y)\ \dee{A} =\int_0^1\dee{y} \int_{-\sqrt{1-y}}^{\sqrt{1-y}}\dee{x}\ f(x,y) =\int_0^1 \int_{-\sqrt{1-y}}^{\sqrt{1-y}} f(x,y)\ \dee{x}\,\dee{y} \end{equation*}

Figure from prob_s3.1, line 1544

Figure from prob_s3.1, line 1544

Figure from prob_s3.1, line 1544

Figure from prob_s3.1, line 1544

Looking at the figure on the right above, we see that we can write the domain

D={ (x,y) ∣ −1≤x≤1, 0≤y≤1−x2 }\begin{equation*} D = \Set{(x,y)}{-1\le x\le 1,\ 0\le y\le 1-x^2} \end{equation*}

So

∬Df(x,y) dA=∫−11dx∫01−x2dy f(x,y)=∫−11∫01−x2f(x,y) dy dx\begin{equation*} \dblInt_D f(x,y)\ \dee{A} =\int_{-1}^1\dee{x} \int_0^{1-x^2}\dee{y}\ f(x,y) =\int_{-1}^1 \int_0^{1-x^2} f(x,y)\ \dee{y}\,\dee{x} \end{equation*}

(c) Using the second form from part (b),

∬Dex−(x3/3) dA=∫−11dx∫01−x2dy ex−(x3/3)=∫−11(1−x2)ex−(x3/3) dx=∫−2/32/3eu duwith u=x−x33, du=(1−x2) dx=e2/3−e−2/3\begin{align*} \dblInt_D e^{x-(x^3/3)}\ \dee{A} &=\int_{-1}^1\dee{x} \int_0^{1-x^2}\dee{y}\ e^{x-(x^3/3)} \\ &=\int_{-1}^1 (1-x^2) e^{x-(x^3/3)}\ \dee{x} \\ &=\int_{-2/3}^{2/3} e^u\ \dee{u} \qquad\text{with } u=x -\frac{x^3}{3},\ \dee{u} = \big(1-x^2)\,\dee{x} \\ &= e^{2/3}-e^{-2/3} \end{align*}
Q22Stage 2Past exam · M200 2016D

Let I=∫01∫x21x3 sin⁡(y3) dy dxI=\int_0^1 \int_{x^2}^1 x^3\ \sin(y^3)\ \dee{y}\ \dee{x}.

  1. Sketch the region of integration in the xyxy–plane. Label your sketch sufficiently well that one could use it to determine the limits of double integration.

  2. Evaluate II.

Hint

The inside integral, over yy, looks pretty nasty because sin⁡(y3)\sin(y^3) does not have an obvious antiderivative. So try reversing the order of integration.

Answer

(a)

Figure from prob_s3.1, line 1608

Figure from prob_s3.1, line 1608

(b) 1−cos⁡(1)12\frac{1-\cos(1)}{12}

Full solution

(a) On the domain of integration,

  • xx runs from 00 to 11 and

  • for each fixed xx in that range, yy runs from x2x^2 to 11.

The figure on the left below is a sketch of that domain, together with a generic vertical strip as was used in setting up the integral.

Figure from prob_s3.1, line 1616

Figure from prob_s3.1, line 1616

Figure from prob_s3.1, line 1616

Figure from prob_s3.1, line 1616

(b) As it stands, the inside integral, over yy, looks pretty nasty because sin⁡(y3)\sin(y^3) does not have an obvious antiderivative. So let's reverse the order of integration. The given integral was set up using vertical strips. So, to reverse the order of integration, we use horizontal strips as in the figure on the right above. Looking at that figure we see that, on the domain of integration,

  • yy runs from 00 to 11 and

  • for each fixed yy in that range, xx runs from 00 to y\sqrt{y}.

So

I=∫01dy∫0ydx x3 sin⁡(y3)=∫01dy sin⁡(y3)[x44]0y=14∫01dy y2sin⁡(y3)=14[−cos⁡(y3)3]01=1−cos⁡(1)12\begin{align*} I&=\int_0^1\dee{y} \int_0^{\sqrt{y}}\dee{x}\ x^3\ \sin(y^3) \\ &=\int_0^1\dee{y}\ \sin(y^3)\left[\frac{x^4}{4}\right]_0^{\sqrt{y}} \\ &=\frac{1}{4}\int_0^1\dee{y}\ y^2\sin(y^3) \\ &=\frac{1}{4}\left[-\frac{\cos(y^3)}{3}\right]_0^1 \\ &=\frac{1-\cos(1)}{12} \end{align*}
Q23Stage 2Past exam · M200 2003D

Consider the solid under the surface z=6−xyz=6-xy, bounded by the five planes x=0x=0, x=3x=3, y=0y=0, y=3y=3, z=0z=0. Note that no part of the solid lies below the xx–yy plane.

  1. Sketch the base of the solid in the xyxy–plane. Note that it is not a square!

  2. Compute the volume of the solid.

Answer

(a)

Figure from prob_s3.1, line 1672

Figure from prob_s3.1, line 1672

(b) 27+18ln⁡32≈34.3027+18\ln\frac{3}{2}\approx 34.30

Full solution

(a) The solid is the set of all (x,y,z)(x,y,z) obeying 0≤x≤30\le x\le 3, 0≤y≤30\le y\le 3 and 0≤z≤6−xy0\le z\le 6-xy. The base of this region is the set of all (x,y)(x,y) for which there is a zz such that (x,y,z)(x,y,z) is in the solid. So the base is the set of all (x,y)(x,y) obeying 0≤x≤30\le x\le 3, 0≤y≤30\le y\le 3 and 6−xy≥06-xy\ge 0, i.e. xy≤6xy\le 6. This region is sketched in the figure on the left below.

Figure from prob_s3.1, line 1672

Figure from prob_s3.1, line 1672

Figure from prob_s3.1, line 1682

Figure from prob_s3.1, line 1682

(b) We'll deompose the base region into vertical strips as in the figure on the right above. Observe that the line y=3y=3 intersects the curve xy=6xy=6 at the point (2,3)(2,3) and that on the base

  • xx runs from 00 to 33 and that

  • for each fixed xx between 00 and 22, yy runs from 00 to 33, while

  • for each fixed xx between 22 and 33, yy runs from 00 to 6/x6/x

and that, for each (x,y)(x,y) in the base, zz runs from 00 to 6−xy6-xy. So the

Volume=∫02dx∫03dy (6−xy)+∫23dx∫06/xdy (6−xy)=∫02dx [6y−12xy2]03+∫23dx [6y−12xy2]06/x=∫02dx [18−92x]+∫23dx [36x−18x]=[18x−94x2]02+[18ln⁡x]23=27+18ln⁡32≈34.30\begin{align*} \text{Volume}&=\int_0^2\dee{x}\int_0^3\dee{y}\ (6-xy)+ \int_2^3\dee{x}\int_0^{6/x} \dee{y}\ (6-xy) \\ &=\int_0^2\dee{x}\ \left[6y-\frac{1}{2} xy^2\right]_0^3 +\int_2^3\dee{x}\ \left[6y-\frac{1}{2} xy^2\right]_0^{6/x} \\ &=\int_0^2\dee{x}\ \left[18-\frac{9}{2}x\right] +\int_2^3\dee{x}\ \left[\frac{36}{x}-\frac{18}{x}\right] \\ &=\left[18x-\frac{9}{4}x^2\right]_0^2+\Big[18\ln x\Big]_2^3 =27+18\ln\frac{3}{2}\approx 34.30 \end{align*}
Q24Stage 2Past exam · M200 2002D

Evaluate the following integral:

∫−22∫x24cos⁡(y3/2) dy dx\begin{equation*} \int_{-2}^2\int_{x^2}^4\cos\big(y^{3/2}\big)\ \dee{y}\,\dee{x} \end{equation*}
Hint

The inside integral, ∫x24cos⁡(y3/2) dy\int_{x^2}^4\cos\big(y^{3/2}\big)\ \dee{y}, in the given integral looks really nasty. So try exchanging the order of integration.

Answer

43sin⁡8≈1.319\frac{4}{3}\sin 8\approx 1.319

Full solution

In the given integral

  • xx runs from −2-2 to 22 and

  • for each fixed xx between −2-2 and 22, yy runs from x2x^2 to 44

So the domain of integration is

D={ (x,y) ∣ −2≤x≤2, x2≤y≤4 }\begin{equation*} D = \Set{(x,y)}{-2\le x\le 2,\ x^2\le y\le 4} \end{equation*}

This is sketched below.

Figure from prob_s3.1, line 1739

Figure from prob_s3.1, line 1739

The inside integral, ∫x24cos⁡(y3/2) dy\int_{x^2}^4\cos\big(y^{3/2}\big)\ \dee{y}, in the given integral looks really nasty. So let's try exchanging the order of integration. The given integral was formed by decomposing the domain of integration DD into horizontal strips, like the blue strip in the figure above. To exchange the order of integration we instead decompose the domain of integration DD into vertical strips, like the pink strip in the figure above. To do so, we observe that, on DD,

  • yy runs from 00 to 44 and

  • for each fixed yy between 00 and 44, xx runs from −y-\sqrt{y} to y\sqrt{y}.

That is, we reexpress the domain of integration as

D={ (x,y) ∣ 0≤y≤4, −y≤x≤y }\begin{equation*} D = \Set{(x,y)}{0\le y\le 4,\ -\sqrt{y}\le x\le \sqrt{y}} \end{equation*}

and the given integral as

∫−22∫x24cos⁡(y3/2) dy dx=∫04dy∫−yydx cos⁡(y3/2)=∫04dy 2ycos⁡(y3/2)=43∫08dt cos⁡twhere t=y3/2, dt=32y dy=43sin⁡t∣08=43sin⁡8≈1.319\begin{align*} \int_{-2}^2\int_{x^2}^4\cos\big(y^{3/2}\big)\ \dee{y}\,\dee{x} &=\int_0^4\dee{y} \int_{-\sqrt{y}}^{\sqrt{y}}\dee{x}\ \cos\big(y^{3/2}\big)\cr &=\int_0^4\dee{y} \ 2\sqrt{y}\cos\big(y^{3/2}\big)\cr &=\frac{4}{3}\int_0^8\dee{t}\ \cos t\quad\hbox{where } t=y^{3/2},\ \dee{t}=\frac{3}{2}\sqrt{y}\ \dee{y}\cr &=\frac{4}{3}\sin t\Big|_0^8 =\frac{4}{3}\sin 8\approx 1.319 \end{align*}
Q25Stage 2Past exam · M200 2002A

Consider the volume above the xyxy-plane that is inside the circular cylinder x2+y2=2yx^2+y^2=2y and underneath the surface z=8+2xyz=8+2xy.

  1. Express this volume as a double integral II, stating clearly the domain over which I is to be taken.

  2. Express in Cartesian coordinates, the double integral II as an iterated intergal in two different ways, indicating clearly the limits of integration in each case.

  3. How much is this volume?

Answer

(a) I=∬D(8+2xy) dxdy\dst I=\dblInt_D (8+2xy)\ \dee{x}\dee{y} where D={ (x,y) ∣ x2+(y−1)2≤1 }D=\Set{(x,y)}{x^2+(y-1)^2\le 1}

(b) I=∫02dy∫−2y−y22y−y2dx (8+2xy)=∫−11dx∫1−1−x21+1−x2dy (8+2xy)\dst I =\int_0^2 \dee{y}\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}\dee{x}\ (8+2xy) =\int_{-1}^1 \dee{x}\int_{1-\sqrt{1-x^2}}^{1+\sqrt{1-x^2}}\dee{y}\ (8+2xy)

(c) 8π8\pi

Full solution

(a) We may rewrite the equation x2+y2=2yx^2+y^2=2y of the cylinder as x2+(y−1)2=1x^2+(y-1)^2=1. We are (in part (c)) to find the volume of the set

V={ (x,y,z) ∣ x2+(y−1)2≤1, 0≤z≤8+2xy }\begin{equation*} V=\Set{(x,y,z)}{x^2+(y-1)^2\le 1,\ 0\le z\le 8+2xy} \end{equation*}

When we look at this solid from far above (so that we can't see zz) we see the set of points (x,y)(x,y) that obey x2+(y−1)2≤1x^2+(y-1)^2\le 1 and 8+2xy≥08+2xy\ge 0 (so that there is at least one allowed zz for that (x,y)(x,y)). All points in x2+(y−1)2≤1x^2+(y-1)^2\le 1 have −1≤x≤1-1\le x\le 1 and 0≤y≤20\le y\le 2 and hence −2≤xy≤2-2\le xy\le 2 and 8+2xy≥08+2xy\ge 0. So the domain of integration consists of the full disk

D={ (x,y) ∣ x2+(y−1)2≤1 }\begin{equation*} D = \Set{(x,y)}{x^2+(y-1)^2\le 1} \end{equation*}

The volume is

I=∬D(8+2xy) dxdy\begin{equation*} I=\dblInt_D (8+2xy)\ \dee{x}\dee{y} \end{equation*}

(b) We can express the double integral over DD as iterated integrals by decomposing DD into horizontal strips, like the pink strip in the figure below, and also by decomposing DD into blue strips, like the blue strip in the figure below.

Figure from prob_s3.1, line 1817

Figure from prob_s3.1, line 1817

For horizontal strips, we use that, on DD

  • yy runs from 00 to 22 and,

  • for each fixed yy between 00 and 22, xx runs from −2y−y2-\sqrt{2y-y^2} to 2y−y2\sqrt{2y-y^2}

so that

D={ (x,y) ∣ 0≤y≤2, −2y−y2≤x≤2y−y2 }\begin{equation*} D=\Set{(x,y)}{0\le y\le 2,\ -\sqrt{2y-y^2}\le x\le \sqrt{2y-y^2}} \end{equation*}

For vertical strips, we use that, on DD

  • xx runs from −1-1 to 11 and,

  • for each fixed xx between −1-1 and 11, yy runs from 1−1−x21-\sqrt{1-x^2} to 1+1−x21+\sqrt{1-x^2}

so that

D={ (x,y) ∣ −1≤x≤1, 1−1−x2≤y≤1+1−x2 }\begin{equation*} D=\Set{(x,y)}{-1\le x\le 1,\ 1-\sqrt{1-x^2}\le y\le 1+\sqrt{1-x^2}} \end{equation*}

Thus

I=∫02dy∫−2y−y22y−y2dx (8+2xy)=∫−11dx∫1−1−x21+1−x2dy (8+2xy)\begin{align*} I&=\int_0^2 \dee{y}\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}\dee{x}\ (8+2xy) \\ &=\int_{-1}^1 \dee{x}\int_{1-\sqrt{1-x^2}}^{1+\sqrt{1-x^2}}\dee{y}\ (8+2xy) \end{align*}

(c) Since ∬D8 dxdy\dblInt_D 8\ \dee{x}\dee{y} is just 88 times the area of DD, which is π\pi,

Volume=8π+∫02dy∫−2y−y22y−y2dx 2xy=8π+2∫02dy y∫−2y−y22y−y2dx x=8π\begin{align*} \text{Volume}&=8\pi+\int_0^2 \dee{y}\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}\dee{x}\ 2xy =8\pi+2\int_0^2 \dee{y}\ y\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}\dee{x}\ x \\ &=8\pi \end{align*}

because ∫−2y−y22y−y2dx x=0\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}\dee{x}\ x=0 for all yy, because the integrand is odd and the domain of integration is even.

Q26Stage 2Past exam · M200 2001D

Evaluate the following integral:

∫09∫y3sin⁡(πx3) dxdy\begin{equation*} \int_0^9\int_{\sqrt{y}}^3\sin(\pi x^3)\ \dee{x}\dee{y} \end{equation*}
Hint

The inside integral, ∫y3sin⁡(πx3) dx\dst\int_{\sqrt{y}}^3\sin\big(\pi x^3\big)\ \dee{x}, in the given integral looks really nasty. So try exchanging the order of integration.

Answer

23π≈0.212\frac{2}{3\pi}\approx 0.212

Full solution

In the given integral

  • yy runs from 00 to 99 and

  • for each fixed yy between 00 and 99, xx runs from y\sqrt{y} to 33

So the domain of integration is

D={ (x,y) ∣ 0≤y≤9, y≤x≤3 }\begin{equation*} D = \Set{(x,y)}{0\le y\le 9,\ \sqrt{y}\le x\le 3} \end{equation*}

This is sketched below.

Figure from prob_s3.1, line 1903

Figure from prob_s3.1, line 1903

The inside integral, ∫y3sin⁡(πx3) dx\int_{\sqrt{y}}^3\sin\big(\pi x^3\big)\ \dee{x}, in the given integral looks really nasty. So let's try exchanging the order of integration. The given integral was formed by decomposing the domain of integration DD into horizontal strips, like the blue strip in the figure above. To exchange the order of integration we instead decompose the domain of integration DD into vertical strips, like the pink strip in the figure above. To do so, we observe that, on DD,

  • xx runs from 00 to 33 and

  • for each fixed xx between 00 and 33, yy runs from 00 to x2x^2.

That is, we reexpress the domain of integration as

D={ (x,y) ∣ 0≤x≤3, 0≤y≤x2 }\begin{equation*} D = \Set{(x,y)}{0\le x\le 3,\ 0\le y\le x^2} \end{equation*}

and the given integral as

∫09∫y3sin⁡(πx3) dxdy=∫03dx∫0x2dy sin⁡(πx3)=∫03dx x2sin⁡(πx3)=13π∫027πdt sin⁡twhere t=πx3, dt=3πx2 dx=−13πcos⁡t∣027π=−13πcos⁡t∣0π=23π≈0.212\begin{align*} \int_0^9\int_{\sqrt{y}}^3\sin(\pi x^3)\ \dee{x}\dee{y} &=\int_0^3\dee{x} \int_0^{x^2}\dee{y}\ \sin(\pi x^3) \cr &=\int_0^3\dee{x} \ x^2\sin(\pi x^3)\cr &=\frac{1}{3\pi}\int_0^{27\pi}\dee{t}\ \sin t\quad\hbox{where } t=\pi x^3,\ \dee{t}=3\pi x^2\,\dee{x}\cr &=-\frac{1}{3\pi}\cos t\Big|_0^{27\pi}=-\frac{1}{3\pi}\cos t\Big|_0^\pi =\frac{2}{3\pi}\approx 0.212 \end{align*}
Q27Stage 2Past exam · M200 2000D

The iterated integral

I=∫01[∫−xxsin⁡(y3−3y) dy] dx\begin{equation*} I=\int_0^1\bigg[\int_{-\sqrt{x}}^{\sqrt{x}} \sin\big(y^3-3y\big)\,\dee{y}\bigg] \ \dee{x} \end{equation*}

is equal to ∬Rsin⁡(y3−3y) dA\dblInt_R\sin\big(y^3-3y)\ dA for a suitable region RR in the xyxy-plane.

  1. Sketch the region RR.

  2. Write the integral II with the orders of integration reversed, and with suitable limits of integration.

  3. Find II.

Answer

(b) ∫−11[∫y21sin⁡(y3−3y) dx] dy\dst\int_{-1}^1\bigg[\int_{y^2}^1 \sin\big(y^3-3y\big)\,\dee{x}\bigg]\ \dee{y} (c) 00

Full solution

(a) In the given integral

  • xx runs from 00 to 11, and

  • for each fixed xx between 00 and 11, yy runs from −x-\sqrt{x} to x\sqrt{x}.

So the region

R={ (x,y) ∣ 0≤x≤1, −x≤y≤x }\begin{equation*} R=\Set{(x,y)}{0\le x\le 1,\ -\sqrt{x}\le y\le \sqrt{x}} \end{equation*}

It is sketched below.

Figure from prob_s3.1, line 1980

Figure from prob_s3.1, line 1980

(b) The given integral was formed by decomposing the domain of integration RR into vertical strips, like the pink strip in the figure above. To exchange the order of integration we instead decompose the domain of integration RR into horizontal strips, like the blue strip in the figure above. To do so, we observe that, on RR,

  • yy runs from −1-1 to 11, and

  • for each fixed yy between −1-1 and 11, xx runs from y2y^2 to 11.

So

I=∫−11[∫y21sin⁡(y3−3y) dx] dy\begin{equation*} I=\int_{-1}^1\bigg[\int_{y^2}^1 \sin\big(y^3-3y\big)\,\dee{x}\bigg]\ \dee{y} \end{equation*}

(c) The easy way to evaluate II is to observe that, since sin⁡(y3−3y)\sin\big(y^3-3y\big) is odd under y→−yy\rightarrow -y, the integral

∫−xxsin⁡(y3−3y) dy=0\begin{equation*} \int_{-\sqrt{x}}^{\sqrt{x}} \sin\big(y^3-3y\big)\,\dee{y}=0 \end{equation*}

for all xx. Hence I=0I=0. The hard way is

I=∫−11[∫y21sin⁡(y3−3y) dx] dy=∫−11(1−y2)sin⁡(y3−3y) dy=∫2−2sin⁡t dt−3 where t=y3−3y, dt=3(y2−1) dy=13cos⁡t ∣2−2=0\begin{align*} I&=\int_{-1}^1\bigg[\int_{y^2}^1 \sin\big(y^3-3y\big)\,\dee{x}\bigg]\ \dee{y}\\ &=\int_{-1}^1 (1-y^2) \sin\big(y^3-3y\big)\ \dee{y}\\ &=\int_2^{-2} \sin t\ \frac{\dee{t}}{-3} \qquad\hbox{ where }t=y^3-3y,\ \dee{t} =3(y^2-1)\,\dee{y} \\ &=\frac{1}{3}\cos t\ \Big|_2^{-2}=0 \end{align*}

again, since cos⁡\cos is even.

Q28Stage 2Past exam · M200 2000A

Find the double integral of the function f(x,y)=xyf(x, y) = xy over the region bounded by y=x−1y = x - 1 and y2=2x+6y^2 = 2x + 6.

Answer

3636

Full solution

The parabola y2=2x+6y^2=2x+6 and the line y=x−1y=x-1 meet when x=y+1x=y+1 with y2=2(y+1)+6y^2=2(y+1)+6 or y2−2y−8=(y−4)(y+2)=0y^2-2y-8=(y-4)(y+2)=0. So they meet at (−1,−2)(-1,-2) and (5,4)(5,4). The domain of integration is sketched below.

Figure from prob_s3.1, line 2044

Figure from prob_s3.1, line 2044

On this domain

  • yy runs from −2-2 to 44, and

  • for each fixed yy between −2-2 and 44, xx runs from y22−3\frac{y^2}{2}-3 to y+1y+1.

So the integral is

∫−24dy∫y2/2−3y+1dx xy=∫−24dy 12x2y∣y2/2−3y+1=12∫−24dy [y3+2y2+y−14y5+3y3−9y]=12∫−24dy [−8y+2y2+4y3−14y5]=[−2y2+13y3+12y4−148y6]−24=−2(16−4)+13(64+8)+12(256−16)−148(4096−64)=−24+24+120−84=36\begin{align*} \int_{-2}^4\dee{y}\int_{y^2/2-3}^{y+1}\dee{x}\ xy &=\int_{-2}^4\dee{y}\ \frac{1}{2} x^2y\bigg|_{y^2/2-3}^{y+1} \\ &=\frac{1}{2}\int_{-2}^4\dee{y}\ \left[y^3+2y^2+y-\frac{1}{4}y^5+3y^3-9y\right] \\ &=\frac{1}{2}\int_{-2}^4\dee{y}\ \left[-8y+2y^2+4y^3-\frac{1}{4}y^5\right] \\ &=\left[-2y^2+\frac{1}{3}y^3+\frac{1}{2}y^4-\frac{1}{48}y^6\right]_{-2}^4 \\ &=-2(16-4)+\frac{1}{3}(64+8)+\frac{1}{2}(256-16) -\frac{1}{48}(4096-64)\cr &=-24+24+120-84=36 \end{align*}

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q29Stage 3

Find the volume of the solid inside the cylinder x2+2y2=8x^2+2y^2=8, above the plane z=y−4z=y-4 and below the plane z=8−xz=8-x.

Answer

482 π48\sqrt{2}\,\pi

Full solution

Looking down from the top, we see the cylinder x2+2y2≤8x^2+2y^2\le 8. That gives the base region. The top of the solid, above any fixed (x,y)(x,y) in the base region, is at z=8−xz=8-x (this is always positive because xx never gets bigger than 8\sqrt{8}) . The bottom of the solid, below any fixed (x,y)(x,y) in the base region, is at z=y−4z=y-4 (this is always negative because yy is always smaller than 22). So the height of the solid at any (x,y)(x,y) is

ztop−zbottom=(8−x)−(y−4)=12−x−y\begin{equation*} z_{\rm top}-z_{\rm bottom} =(8-x)-(y-4)=12-x-y \end{equation*}

The volume is

∫−22dy∫−8−2y28−2y2dx (12−x−y)\begin{equation*} \int_{-2}^{2}\dee{y}\int_{-\sqrt{8-2y^2}}^{\sqrt{8-2y^2}}\dee{x}\ (12-x-y) \end{equation*}

Recall, from Theorem 1.2.11 in the CLP-2 text, that if f(x)f(x) is an odd function (meaning that f(−x)=−f(x)f(-x)=-f(x) for all xx), then ∫−aaf(x) dx=0\int_{-a}^a f(x)\ \dee{x}=0 (because the two integrals ∫0af(x) dx\int_0^a f(x)\ \dee{x} and ∫−a0f(x) dx\int_{-a}^0 f(x)\ \dee{x} have the same magnitude but opposite signs). Applying this twice gives

∫−8−2y28−2y2dx x=0 and ∫−22dy∫−8−2y28−2y2dx y=∫−22dy 2y8−2y2=0\begin{equation*} \int_{-\sqrt{8-2y^2}}^{\sqrt{8-2y^2}}\dee{x}\ x=0\text{ and } \int_{-2}^{2}\dee{y}\int_{-\sqrt{8-2y^2}}^{\sqrt{8-2y^2}}\dee{x}\ y =\int_{-2}^{2}\dee{y}\ 2y\sqrt{8-2y^2}=0 \end{equation*}

since xx and y8−2y2y\sqrt{8-2y^2} are both odd. Thus

∫−22dy∫−8−2y28−2y2dx (−x−y)=0  ⟹  Volume=∫−22dy∫−8−2y28−2y2dx 12\begin{equation*} \int_{-2}^{2}\dee{y}\int_{-\sqrt{8-2y^2}}^{\sqrt{8-2y^2}}\dee{x}\ (-x-y)=0 \implies \text{Volume} = \int_{-2}^{2}\dee{y}\int_{-\sqrt{8-2y^2}}^{\sqrt{8-2y^2}}\dee{x}\ 12 \end{equation*}

so that the volume is just 12 times the area of the ellipse x2+2y2=8x^2+2y^2=8, which is

12(π 8 2)=482 π\begin{equation*} 12\big(\pi\,\sqrt{8}\,2\big)= 48\sqrt{2}\,\pi \end{equation*}

My list

nothing marked yet

Loading…

Open the whole list →

Your tutor can open this list with you. It follows your account, so it is there on whichever device you study on.

From the UBC Math 200 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.