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Multiple Integrals

3.5 Triple Integrals

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

1Stage 1Conceptual

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

Q1Stage 1

Evaluate the integral

∬Rb2−y2 dx dy  where R is the rectangle 0≤x≤a, 0≤y≤b\begin{equation*} \dblInt_R \sqrt{b^2-y^2}\ \dee{x}\,\dee{y}\text{\ \ where }R\text{ is the rectangle }0\le x\le a,\ 0\le y\le b \end{equation*}

without using iteration. Instead, interpret the integral geometrically.

Answer

14πab2\frac{1}{4}\pi ab^2

Full solution

∬Rb2−y2 dx dy=∭Vdx dy dz\dblInt_R \sqrt{b^2-y^2}\,\dee{x}\,\dee{y} =\tripInt_V \dee{x}\,\dee{y}\,\dee{z}, where

V={ (x,y,z) ∣ 0≤z≤b2−y2, 0≤x≤a, 0≤y≤b }={ (x,y,z) ∣ y2+z2≤b2, 0≤x≤a, y≥0, z≥0 }\begin{align*} V&=\Set{(x,y,z)} {0\le z\le\sqrt{b^2-y^2},\ 0\le x\le a,\ 0\le y\le b} \\ &=\Set{(x,y,z)} {y^2+z^2\le b^2,\ 0\le x\le a,\ y\ge 0,\ z\ge 0} \end{align*}

Now y2+z2≤b2y^2+z^2\le b^2 is a cylinder of radius bb centered on the xx–axis and the part of y2+z2≤b2y^2+z^2\le b^2, with y≥0, z≥0y\ge0,\ z\ge0 is one quarter of this cylinder. It has cross–sectional area 14πb2\frac{1}{4}\pi b^2. VV is the part of this quarter–cylinder with 0≤x≤a0\le x\le a. It has length aa and cross–sectional area 14πb2\frac{1}{4}\pi b^2. So, ∬Rb2−y2 dx dy=14πab2\dblInt_R \sqrt{b^2-y^2}\,\dee{x}\,\dee{y}=\frac{1}{4}\pi ab^2.

Q2Stage 1Past exam · M200 2013D

Find the total mass of the rectangular box [0,1]×[0,2]×[0,3][0, 1] \times [0, 2] \times [0, 3] (that is, the box defined by the inequalities 0≤x≤10 \le x \le 1, 0≤y≤20 \le y \le 2, 0≤z≤30 \le z \le 3), with density function h(x,y,z)=xh(x, y, z) = x.

Answer

33

Full solution

The mass is

∫01dx∫02dy∫03dz x=6∫01dx x=3\begin{align*} \int_0^1\dee{x} \int_0^2\dee{y}\int_0^3\dee{z}\ x &= 6\int_0^1 \dee{x}\ x =3 \end{align*}

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q3Stage 2

Evaluate ∭Rx dV\dst\tripInt_R x\ \dee{V} where RR is the tetrahedron bounded by the coordinate planes and the plane xa+yb+zc=1\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.

Answer

a2bc24\frac{a^2bc}{24}

Full solution

The domain of integration is

V={ (x,y,z) ∣ x,y,z≥0, xa+yb+zc≤1 }\begin{equation*} V=\Set{(x,y,z)}{x,y,z\ge 0,\ \tfrac{x}{a}+\tfrac{y}{b}+\tfrac{z}{c}\le 1} \end{equation*}

  • In VV, zc≤1−xa−yb\tfrac{z}{c}\le 1-\tfrac{x}{a}-\tfrac{y}{b} and x,y≥0x,y\ge0, so the biggest value of zz in VV is achieved when x=y=0x=y=0 and is cc. Thus, in VV, zz runs from 00 to cc.

  • For each fixed 0≤z≤c0\le z\le c, (x,y)(x,y) takes all values in

    Dz={ (x,y) ∣ x,y≥0, xa+yb≤1−zc }\begin{equation*} D_z=\Set{(x,y)}{x,y\ge 0,\ \tfrac{x}{a}+\tfrac{y}{b}\le 1-\tfrac{z}{c}} \end{equation*}

    The biggest value of yy on DzD_z is achieved when x=0x=0 and is b(1−zc)b\left(1-\tfrac{z}{c}\right). Thus, on DzD_z, yy runs from 00 to b(1−zc)b\left(1-\tfrac{z}{c}\right).

  • For each fixed 0≤z≤c0\le z\le c and 0≤y≤b(1−zc)0\le y \le b\left(1-\tfrac{z}{c}\right), xx runs over

    Dy,z={ x ∣ 0≤x≤a(1−yb−zc) }\begin{equation*} D_{y,z}=\Set{x}{0\le x \le a\left(1-\tfrac{y}{b}-\tfrac{z}{c}\right)} \end{equation*}

    This is pictured in the figure on the right below.

Figure from prob_s3.5, line 90

Figure from prob_s3.5, line 90

Figure from prob_s3.5, line 90

Figure from prob_s3.5, line 90

So the specified integral is

∭Rx dV=∫0cdz∬Dzdx dy x=∫0cdz∫0b(1−zc)dy∫Dy,zdx x=∫0cdz∫0b(1−zc)dy∫0a(1−yb−zc)dx x=∫0cdz∫0b(1−zc)dy a22(1−yb−zc)2=∫0cdz [−a2b6(1−yb−zc)3]0b(1−zc)=∫0cdz a2b6(1−zc)3=[−a2bc24(1−zc)4]0c=a2bc24\begin{align*} \tripInt_R x\ \dee{V} &=\int_0^c \dee{z} \dblInt_{D_z} \dee{x}\,\dee{y}\ x = \int_0^c \dee{z}\int_0^{b(1-{z\over c})}\dee{y}\int_{D_{y,z}}\dee{x}\ x\\ &=\int_0^c \dee{z}\int_0^{b(1-{z\over c})}\dee{y} \int_0^{a(1-{y\over b}-{z\over c})} \dee{x}\ x =\int_0^c \dee{z}\int_0^{b(1-{z\over c})} \dee{y}\ \frac{a^2}{2} \left(1-\frac{y}{b}-\frac{z}{c}\right)^2 \\ &=\int_0^c \dee{z}\ \left[-\frac{a^2b}{6} \left(1-\frac{y}{b}-\frac{z}{c}\right)^3 \right]_0^{b(1-{z\over c})} =\int_0^c \dee{z}\ \frac{a^2b}{6}\left(1-\frac{z}{c}\right)^3 \\ &=\left[-\frac{a^2bc}{24}\left(1-\frac{z}{c}\right)^4 \right]_0^c =\frac{a^2bc}{24} \end{align*}
Q4Stage 2

Evaluate ∭Ry dV\dst\tripInt_R y\ \dee{V} where RR is the portion of the cube 0≤x,y,z≤10\le x,y,z\le 1 lying above the plane y+z=1y+z=1 and below the plane x+y+z=2x+y+z=2.

Answer

524\frac{5}{24}

Full solution

The domain of integration is

R={ (x,y,z) ∣ 0≤x,y,z≤1, z≥1−y, z≤2−x−y }\begin{equation*} R = \Set{(x,y,z)}{0\le x,y,z\le 1,\ z\ge 1-y,\ z\le2-x-y} \end{equation*}

In the figure on the below, the more darkly shaded region is part of z=1−yz=1-y and the more lightly shaded region is part of z=2−x−yz=2-x-y.

Figure from prob_s3.5, line 154

Figure from prob_s3.5, line 154

  • In RR, zz runs from 00 (for example (0,1,0)(0,1,0) is in RR) to 11 (for example (0,0,1)(0,0,1) is in RR).

  • For each fixed 0≤z≤10\le z\le 1, (x,y)(x,y) runs over

    Dz={ (x,y) ∣ 0≤x,y≤1, y≥1−z, x+y≤2−z }\begin{equation*} D_z = \Set{(x,y)}{0\le x,y\le 1,\ y\ge 1-z,\ x+y\le2-z} \end{equation*}

    Here is a sketch of a top view of DzD_z.

    Figure from prob_s3.5, line 154

    Figure from prob_s3.5, line 154

    On DzD_z, yy runs from 1−z1-z to 11.

  • For each fixed 0≤z≤10\le z\le 1 and 1−z≤y≤11-z\le y\le 1, xx runs from 00 to 2−y−z2-y-z.

So the specified integral is

∭Ry dV=∫01dz∬Dzdxdy y=∫01dz∫1−z1dy∫02−y−zdx y=∫01dz∫1−z1dy y(2−y−z)=−∫01dz∫z0du (1−u)(1+u−z) where u=1−y=∫01dz∫0zdu (1−u2−z+uz)=∫01dz (z−z33−z2+z32)=12−112−13+18=524\begin{align*} \tripInt_R y\ \dee{V} &=\int_0^1\dee{z}\dblInt_{D_z} \dee{x}\dee{y}\ y =\int_0^1 \dee{z}\int_{1-z}^1 \dee{y}\int_0^{2-y-z} \dee{x}\ y =\int_0^1 \dee{z}\int_{1-z}^1 \dee{y}\ y(2-y-z)\cr &=-\int_0^1 \dee{z}\int_z^0 du\ (1-u)(1+u-z)\qquad\text{ where }u=1-y\\ &=\int_0^1 \dee{z}\int_0^z du\ (1-u^2-z+uz) =\int_0^1 \dee{z}\ \big(z-\frac{z^3}{3}-z^2+\frac{z^3}{2})\\ &=\frac{1}{2}-\frac{1}{12}-\frac{1}{3}+\frac{1}{8} =\frac{5}{24} \end{align*}
Q5Stage 2

For each of the following, express the given iterated integral as an iterated integral in which the integrations are performed in the order: first zz, then yy, then xx.

  1. ∫01dz∫01−zdy∫01−zdx f(x,y,z)\dst\int_0^1\dee{z}\int_0^{1-z}\dee{y}\int_0^{1-z} \dee{x}\ f(x,y,z)

  2. ∫01dz∫z1dy∫0ydx f(x,y,z)\dst\int_0^1\dee{z}\int_{\sqrt z}^1\dee{y}\int_0^y \dee{x}\ f(x,y,z)

Answer

(a) ∫01dx∫0xdy∫01−xdzf(x,y,z)+∫01dx∫x1dy∫01−ydzf(x,y,z)\dst\int_0^1\dee{x}\int_0^{x}\dee{y}\int_0^{1-x}\hskip-10pt \dee{z}\hskip3pt f(x,y,z) +\int_0^1\dee{x}\int_x^{1}\dee{y}\int_0^{1-y}\hskip-10pt \dee{z}\hskip3pt f(x,y,z)

(b) ∫01dx∫x1dy∫0y2 ⁣ ⁣dz f(x,y,z)\dst\int_0^1\dee{x}\int_x^{1}\dee{y}\int_0^{y^2}\!\! \dee{z}\ f(x,y,z)

Full solution

(a) The domain of integration is

V={ (x,y,z) ∣ 0≤z≤1, 0≤y≤1−z, 0≤x≤1−z }={ (x,y,z) ∣ x,y,z≥0, x+z≤1, y+z≤1 }\begin{align*} V &= \Set{(x,y,z)}{0\le z\le 1,\ 0\le y\le 1-z,\ 0\le x\le 1-z} \\ &= \Set{(x,y,z)}{ x,y,z\ge 0,\ x+z\le 1,\ y+z\le 1 } \end{align*}

This is sketched in the figure below. The front face is x+z=1x+z=1 and the lightly shaded right face is y+z=1y+z=1.

Figure from prob_s3.5, line 221

Figure from prob_s3.5, line 221

In VV,

  • xx takes all values between 00 and 11.

  • For each fixed 0≤x≤10\le x\le 1, (y,z)(y,z) takes all values in

    Dx={ (y,z) ∣ y,z≥0, z≤1−x, y+z≤1 }\begin{equation*} D_x=\Set{(y,z)}{y,z\ge 0,\ z\le 1-x,\ y+z\le 1} \end{equation*}

    Here is a sketch of DxD_x.

    Figure from prob_s3.5, line 221

    Figure from prob_s3.5, line 221

  • Looking at the sketch above, we see that, on DxD_x, yy runs from 00 to 11 and

    • for each fixed yy between 00 and xx, zz runs from 00 to 1−x1-x and

    • for each fixed yy between xx and 11, zz runs from 00 to 1−y1-y

So the integral is, in the new order,

∭Vf(x,y,z) dV=∫01dx∬Dxdy dz f(x,y,z)=∫01dx∫0xdy∫01−xdzf(x,y,z)+∫01dx∫x1dy∫01−ydzf(x,y,z)\begin{align*} \tripInt_V f(x,y,z)\ \dee{V} &= \int_0^1\dee{x}\dblInt_{D_x} \dee{y}\,\dee{z}\ f(x,y,z) \\ &= \int_0^1\dee{x}\int_0^{x}\dee{y}\int_0^{1-x}\hskip-10pt \dee{z}\hskip3pt f(x,y,z) +\int_0^1\dee{x}\int_x^{1}\dee{y}\int_0^{1-y}\hskip-10pt \dee{z}\hskip3pt f(x,y,z) \end{align*}

(b) The domain of integration is

V={ (x,y,z) ∣ 0≤z≤1, z≤y≤1, 0≤x≤y }={ (x,y,z) ∣ 0≤z≤y2, 0≤x≤y≤1 }\begin{align*} V &= \Set{(x,y,z)}{0\le z\le 1,\ \sqrt{z}\le y\le 1,\ 0\le x\le y} \\ &= \Set{(x,y,z)}{0\le z\le y^2,\ 0\le x\le y\le 1} \end{align*}

In this region, xx takes all values between 00 and 11. For each fixed xx between 00 and 11, (y,z)(y,z) takes all values in

Dx={ (y,z) ∣ 0≤z≤y2, x≤y≤1 }\begin{equation*} D_x=\Set{(y,z)}{0\le z\le y^2,\ x\le y\le 1} \end{equation*}

Here is a sketch of DxD_x.

Figure from prob_s3.5, line 221

Figure from prob_s3.5, line 221

In the new order, the integral is

∫01dx∬Dxdy dz f(x,y,z)=∫01dx∫x1dy∫0y2 ⁣ ⁣dz f(x,y,z)\begin{equation*} \int_0^1\dee{x}\dblInt_{D_x} \dee{y}\,\dee{z}\ f(x,y,z) =\int_0^1\dee{x}\int_x^{1}\dee{y}\int_0^{y^2}\!\! \dee{z}\ f(x,y,z) \end{equation*}
Q6Stage 2Past exam · M200 2005D

A triple integral ∭Ef dV\dst\tripInt_E f\ \dee{V} is given in iterated form by

∫y=−1y=1∫z=0z=1−y2∫x=02−y−zf(x,y,z) dx dz dy\begin{equation*} \int_{y=-1}^{y=1} \int_{z=0}^{z=1-y^2} \int_{x=0}^{2-y-z} f(x,y,z) \ \dee{x}\,\dee{z}\,\dee{y} \end{equation*}

  1. Draw a reasonably accurate picture of EE in 3–dimensions.

  2. Rewrite the triple integral ∭Ef dV\tripInt_E f\ \dee{V} as one or more iterated triple integrals in the order

    ∫y=y=∫x=x=∫z=z=f(x,y,z) dz dx dy\begin{equation*} \int_{y=}^{y=} \int_{x=}^{x=} \int_{z=}^{z=} f(x,y,z) \ \dee{z}\,\dee{x}\,\dee{y} \end{equation*}
Answer

(a)

Figure from prob_s3.5, line 311

Figure from prob_s3.5, line 311

(b)

∫y=−1y=1∫x=0x=1+y2−y∫z=0z=1−y2f(x,y,z) dz dx dy+∫y=−1y=1∫x=1+y2−yx=2−y∫z=0z=2−x−yf(x,y,z) dz dx dy\begin{align*} &\int_{y=-1}^{y=1} \int_{x=0}^{x=1+y^2-y} \int_{z=0}^{z=1-y^2} f(x,y,z) \ \dee{z}\,\dee{x}\,\dee{y} \\ &\hskip1in +\int_{y=-1}^{y=1} \int_{x=1+y^2-y}^{x=2-y} \int_{z=0}^{z=2-x-y} f(x,y,z) \ \dee{z}\,\dee{x}\,\dee{y} \end{align*}
Full solution

(a) In the domain of integration for the given integral

  • yy runs from −1-1 to 11, and

  • for each fixed yy in that range zz runs from 00 to 1−y21-y^2, and

  • for each fixed yy and zz as above, xx runs from 00 to 2−y−z2-y-z.

That is,

E={ (x,y,z) ∣ −1≤y≤1, 0≤z≤1−y2, 0≤x≤2−y−z }\begin{equation*} E = \Set{(x,y,z)}{-1\le y\le 1,\ 0\le z\le 1-y^2,\ 0\le x\le 2-y-z} \end{equation*}

  • Each constant xx cross–section of the surface z=1−y2z=1-y^2 is an upside down parabola. So the surface z=1−y2z=1-y^2 consists of a bunch of copies of the parabola z=1−y2z=1-y^2 stacked front to back. The figure of the left below provides a sketch of z=1−y2z=1-y^2.

  • The surface x=2−y−zx = 2-y-z, or equivalently, x+y+z=2x+y+z=2 is a plane. It passes through the points (2,0,0)(2,0,0), (0,2,0)(0,2,0) and (0,0,2)(0,0,2). It is sketched in the figure on the right below. We know that our domain of integration extends to y=−1y=-1, so we have chosen to include in the sketch the part of the plane in x≥0x\ge 0, y≥−1y\ge-1, z≥0z\ge 0.

Figure from prob_s3.5, line 327

Figure from prob_s3.5, line 327

Figure from prob_s3.5, line 327

Figure from prob_s3.5, line 327

The domain EE is constructed by using the plane x+y+z=2x+y+z=2 to chop the front off of the “tunnel” 0≤z≤1−y20\le z\le 1-y^2. It is outlined in red in the figure below.

Figure from prob_s3.5, line 311

Figure from prob_s3.5, line 311

(b) We are to change the order of integration so that the outside integral is over yy (the same as the given integral), the middle integral is over xx, and the inside integral is over over zz.

  • We still have yy running from −1-1 to 11.

  • For each fixed yy in that range, (x,z)(x,z) runs over

    Ey={ (x,z) ∣ 0≤z≤1−y2, 0≤x+z≤2−y }\begin{equation*} E_y=\Set{(x,z)}{0\le z\le 1-y^2,\ 0\le x+z\le 2-y} \end{equation*}

  • The biggest value of xx in EyE_y is 2−y2-y. It is achieved when z=0z=0. You can also see this in the figure below. The shaded region in that figure is EyE_y.

  • For each fixed xx and yy as above, zz runs over

    Ex,y={ z ∣ 0≤z≤1−y2, 0≤z≤2−x−y }\begin{equation*} E_{x,y} = \Set{z}{0\le z\le 1-y^2,\ 0\le z\le 2-x-y} \end{equation*}

    That is, zz runs from 00 to the smaller of 1−y21-y^2 and 2−x−y2-x-y. Note that 1−y2≤2−x−y1-y^2\le 2-x-y if and only if x≤1+y2−yx\le 1+y^2-y.

  • So if 0≤x≤1+y2−y0\le x\le 1+y^2-y, zz runs from 00 to 1−y21-y^2 and if 1+y2−y≤x≤2−y1+y^2-y\le x\le 2-y, zz runs from 00 to 2−x−y2-x-y.

Figure from prob_s3.5, line 327

Figure from prob_s3.5, line 327

So the integral is

∫y=−1y=1∫x=0x=1+y2−y∫z=0z=1−y2f(x,y,z) dz dx dy+∫y=−1y=1∫x=1+y2−yx=2−y∫z=0z=2−x−yf(x,y,z) dz dx dy\begin{align*} &\int_{y=-1}^{y=1} \int_{x=0}^{x=1+y^2-y} \int_{z=0}^{z=1-y^2} f(x,y,z) \ \dee{z}\,\dee{x}\,\dee{y} \\ &\hskip1in +\int_{y=-1}^{y=1} \int_{x=1+y^2-y}^{x=2-y} \int_{z=0}^{z=2-x-y} f(x,y,z) \ \dee{z}\,\dee{x}\,\dee{y} \end{align*}
Q7Stage 2Past exam · M200 2006D

A triple integral ∭Ef(x,y,z) dV\tripInt_E f(x,y,z)\ \dee{V} is given in the iterated form

J=∫01∫01−x2∫04−2x−4zf(x,y,z) dy dz dx\begin{equation*} J = \int_0^1 \int_0^{1-\frac{x}{2}} \int_0^{4-2x-4z} f(x,y,z) \ \dee{y}\,\dee{z}\,\dee{x} \end{equation*}

  1. Sketch the domain EE in 3–dimensions.

  2. Rewrite the integral as one or more iterated integrals in the form

    J=∫y=y=∫x=x=∫z=z=f(x,y,z) dz dx dy\begin{equation*} J = \int_{y=}^{y=} \int_{x=}^{x=} \int_{z=}^{z=} f(x,y,z) \ \dee{z}\,\dee{x}\,\dee{y} \end{equation*}
Answer

(a)

Figure from prob_s3.5, line 436

Figure from prob_s3.5, line 436

(b)

J=∫y=0y=2∫x=0x=1∫z=0z=4−2x−y4 ⁣ ⁣ ⁣ ⁣f(x,y,z) dz dx dy+∫y=2y=4∫x=0x=4−y2∫z=0z=4−2x−y4 ⁣ ⁣ ⁣ ⁣f(x,y,z) dz dx dy\begin{align*} J = \int_{y=0}^{y=2} \int_{x=0}^{x=1} \int_{z=0}^{z=\frac{4-2x-y}{4}} \!\!\!\! f(x,y,z) \ \dee{z}\,\dee{x}\,\dee{y} +\int_{y=2}^{y=4} \int_{x=0}^{x=\frac{4-y}{2}} \int_{z=0}^{z=\frac{4-2x-y}{4}} \!\!\!\! f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y} \end{align*}
Full solution

(a) In the given integral JJ,

  • xx runs from 00 to 11,

  • for each fixed xx in that range, zz runs from 0 to 1−x21-\frac{x}{2}, and

  • for each fixed xx and zz as above, yy runs from 00 to 4−2x−4z4-2x-4z.

So

E={ (x,y,z) ∣ 0≤x≤1, 0≤z≤1−x2, 0≤y≤4−2x−4z }\begin{align*} E = \Set{(x,y,z)}{0\le x\le 1,\ 0\le z\le 1-\tfrac{x}{2},\ 0 \le y\le 4-2x-4z} \end{align*}

Notice that the condition y≤4−2x−4zy\le 4-2x-4z can be rewritten as z≤1−x2−y4z\le 1-\frac{x}{2} -\frac{y}{4}. When y≥0y\ge 0, this implies that z≤1−x2z\le 1-\frac{x}{2}, so that we can drop the condition z≤1−x2z\le 1-\frac{x}{2} from our description of EE:

E={ (x,y,z) ∣ 0≤x≤1, 0≤y≤4−2x−4z, z≥0 }\begin{align*} E = \Set{(x,y,z)}{0\le x\le 1,\ 0 \le y\le 4-2x-4z,\ z\ge 0} \end{align*}

First, we figure out what EE looks like. The plane 2x+y+4z=42x+y+4z=4 intersects the xx–, yy– and zz–axes at (2,0,0)(2,0,0), (0,4,0)(0,4,0) and (0,0,1)(0,0,1), respectively. That plane is shown in the sketch on the left below. The set of points { (x,y,z) ∣ x,y,z≥0, y≤4−2x−4z }\Set{(x,y,z)}{x,y,z\ge 0,\ y\le 4-2x-4z} is outlined with heavy lines.

Figure from prob_s3.5, line 453

Figure from prob_s3.5, line 453

Figure from prob_s3.5, line 453

Figure from prob_s3.5, line 453

So it only remains to impose the condtion x≤1x\le 1, which chops off the front bit of the tetrahedron. This is done in the sketch on the right above. Here is a cleaned up sketch of EE.

Figure from prob_s3.5, line 436

Figure from prob_s3.5, line 436

(b) We are to reorder the integration so that the outside integral is over yy, the middle integral is over xx, and the inside integral is over zz. Looking at the figure below,

Figure from prob_s3.5, line 453

Figure from prob_s3.5, line 453

we see that

  • yy runs from 00 to 44, and

  • for each fixed yy in that range, (x,z)(x,z) runs over

    { (x,z) ∣ 0≤x≤1, 2x+4z≤4−y, z≥0 }\begin{equation*} \Set{(x,z)}{0\le x\le 1,\ 2x+4z\le 4-y,\ z\ge 0} \end{equation*}

  • for each fixed yy between 00 and 22 (as in the left hand shaded bit in the figure above)

    • xx runs from 00 to 11, and then

    • for each fixed xx in that range, zz runs from 00 to 4−2x−y4\frac{4-2x-y}{4}.

  • for each fixed yy between 22 and 44 (as in the right hand shaded bit in the figure above)

    • xx runs from 00 to 4−y2\frac{4-y}{2} (the line of intersection of the plane 2x+y+4z=42x+y+4z=4 and the xyxy–plane is z=0z=0, 2x+y=42x+y=4), and then

    • for each fixed xx in that range, zz runs from 00 to 4−2x−y4\frac{4-2x-y}{4}.

So the integral

J=∫y=0y=2∫x=0x=1∫z=0z=4−2x−y4 ⁣ ⁣ ⁣ ⁣f(x,y,z) dz dx dy+∫y=2y=4∫x=0x=4−y2∫z=0z=4−2x−y4 ⁣ ⁣ ⁣ ⁣f(x,y,z) dz dx dy\begin{align*} J = \int_{y=0}^{y=2} \int_{x=0}^{x=1} \int_{z=0}^{z=\frac{4-2x-y}{4}} \!\!\!\! f(x,y,z) \ \dee{z}\,\dee{x}\,\dee{y} +\int_{y=2}^{y=4} \int_{x=0}^{x=\frac{4-y}{2}} \int_{z=0}^{z=\frac{4-2x-y}{4}} \!\!\!\! f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y} \end{align*}
Q8Stage 2Past exam · M200 2008A

Write the integral given below 55 other ways, each with a different order of integration.

I=∫01∫x1∫01−yf(x,y,z) dz dy dx\begin{equation*} I=\int_0^1 \int_{\sqrt{x}}^1 \int_0^{1-y} f(x,y,z)\,\dee{z}\,\dee{y}\,\dee{x} \end{equation*}
Answer
I=∫01∫x1∫01−yf(x,y,z) dz dy dx=∫01∫01−x∫x1−z f(x,y,z) dy dz dx=∫01∫0y2∫01−yf(x,y,z) dz dx dy=∫01∫01−y∫0y2f(x,y,z) dx dz dy=∫01∫0(1−z)2∫x1−zf(x,y,z) dy dx dz=∫01∫01−z∫0y2f(x,y,z) dx dy dz\begin{alignat*}{3} I&= \int_0^1 \int_{\sqrt{x}}^1 \int_0^{1-y} f(x,y,z)\,\dee{z}\,\dee{y}\,\dee{x} &&=\int_0^1 \int_0^{1-\sqrt{x}} \int_{\sqrt{x}}^{1-z} \ f(x,y,z)\ \dee{y}\,\dee{z}\,\dee{x} \\ &=\int_0^1 \int_0^{y^2}\int_0^{1-y}f(x,y,z)\ \dee{z}\,\dee{x}\, \dee{y} &&=\int_0^1 \int_0^{1-y} \int_0^{y^2} f(x,y,z)\ \dee{x}\, \dee{z}\,\dee{y} \\ &=\int_0^1 \int_0^{(1-z)^2} \int_{\sqrt{x}}^{1-z}f(x,y,z)\ \dee{y}\,\dee{x}\, \dee{z} &&=\int_0^1 \int_0^{1-z} \int_0^{y^2} f(x,y,z)\ \dee{x}\, \dee{y}\,\dee{z} \end{alignat*}
Full solution

Let's use VV to denote the domain of integration for the given integral. On VV

  • xx runs from 00 to 11, and

  • for each fixed xx in that range, yy runs from x\sqrt{x} to 11. In particular 0≤y≤10\le y\le 1. We can rewrite y=xy=\sqrt{x} as x=y2x=y^2 (with y≥0y\ge 0).

  • For each fixed xx and yy as above, zz runs from 00 to 1−y1-y.

So

V={ (x,y,z) ∣ 0≤x≤1, x≤y≤1, 0≤z≤1−y }={ (x,y,z) ∣ x,z≥0, x≤1, y≥x, y≤1, z≤1−y }\begin{align*} V &= \Set{(x,y,z)}{0\le x\le 1,\ \sqrt{x}\le y\le 1,\ 0\le z\le 1-y} \\ &= \Set{(x,y,z)}{x,z\ge 0,\ x\le 1,\ y\ge \sqrt{x},\ y\le 1,\ z\le 1-y} \end{align*}

Outside integral is with respect to xx:
We have already seen that 0≤x≤10\le x\le 1 and that, for each fixed xx in that range, (y,z)(y,z) runs over

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

Here are two sketches of VxV_x. The sketch on the left shows a vertical strip as was used in setting up the integral given in the statement of this problem.

Figure from prob_s3.5, line 571

Figure from prob_s3.5, line 571

To reverse the order of the yy– and zz–integrals we use horizontal strips as in the figure on the right above. Looking at that figure, we see that, on VxV_x,

  • zz runs from 00 to 1−x1-\sqrt{x}, and

  • for each fixed zz in that range, yy runs from x\sqrt{x} to 1−z1-z.

So

I=∫01dx∫01−xdz∫x1−zdy f(x,y,z)=∫01∫01−x∫x1−z f(x,y,z) dy dz dx\begin{align*} I = \int_0^1\dee{x} \int_0^{1-\sqrt{x}}\dee{z} \int_{\sqrt{x}}^{1-z}\dee{y} \ f(x,y,z) = \int_0^1 \int_0^{1-\sqrt{x}} \int_{\sqrt{x}}^{1-z} \ f(x,y,z)\ \dee{y}\,\dee{z}\,\dee{x} \end{align*}

Outside integral is with respect to yy:
Looking at the figures above we see that, for each 0≤x≤10\le x\le 1, yy runs from x\sqrt{x} to 11 on VxV_x. As xx runs from 00 to 11 in VV, we have that x\sqrt{x} also runs from 00 to 11 on VV, so that yy runs from 00 to 11 on VV. Reviewing the definition of VV, we see that, for each fixed 0≤y≤10\le y\le 1, (x,z)(x,z) runs over

Vy={ (x,z) ∣ 0≤x≤y2, 0≤z≤1−y }\begin{equation*} V_y= \Set{(x,z)}{0\le x\le y^2,\ 0\le z\le 1-y} \end{equation*}

Here are two sketches of VyV_y.

Figure from prob_s3.5, line 571

Figure from prob_s3.5, line 571

Looking at the figure on the left (with the vertical strip), we see that, on VyV_y,

  • xx runs from 00 to y2y^2, and

  • for each fixed xx in that range, zz runs from 00 to 1−y1-y.

So

I=∫01dy∫0y2dx∫01−ydz f(x,y,z)=∫01∫0y2∫01−yf(x,y,z) dz dx dy\begin{equation*} I = \int_0^1 \dee{y} \int_0^{y^2}\dee{x} \int_0^{1-y}\dee{z}\ f(x,y,z) = \int_0^1 \int_0^{y^2}\int_0^{1-y}f(x,y,z)\ \dee{z}\,\dee{x}\, \dee{y} \end{equation*}

Looking at the figure on the right above (with the horizontal strip), we see that, on VyV_y,

  • zz runs from 00 to 1−y1-y.

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

So

I=∫01dy∫01−ydz∫0y2dx f(x,y,z)=∫01∫01−y∫0y2f(x,y,z) dx dz dy\begin{equation*} I = \int_0^1 \dee{y} \int_0^{1-y}\dee{z} \int_0^{y^2}\dee{x}\ f(x,y,z) = \int_0^1 \int_0^{1-y} \int_0^{y^2} f(x,y,z)\ \dee{x}\, \dee{z}\,\dee{y} \end{equation*}

Outside integral is with respect to zz:
Looking at the sketches of VxV_x above we see that, for each 0≤x≤10\le x\le 1, zz runs from 00 to 1−x1-\sqrt{x} on VxV_x. As xx runs from 00 to 11 in VV, 1−x1-\sqrt{x} also runs between 00 to 11 on VV, so that zz runs from 00 to 11 on VV. Reviewing the definition of VV, we see that, for each fixed 0≤z≤10\le z\le 1, (x,y)(x,y) runs over

Vz={ (x,y) ∣ 0≤x≤y2, x≤y≤1−z }\begin{equation*} V_z= \Set{(x,y)}{0\le x\le y^2,\ \sqrt{x}\le y\le 1-z} \end{equation*}

Here are two sketches of VzV_z.

Figure from prob_s3.5, line 571

Figure from prob_s3.5, line 571

Looking at the figure on the left (with the vertical strip), we see that, on VzV_z,

  • xx runs from 00 to (1−z)2(1-z)^2, and

  • for each fixed xx in that range, yy runs from x\sqrt{x} to 1−z1-z.

So

I=∫01dz∫0(1−z)2dx∫x1−zdy f(x,y,z)=∫01∫0(1−z)2∫x1−zf(x,y,z) dy dx dz\begin{equation*} I = \int_0^1 \dee{z} \int_0^{(1-z)^2}\dee{x} \int_{\sqrt{x}}^{1-z}\dee{y}\ f(x,y,z) = \int_0^1 \int_0^{(1-z)^2} \int_{\sqrt{x}}^{1-z}f(x,y,z)\ \dee{y}\,\dee{x}\, \dee{z} \end{equation*}

Looking at the figure on the right above (with the horizontal strip), we see that, on VzV_z,

  • yy runs from 00 to 1−z1-z.

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

So

I=∫01dz∫01−zdy∫0y2dx f(x,y,z)=∫01∫01−z∫0y2f(x,y,z) dx dy dz\begin{equation*} I = \int_0^1 \dee{z} \int_0^{1-z}\dee{y} \int_0^{y^2}\dee{x}\ f(x,y,z) = \int_0^1 \int_0^{1-z} \int_0^{y^2} f(x,y,z)\ \dee{x}\, \dee{y}\,\dee{z} \end{equation*}

Summary:
We have found that

I=∫01∫x1∫01−yf(x,y,z) dz dy dx=∫01∫01−x∫x1−z f(x,y,z) dy dz dx=∫01∫0y2∫01−yf(x,y,z) dz dx dy=∫01∫01−y∫0y2f(x,y,z) dx dz dy=∫01∫0(1−z)2∫x1−zf(x,y,z) dy dx dz=∫01∫01−z∫0y2f(x,y,z) dx dy dz\begin{alignat*}{3} I&= \int_0^1 \int_{\sqrt{x}}^1 \int_0^{1-y} f(x,y,z)\,\dee{z}\,\dee{y}\,\dee{x} &&=\int_0^1 \int_0^{1-\sqrt{x}} \int_{\sqrt{x}}^{1-z} \ f(x,y,z)\ \dee{y}\,\dee{z}\,\dee{x} \\ &=\int_0^1 \int_0^{y^2}\int_0^{1-y}f(x,y,z)\ \dee{z}\,\dee{x}\, \dee{y} &&=\int_0^1 \int_0^{1-y} \int_0^{y^2} f(x,y,z)\ \dee{x}\, \dee{z}\,\dee{y} \\ &=\int_0^1 \int_0^{(1-z)^2} \int_{\sqrt{x}}^{1-z}f(x,y,z)\ \dee{y}\,\dee{x}\, \dee{z} &&=\int_0^1 \int_0^{1-z} \int_0^{y^2} f(x,y,z)\ \dee{x}\, \dee{y}\,\dee{z} \end{alignat*}
Q9Stage 2Past exam · M200 2009A

Let I=∭Ef(x,y,z) dV\displaystyle I = \tripInt_E f(x,y,z)\ \dee{V} where EE is the tetrahedron with vertices (−1,0,0)(-1, 0, 0), (0,0,0)(0, 0, 0), (0,0,3)(0, 0, 3) and (0,−2,0)(0, -2, 0).

  1. Rewrite the integral I in the form

    I=∫x=x=∫y=y=∫z=z=f(x,y,z) dz dy dx\begin{equation*} I = \int_{x=}^{x=}\int_{y=}^{y=}\int_{z=}^{z=} f(x,y,z)\ \dee{z}\,\dee{y}\,\dee{x} \end{equation*}

  2. Rewrite the integral I in the form

    I=∫z=z=∫x=x=∫y=y=f(x,y,z) dy dx dz\begin{equation*} I = \int_{z=}^{z=}\int_{x=}^{x=}\int_{y=}^{y=} f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z} \end{equation*}
Answer

(a) I=∫x=−1x=0∫y=−2(1+x)y=0∫z=0z=3(1+x+y/2)f(x,y,z) dz dy dxI = \int_{x=-1}^{x=0}\int_{y=-2(1+x)}^{y=0}\int_{z=0}^{z=3(1+x+y/2)} f(x,y,z)\ \dee{z}\,\dee{y}\,\dee{x}

(b) I=∫z=0z=3∫x=−(1−z/3)x=0∫y=−2(1+x−z/3)y=0f(x,y,z) dy dx dzI = \int_{z=0}^{z=3}\int_{x=-(1-z/3)}^{x=0}\int_{y=-2(1+x-z/3)}^{y=0} f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z}

Full solution

First we have to get some idea as to what EE looks like. Here is a sketch.

Figure from prob_s3.5, line 773

Figure from prob_s3.5, line 773

We are going to need the equation of the plane that contains the points (−1,0,0)(-1,0,0), (0,−2,0)(0,-2,0) and (0,0,3)(0,0,3). This plane does not contain the origin and so has an equation of the form ax+by+cz=1ax+by+cz=1.

  • (−1,0,0)(-1,0,0) lies on the plane ax+by+cz=1ax+by+cz=1 if and only if a(−1)+b(0)+c(0)=1a(-1)+b(0)+c(0)=1. So a=−1a=-1.

  • (0,−2,0)(0,-2,0) lies on the plane ax+by+cz=1ax+by+cz=1 if and only if a(0)+b(−2)+c(0)=1a(0)+b(-2)+c(0)=1. So b=−12b=-\frac{1}{2}.

  • (0,0,3)(0,0,3) lies on the plane ax+by+cz=1ax+by+cz=1 if and only if a(0)+b(0)+c(3)=1a(0)+b(0)+c(3)=1. So c=13c=\frac{1}{3}.

So the plane that contains the points (−1,0,0)(-1,0,0), (0,−2,0)(0,-2,0) and (0,0,3)(0,0,3) is −x−y2+z3=1-x-\frac{y}{2}+\frac{z}{3}=1.

We can now get a detailed mathematical description of EE. A point (x,y,z)(x,y,z) is in EE if and only if

  • (x,y,z)(x,y,z) lies above the xyxy–plane, i.e. z≥0z\ge 0, and

  • (x,y,z)(x,y,z) lies to the left of the xzxz–plane, i.e. y≤0y\le 0, and

  • (x,y,z)(x,y,z) lies behind the yzyz–plane, i.e. x≤0x\le 0, and

  • (x,y,z)(x,y,z) lies on the same side of the plane −x−y2+z3=1-x-\frac{y}{2}+\frac{z}{3}=1 as the origin. That is −x−y2+z3≤1-x-\frac{y}{2}+\frac{z}{3}\le 1. (Go ahead and check that (0,0,0)(0,0,0) obeys this inequality.)

So

E={ (x,y,z) ∣ x≤0, y≤0, z≥0, −x−y2+z3≤1 }\begin{equation*} E=\Set{(x,y,z)}{x\le 0,\ y\le 0,\ z\ge 0,\ -x-\tfrac{y}{2}+\tfrac{z}{3}\le 1} \end{equation*}

(a) Note that we want the outside integral to be the xx–integral. On EE

  • xx runs from −1-1 to 00 and

  • for each fixed xx in that range (y,z)(y,z) runs over

    Ex={ (y,z) ∣ y≤0, z≥0, −y2+z3≤1+x }\begin{align*} E_x=\Set{(y,z)}{y\le 0,\ z\ge 0,\ -\tfrac{y}{2}+\tfrac{z}{3}\le 1+x} \end{align*}

    Here is a sketch of ExE_x.

    Figure from prob_s3.5, line 773

    Figure from prob_s3.5, line 773

  • On ExE_x, yy runs from −2(1+x)-2(1+x) to 00 and

  • for each fixed such yy, zz runs from 00 to 3(1+x+y/2)3(1+x+y/2)

So

I=∫x=−1x=0∫y=−2(1+x)y=0∫z=0z=3(1+x+y/2)f(x,y,z) dz dy dx\begin{align*} I = \int_{x=-1}^{x=0}\int_{y=-2(1+x)}^{y=0}\int_{z=0}^{z=3(1+x+y/2)} f(x,y,z)\ \dee{z}\,\dee{y}\,\dee{x} \end{align*}

(b) This time we want the outside integral to be the zz–integral. Looking back at the sketch of EE, we see that, on EE,

  • zz runs from 00 to 33 and

  • for each fixed zz in that range (x,y)(x,y) runs over

    Ez={ (x,y) ∣ x≤0, y≤0, −x−y2≤1−z3 }\begin{align*} E_z=\Set{(x,y)}{x\le 0,\ y\le 0,\ -x-\tfrac{y}{2}\le 1-\tfrac{z}{3}} \end{align*}

    Here is a sketch of EzE_z.

    Figure from prob_s3.5, line 773

    Figure from prob_s3.5, line 773

  • On EzE_z, xx runs from −(1−z/3)-(1-z/3) to 00 and

  • for each fixed such xx, yy runs from −2(1+x−z/3)-2(1+x-z/3) to 00

So

I=∫z=0z=3∫x=−(1−z/3)x=0∫y=−2(1+x−z/3)y=0f(x,y,z) dy dx dz\begin{align*} I = \int_{z=0}^{z=3}\int_{x=-(1-z/3)}^{x=0}\int_{y=-2(1+x-z/3)}^{y=0} f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z} \end{align*}
Q10Stage 2Past exam · M200 2010D

Let TT denote the tetrahedron bounded by the coordinate planes x=0x = 0, y=0y = 0, z=0z = 0 and the plane x+y+z=1x + y + z = 1. Compute

K=∭T1(1+x+y+z)4 dV\begin{equation*} K = \tripInt_T \frac{1}{ (1 + x + y + z)^4}\ \dee{V} \end{equation*}
Answer

148\frac{1}{48}

Full solution

The plane x+y+z=1x+y+z=1 intersects the coordinate plane z=0z=0 along the line x+y=1x+y=1, z=0z=0. So

T={ (x,y,z) ∣ x≥0, y≥0, x+y≤1, 0≤z≤1−x−y }={ (x,y,z) ∣ 0≤x≤1, 0≤y≤1−x, 0≤z≤1−x−y }\begin{align*} T&=\Set{(x,y,z)}{x\ge 0,\ y\ge0,\ x+y\le 1,\ 0\le z\le 1-x-y} \\ &=\Set{(x,y,z)}{0\le x\le 1,\ 0\le y\le 1-x,\ 0\le z\le 1-x-y} \end{align*}

Figure from prob_s3.5, line 887

Figure from prob_s3.5, line 887

and

K=∫01dx∫01−xdy∫01−x−ydz 1(1+x+y+z)4=∫01dx∫01−xdy [−13(1+x+y+z)3]z=0z=1−x−y=13∫01dx∫01−xdy [1(1+x+y)3−123]=13∫01dx [−12(1+x+y)2−y2(4)]y=0y=1−x=16∫01dx [1(1+x)2−122−1−x4]=16∫01dx [1(1+x)2−12+x4]=16[−11+x−x2+x28]x=0x=1=16[1−12−12+18]=148\begin{align*} K&=\int_0^1\dee{x}\int_0^{1-x}\dee{y}\int_0^{1-x-y}\dee{z}\ \frac{1}{ (1 + x + y + z)^4} \\[0.1in] &=\int_0^1\dee{x}\int_0^{1-x}\dee{y}\ \left[-\frac{1}{ 3(1 + x + y + z)^3}\right]_{z=0}^{z=1-x-y} \\[0.1in] &=\frac{1}{3}\int_0^1\dee{x}\int_0^{1-x}\dee{y}\ \left[\frac{1}{(1 + x + y)^3} - \frac{1}{2^3}\right] \\[0.1in] &=\frac{1}{3}\int_0^1\dee{x}\ \left[-\frac{1}{2(1 + x + y)^2} - \frac{y}{2(4)}\right]_{y=0}^{y=1-x} \\[0.1in] &=\frac{1}{6}\int_0^1\dee{x}\ \left[\frac{1}{(1 + x)^2}-\frac{1}{2^2}- \frac{1-x}{4}\right] =\frac{1}{6}\int_0^1\dee{x}\ \left[\frac{1}{(1 + x)^2} -\frac{1}{2} + \frac{x}{4}\right] \\[0.1in] &=\frac{1}{6} \left[-\frac{1}{1 + x} -\frac{x}{2} + \frac{x^2}{8}\right]_{x=0}^{x=1} =\frac{1}{6} \left[1-\frac{1}{2} -\frac{1}{2} + \frac{1}{8}\right] \\[0.1in] &=\frac{1}{48} \end{align*}
Q11Stage 2Past exam · M200 2011A

Let EE be the portion of the first octant which is above the plane z=x+yz = x + y and below the plane z=2z = 2. The density in EE is ρ(x,y,z)=z\rho(x, y, z) = z. Find the mass of EE.

Answer

22

Full solution

Note that the planes z=x+yz = x + y and z=2z = 2 intersect along the line x+y=2x+y=2, z=2z=2.

Figure from prob_s3.5, line 933

Figure from prob_s3.5, line 933

So

E={ (x,y,z) ∣ x≥0, y≥0, x+y≤2, x+y≤z≤2 }={ (x,y,z) ∣ 0≤x≤2, 0≤y≤2−x, x+y≤z≤2 }\begin{align*} E &=\Set{(x,y,z)}{x\ge 0,\ y\ge 0,\ x+y\le 2,\ x+y\le z\le 2} \\ &= \Set{(x,y,z)}{0\le x\le 2,\ 0\le y\le 2-x,\ x+y\le z\le 2} \end{align*}

and the mass of EE is

∭Eρ(x,y,z) dV=∫02dx∫02−xdy∫x+y2dz z=12∫02dx∫02−xdy [4−(x+y)2]=12∫02dx [4(2−x)−(x+(2−x))3−x33]=12[4(2)(2)−2(2)2−83(2)+2412]=12[8−163+43]=2\begin{align*} \tripInt_E \rho(x,y,z)\,\dee{V} &=\int_0^2\dee{x} \int_0^{2-x}\dee{y} \int_{x+y}^2\dee{z}\ z \\ &=\frac{1}{2}\int_0^2\dee{x} \int_0^{2-x}\dee{y} \ \big[4-(x+y)^2\big] \\ &=\frac{1}{2}\int_0^2\dee{x}\ \left[4(2-x)-\frac{\big(x+(2-x)\big)^3-x^3}{3}\right] \\ &=\frac{1}{2} \left[4(2)(2)-2(2)^2 -\frac{8}{3}(2)+\frac{2^4}{12} \right] =\frac{1}{2} \left[8 -\frac{16}{3}+\frac{4}{3} \right] \\ &=2 \end{align*}
Q12Stage 2Past exam · M200 2011D

Evaluate the triple integral ∭Ex dV\tripInt_E x\ \dee{V}, where EE is the region in the first octant bounded by the parabolic cylinder y=x2y = x^2 and the planes y+z=1y + z = 1, x=0x = 0, and z=0z = 0.

Answer

112\frac{1}{12}

Full solution

First, we need to develop an understanding of what EE looks like. Here are sketches of the parabolic cylinder y=x2y=x^2, on the left, and the plane y+z=1y+z=1, on the right.

Figure from prob_s3.5, line 976

Figure from prob_s3.5, line 976

Figure from prob_s3.5, line 976

Figure from prob_s3.5, line 976

EE is constructed by using the plane y+z=1y+z=1 to chop the top off of the parabolic cylinder y=x2y=x^2. Here is a sketch.

Figure from prob_s3.5, line 976

Figure from prob_s3.5, line 976

So

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

and the integral

∭Ex dV=∫01dx∫x21dy∫01−ydz x=∫01dx∫x21dy x(1−y)=∫01dx x[y−y22]x21=∫01dx [x2−x3+x52]=14−14+112=112\begin{align*} \tripInt_E x\ \dee{V} &=\int_0^1\dee{x}\int_{x^2}^1\dee{y}\int_0^{1-y}\dee{z}\ x \\ &=\int_0^1\dee{x}\int_{x^2}^1\dee{y}\ x(1-y) \\ &=\int_0^1\dee{x}\ x\left[y-\frac{y^2}{2}\right]_{x^2}^1 \\ &=\int_0^1\dee{x}\ \left[\frac{x}{2} -x^3+\frac{x^5}{2}\right] \\ &=\frac{1}{4}-\frac{1}{4}+\frac{1}{12} \\ &=\frac{1}{12} \end{align*}
Q13Stage 2Past exam · M200 2012A

Let EE be the region in the first octant bounded by the coordinate planes, the plane x+y=1x + y = 1 and the surface z=y2z = y^2 . Evaluate ∭Ez dV\tripInt_E z\ \dee{V} .

Answer

160\frac{1}{60}

Full solution

First, we need to develop an understanding of what EE looks like. Here are sketches of the plane x+y=1x+y=1, on the left, and of the “tower” bounded by the coordinate planes x=0x=0, y=0y=0, z=0z=0 and the plane x+y=1x+y=1, on the right.

Figure from prob_s3.5, line 1020

Figure from prob_s3.5, line 1020

Figure from prob_s3.5, line 1020

Figure from prob_s3.5, line 1020

Now here is the parabolic cylinder z=y2z=y^2 on the left. EE is constructed by using the parabolic cylinder z=y2z=y^2 to chop the top off of the tower x≥0x\ge 0, y≥0y\ge 0, z≥0z\ge 0, x+y≤1x+y\le 1. The figure on the right is a sketch.

Figure from prob_s3.5, line 1020

Figure from prob_s3.5, line 1020

Figure from prob_s3.5, line 1020

Figure from prob_s3.5, line 1020

So

E={ (x,y,z) ∣ 0≤x≤1, 0≤y≤1−x, 0≤z≤y2 }\begin{equation*} E=\Set{(x,y,z)}{0\le x\le 1,\ 0\le y\le 1-x,\ 0\le z\le y^2} \end{equation*}

and the integral

∭Ez dV=∫01dx∫01−xdy∫0y2dz z=∫01dx∫01−xdy y42=∫01dx (1−x)510=[−(1−x)660]01=160\begin{align*} \tripInt_E z\ \dee{V} &=\int_0^1\dee{x}\int_0^{1-x}\dee{y}\int_0^{y^2}\dee{z}\ z \\ &=\int_0^1\dee{x}\int_0^{1-x}\dee{y}\ \frac{y^4}{2} \\ &=\int_0^1\dee{x}\ \frac{(1-x)^5}{10} \\ &=\left[-\frac{(1-x)^6}{60}\right]_0^1 \\ &=\frac{1}{60} \end{align*}
Q14Stage 2Past exam · M200 2012D

Evaluate ∭Ryz2e−xyz dV\tripInt_R yz^2 e^{-xyz}\ \dee{V} over the rectangular box

R={ (x,y,z) ∣ 0≤x≤1, 0≤y≤2, 0≤z≤3 }\begin{equation*} R=\Set{(x,y,z)}{0\le x\le 1,\ 0\le y\le 2,\ 0\le z\le 3} \end{equation*}
Answer

132−e−62\frac{13}{2}-\frac{e^{-6}}{2}

Full solution

The integral

∭Ryz2e−xyz dV=∫03dz∫02dy∫01dx yz2e−xyz=∫03dz∫02dy [−ze−xyz]x=0x=1=∫03dz∫02dy [z−ze−yz]=∫03dz [zy+e−yz]y=0y=2=∫03dz [2z+e−2z−1]=[z2−12e−2z−z]03=132−e−62\begin{align*} \tripInt_R yz^2 e^{-xyz}\ \dee{V} &=\int_0^3\dee{z}\int_0^2\dee{y}\int_0^1\dee{x}\ yz^2 e^{-xyz} \\ &=\int_0^3\dee{z}\int_0^2\dee{y}\ \Big[-z e^{-xyz}\Big]_{x=0}^{x=1} =\int_0^3\dee{z}\int_0^2\dee{y}\ \Big[z-z e^{-yz}\Big] \\ &=\int_0^3\dee{z}\ \Big[zy+ e^{-yz}\Big]_{y=0}^{y=2} =\int_0^3\dee{z}\ \Big[2z+ e^{-2z}-1\Big] \\ &=\left[z^2-\frac{1}{2} e^{-2z}-z\right]_0^3 =\frac{13}{2}-\frac{e^{-6}}{2} \end{align*}
Q15Stage 2Past exam · M200 2013D

  1. Sketch the surface given by the equation z=1−x2z = 1 - x^2.

  2. Let EE be the solid bounded by the plane y=0y = 0, the cylinder z=1−x2z = 1 - x^2, and the plane y=zy = z. Set up the integral

    ∭Ef(x,y,z) dV\begin{equation*} \tripInt_E f(x,y,z)\,\dee{V} \end{equation*}

    as an iterated integral.

Answer

(a)

Figure from prob_s3.5, line 1101

Figure from prob_s3.5, line 1101

(b) ∫−11dx∫01−x2dy∫y1−x2dz f(x,y,z)\int_{-1}^1\dee{x}\int_0^{1-x^2}\dee{y}\int_y^{1-x^2}\dee{z}\ f(x,y,z)

Full solution

(a) Each constant yy cross section of z=1−x2z=1-x^2 is an upside down parabola. So the surface is a bunch of upside down parabolas stacked side by side. The figure on the left below is a sketch of the part of the surface with y≥0y\ge 0 and z≥0z\ge 0 (both of which conditions will be required in part (b)).

Figure from prob_s3.5, line 1101

Figure from prob_s3.5, line 1101

Figure from prob_s3.5, line 1110

Figure from prob_s3.5, line 1110

(b) The figure on the right above is a sketch of the plane y=zy=z. It intersects the surface z=1−x2z=1-x^2 in the solid blue sloped parabolic curve in the figure below.

Figure from prob_s3.5, line 1110

Figure from prob_s3.5, line 1110

Observe that, on the curve z=1−x2z=1-x^2, z=yz=y, we have y=1−x2y=1-x^2. So that when one looks at the solid EE from high on the zz–axis, one sees

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

The y=1−x2y=1-x^2 boundary of that region is the dashed blue line in the xyxy–plane in the figure above. So

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

and the integral

∭Ef(x,y,z) dV=∫−11dx∫01−x2dy∫y1−x2dz f(x,y,z)\begin{equation*} \tripInt_E f(x,y,z)\,\dee{V} =\int_{-1}^1\dee{x}\int_0^{1-x^2}\dee{y}\int_y^{1-x^2}\dee{z}\ f(x,y,z) \end{equation*}
Q16Stage 2Past exam · M200 2014A

Let

J=∫01∫0x∫0yf(x,y,z) dz dy dx\begin{equation*} J=\int_0^1 \int_0^x\int_0^y f(x,y,z)\ \dee{z}\,\dee{y}\,\dee{x} \end{equation*}

Express JJ as an integral where the integrations are to be performed in the order xx first, then yy, then zz.

Answer

J=∫01∫z1∫y1f(x,y,z) dx dy dzJ = \int_0^1 \int_z^1\int_y^1 f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z}

Full solution

In the integral JJ,

  • xx runs from 00 to 11. In inequalities, 0≤x≤10\le x\le 1.

  • Then, for each fixed xx in that range, yy runs from 00 to xx. In inequalities, 0≤y≤x0\le y\le x.

  • Then, for each fixed xx and yy in those ranges, zz runs from 00 to yy. In inequalities, 0≤z≤y0\le z\le y.

These inequalties can be combined into

0≤z≤y≤x≤1\begin{equation*} 0\le z\le y\le x\le 1 \tag{$*$} \end{equation*}

We wish to reverse the order of integration so that the zz–integral is on the outside, the yy–integral is in the middle and the xx–integral is on the inside.

  • The smallest zz compatible with (∗)(*) is z=0z=0 and the largest zz compatible with (∗)(*) is z=1z=1 (when x=y=z=1x=y=z=1). So 0≤z≤10\le z\le 1.

  • Then, for each fixed zz in that range, (x,y)(x,y) run over z≤y≤x≤1z\le y\le x\le 1. In particular, the smallest allowed yy is y=zy=z and the largest allowed yy is y=1y=1 (when x=y=1x=y=1). So z≤y≤1z\le y\le 1.

  • Then, for each fixed yy and zz in those ranges, xx runs over y≤x≤1y\le x\le 1.

So

J=∫01∫z1∫y1f(x,y,z) dx dy dz\begin{equation*} J = \int_0^1 \int_z^1\int_y^1 f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z} \end{equation*}
Q17Stage 2Past exam · M200 2014D

Let EE be the region bounded by z=2xz = 2x, z=y2z = y^2, and x=3x = 3. The triple integral ∭f(x,y,z) dV\tripInt f(x,y,z)\,\dee{V} can be expressed as an iterated integral in the following three orders of integration. Fill in the limits of integration in each case. No explanation required.

∫y=y=∫x=x=∫z=z=f(x,y,z) dz dx dy∫y=y=∫z=z=∫x=x=f(x,y,z) dx dz dy∫z=z=∫x=x=∫y=y=f(x,y,z) dy dx dz\begin{align*} &\int_{y=}^{y=}\qquad\int_{x=}^{x=}\qquad\int_{z=}^{z=}\qquad f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y} \\[0.1in] &\int_{y=}^{y=}\qquad\int_{z=}^{z=}\qquad\int_{x=}^{x=}\qquad f(x,y,z)\ \dee{x}\,\dee{z}\,\dee{y} \\[0.1in] &\int_{z=}^{z=}\qquad\int_{x=}^{x=}\qquad\int_{y=}^{y=}\qquad f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z} \end{align*}
Answer

∫y=−6y=6∫x=y2/2x=3∫z=y2z=2xf(x,y,z) dz dx dy\int_{y=-\sqrt{6}}^{y=\sqrt{6}}\int_{x=y^2/2}^{x=3}\int_{z=y^2}^{z=2x} f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y}

∫y=−6y=6∫z=y2z=6∫x=z/2x=3f(x,y,z) dx dz dy\int_{y=-\sqrt{6}}^{y=\sqrt{6}}\int_{z=y^2}^{z=6}\int_{x=z/2}^{x=3} f(x,y,z)\ \dee{x}\,\dee{z}\,\dee{y}

∫z=0z=6∫x=z/2x=3∫y=−zy=zf(x,y,z) dy dx dz\int_{z=0}^{z=6}\int_{x=z/2}^{x=3}\int_{y=-\sqrt{z}}^{y=\sqrt{z}} f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z}

Full solution

The hard part of this problem is figuring out what EE looks like. First here are separate sketches of the plane x=3x=3 and the plane z=2xz=2x followed by a sketch of the two planes together.

Figure from prob_s3.5, line 1240

Figure from prob_s3.5, line 1240

Figure from prob_s3.5, line 1240

Figure from prob_s3.5, line 1240

Figure from prob_s3.5, line 1240

Figure from prob_s3.5, line 1240

Next for the parabolic cylinder z=y2z=y^2. It is a bunch of parabolas z=y2z=y^2 stacked side by side along the xx–axis. Here is a sketch of the part of z=y2z=y^2 in the first octant.

Figure from prob_s3.5, line 1240

Figure from prob_s3.5, line 1240

Finally, here is a sketch of the part of EE in the first octant. EE does have a second half gotten from the sketch by reflecting it in the xzxz–plane, i.e. by replacing y→−yy\rightarrow-y.

Figure from prob_s3.5, line 1240

Figure from prob_s3.5, line 1240

So (The question doesn't specify on which side of the three surfaces EE lies. When in doubt take the finite region bounded by the given surfaces. That's what we have done.)

E={ (x,y,z) ∣ x≤3, −6≤y≤6, y2≤z≤2x }\begin{equation*} E = \Set{(x,y,z)}{x\le 3,\ -\sqrt{6}\le y\le \sqrt{6},\ y^2\le z\le 2x} \end{equation*}

*Order dz dx dy\dee{z}\,\dee{x}\,\dee{y}: * On EE, yy runs from −6-\sqrt{6} to 6\sqrt{6}. For each fixed yy in this range (x,z)(x,z) runs over Ey={ (x,z) ∣ x≤3, y2≤z≤2x }E_y=\Set{(x,z)}{x\le 3,\ y^2\le z\le 2x}. Here is a sketch of EyE_y.

Figure from prob_s3.5, line 1240

Figure from prob_s3.5, line 1240

From the sketch

Ey={ (x,z) ∣ y2/2≤x≤3, y2≤z≤2x }\begin{equation*} E_y = \Set{(x,z)}{y^2/2\le x\le 3,\ y^2\le z\le 2x} \end{equation*}

and the integral is

∫y=−6y=6∫x=y2/2x=3∫z=y2z=2xf(x,y,z) dz dx dy\begin{equation*} \int_{y=-\sqrt{6}}^{y=\sqrt{6}}\int_{x=y^2/2}^{x=3}\int_{z=y^2}^{z=2x} f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y} \end{equation*}

*Order dx dz dy\dee{x}\,\dee{z}\,\dee{y}: * Also from the sketch of EyE_y above

Ey={ (x,z) ∣ y2≤z≤6, z/2≤x≤3 }\begin{equation*} E_y = \Set{(x,z)}{y^2\le z\le 6,\ z/2\le x\le 3} \end{equation*}

and the integral is

∫y=−6y=6∫z=y2z=6∫x=z/2x=3f(x,y,z) dx dz dy\begin{equation*} \int_{y=-\sqrt{6}}^{y=\sqrt{6}}\int_{z=y^2}^{z=6}\int_{x=z/2}^{x=3} f(x,y,z)\ \dee{x}\,\dee{z}\,\dee{y} \end{equation*}

*Order dy dx dz\dee{y}\,\dee{x}\,\dee{z}: * From the sketch of the part of EE in the first octant, we see that, on EE, zz runs from 00 to 66. For each fixed zz in this range (x,y)(x,y) runs over

Ez={ (x,y) ∣ x≤3, −6≤y≤6, y2≤z≤2x }={ (x,y) ∣ z/2≤x≤3, y2≤z }={ (x,y) ∣ z/2≤x≤3, −z≤y≤z }\begin{align*} E_z&=\Set{(x,y)}{x\le 3,\ -\sqrt{6}\le y\le \sqrt{6},\ y^2\le z\le 2x} \\ &=\Set{(x,y)}{z/2\le x\le 3,\ y^2\le z} \\ &=\Set{(x,y)}{z/2\le x\le 3,\ -\sqrt{z}\le y\le \sqrt{z} } \end{align*}

So the integral is

∫z=0z=6∫x=z/2x=3∫y=−zy=zf(x,y,z) dy dx dz\begin{equation*} \int_{z=0}^{z=6}\int_{x=z/2}^{x=3}\int_{y=-\sqrt{z}}^{y=\sqrt{z}} f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z} \end{equation*}
Q18Stage 2Past exam · M200 2015D

Let E be the region inside the cylinder x2+y2=1x^2 + y^2 = 1, below the plane z=yz = y and above the plane z=−1z = -1. Express the integral

∭Ef(x,y,z) dV\begin{equation*} \tripInt_E f(x,y,z)\ \dee{V} \end{equation*}

as three different iterated integrals corresponding to the orders of integration: (a) dz dx dy\dee{z}\, \dee{x}\, \dee{y}, (b) dx dy dz\dee{x}\, \dee{y}\, \dee{z}, and (c) dy dz dx\dee{y}\, \dee{z}\, \dee{x}.

Answer

(a) ∫−11∫−1−y21−y2∫−1yf(x,y,z) dz dx dy\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \int_{-1}^y f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y} (b) ∫−11∫z1∫−1−y21−y2f(x,y,z) dx dy dz\int_{-1}^1\int_z^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z}

(c) ∫−11∫−1−1−x2∫−1−x21−x2f(x,y,z) dy dz dx+∫−11∫−1−x21−x2∫z1−x2f(x,y,z) dy dz dx\int_{-1}^1 \int_{-1}^{-\sqrt{1-x^2}} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} f(x,y,z)\ \dee{y}\,\dee{z}\,\dee{x} +\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_z^{\sqrt{1-x^2}} f(x,y,z)\ \dee{y}\,\dee{z}\,\dee{x}

Full solution

(a) The region EE is

E={ (x,y,z) ∣ x2+y2≤1, −1≤z≤y }\begin{align*} E &= \Set{(x,y,z)}{x^2+y^2\le 1,\ -1\le z\le y} \end{align*}

Here is are sketches, one without axes and one with axes, of the front half of EE, outlined in red.

Figure from prob_s3.5, line 1350

Figure from prob_s3.5, line 1350

Figure from prob_s3.5, line 1350

Figure from prob_s3.5, line 1350

The integral

∭Ef(x,y,z) dV=∫x2+y2≤1dx dy∫−1ydz f(x,y,z)=∫−11dy∫−1−y21−y2dx∫−1ydz f(x,y,z)=∫−11∫−1−y21−y2∫−1yf(x,y,z) dz dx dy\begin{align*} \tripInt_E f(x,y,z)\ \dee{V} &=\int_{x^2+y^2\le 1}\dee{x}\,\dee{y}\int_{-1}^y\dee{z}\ f(x,y,z) \\ &=\int_{-1}^1\dee{y}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\dee{x} \int_{-1}^y\dee{z}\ f(x,y,z) \\ &=\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \int_{-1}^y f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y} \end{align*}

(b) Here is a sketch of (the front half of) a constant zz slice of EE.

Figure from prob_s3.5, line 1350

Figure from prob_s3.5, line 1350

Note that

  • in EE, zz runs from −1-1 to 11.

  • Once zz has been fixed, xx and yy must obey x2+y2≤1x^2+y^2\le 1, z≤y≤1z\le y\le 1

So

E={ (x,y,z) ∣ −1≤z≤1, z≤y≤1, −1−y2≤x≤1−y2 }\begin{align*} E &= \Set{(x,y,z)}{-1\le z\le 1,\ z\le y\le 1,\ -\sqrt{1-y^2}\le x\le\sqrt{1-y^2}} \end{align*}

and

∭Ef(x,y,z) dV=∫−11dz∫z1dy∫−1−y21−y2dx f(x,y,z)=∫−11∫z1∫−1−y21−y2f(x,y,z) dx dy dz\begin{align*} \tripInt_E f(x,y,z)\ \dee{V} &=\int_{-1}^1\dee{z}\int_z^1\dee{y} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\dee{x}\ f(x,y,z) \\ &=\int_{-1}^1\int_z^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z} \end{align*}

(c) Here is a sketch of a constant xx slice of EE.

Figure from prob_s3.5, line 1350

Figure from prob_s3.5, line 1350

Note that

  • in EE, xx runs from −1-1 to 11.

  • Once xx has been fixed, yy and zz must obey

    −1−x2≤y≤1−x2−1≤z≤y\begin{equation*} -\sqrt{1-x^2}\le y\le\sqrt{1-x^2}\qquad -1\le z\le y \end{equation*}

Here is a sketch.

Figure from prob_s3.5, line 1350

Figure from prob_s3.5, line 1350

Note that

  • zz runs from −1-1 to 1−x2\sqrt{1-x^2}.

  • For each zz between −1-1 and −1−x2-\sqrt{1-x^2}, yy runs from −1−x2-\sqrt{1-x^2} to 1−x2\sqrt{1-x^2}, while

  • for each zz between −1−x2-\sqrt{1-x^2} and 1−x2\sqrt{1-x^2}, yy runs from zz to 1−x2\sqrt{1-x^2}.

So

∭Ef(x,y,z) dV=∫−11dx∫−1−1−x2dz∫−1−x21−x2dy f(x,y,z)+∫−11dx∫−1−x21−x2dz∫z1−x2dy f(x,y,z)\begin{align*} \tripInt_E f(x,y,z)\ \dee{V} &=\int_{-1}^1\dee{x} \int_{-1}^{-\sqrt{1-x^2}}\dee{z} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\dee{y} \ f(x,y,z) \\ &\hskip1in +\int_{-1}^1\dee{x} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\dee{z} \int_z^{\sqrt{1-x^2}}\dee{y} \ f(x,y,z) \end{align*}

or

∭Ef(x,y,z) dV=∫−11∫−1−1−x2∫−1−x21−x2f(x,y,z) dy dz dx+∫−11∫−1−x21−x2∫z1−x2f(x,y,z) dy dz dx\begin{align*} \tripInt_E f(x,y,z)\ \dee{V} &=\int_{-1}^1 \int_{-1}^{-\sqrt{1-x^2}} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} f(x,y,z)\ \dee{y}\,\dee{z}\,\dee{x} \\ &\hskip1in +\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_z^{\sqrt{1-x^2}} f(x,y,z)\ \dee{y}\,\dee{z}\,\dee{x} \end{align*}
Q19Stage 2Past exam · M200 2016D

Let EE be the region bounded by the planes y=0y=0, y=2y=2, y+z=3y+z=3 and the surface z=x2z=x^2. Consider the intergal

I=∭Ef(x,y,z) dV\begin{align*} I=\tripInt_E f(x,y,z)\ \dee{V} \end{align*}

Fill in the blanks below. In each part below, you may need only one integral to express your answer. In that case, leave the other blank.

  1. I=∫    ‾    ‾∫    ‾    ‾∫    ‾    ‾f(x,y,z) dz dx dy+∫    ‾    ‾∫    ‾    ‾∫    ‾    ‾f(x,y,z) dz dx dy\displaystyle I=\int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} \int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} \int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y} + \int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} \int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} \int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y}

  2. I=∫    ‾    ‾∫    ‾    ‾∫    ‾    ‾f(x,y,z) dx dy dz+∫    ‾    ‾∫    ‾    ‾∫    ‾    ‾f(x,y,z) dx dy dz\displaystyle I=\int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} \int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} \int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z} + \int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} \int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} \int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z}

  3. I=∫    ‾    ‾∫    ‾    ‾∫    ‾    ‾f(x,y,z) dy dx dz+∫    ‾    ‾∫    ‾    ‾∫    ‾    ‾f(x,y,z) dy dx dz\displaystyle I=\int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} \int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} \int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z} + \int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} \int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} \int_{\underline{\ \ \ \ }}^{\underline{\ \ \ \ }} f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z}

Answer

(a) ∫02∫−3−y3−y∫x23−yf(x,y,z) dz dx dy\int_0^2 \int_{-\sqrt{3-y}}^{\sqrt{3-y}} \int_{x^2}^{3-y} f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y}

(b) ∫01∫02∫−zzf(x,y,z) dx dy dz+∫13∫03−z∫−zzf(x,y,z) dx dy dz\int_0^1 \int_0^2 \int_{-\sqrt{z}}^{\sqrt{z}} f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z} +\int_1^3 \int_0^{3-z} \int_{-\sqrt{z}}^{\sqrt{z}} f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z}

(c) ∫01∫−zz∫02f(x,y,z) dy dx dz+∫13∫−zz∫03−zf(x,y,z) dy dx dz\int_0^1 \int_{-\sqrt{z}}^{\sqrt{z}}\int_0^2 f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z} +\int_1^3 \int_{-\sqrt{z}}^{\sqrt{z}} \int_0^{3-z} f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z}

Full solution

First, we need to develop an understanding of what EE looks like. Note that all of the equations y=0y=0, y=2y=2, y+z=3y+z=3 and z=x2z=x^2 are invariant under x→−xx\rightarrow -x. So EE is invariant under x→−xx\rightarrow -x, i.e. is symmetric about the yzyz–plane. We'll sketch the first octant (i.e. x,y,z≥0x,y,z\ge 0) part of EE. There is also a x≤0x\le 0, y≥0y\ge 0, z≥0z\ge 0 part.

Here are sketches of the plane y=2y=2, on the left, the plane y+z=3y+z=3 in the centre and of the “tunnel” bounded by the coordinate planes x=0x=0, y=0y=0, z=0z=0 and the planes y=2y=2, y+z=3y+z=3, on the right.

Figure from prob_s3.5, line 1521

Figure from prob_s3.5, line 1521

Figure from prob_s3.5, line 1521

Figure from prob_s3.5, line 1521

Figure from prob_s3.5, line 1521

Figure from prob_s3.5, line 1521

Now here is the parabolic cylinder z=x2z=x^2 on the left. EE is constructed by using the parabolic cylinder z=x2z=x^2 to chop the front off of the tunnel x≥0x\ge 0, 0≤y≤20\le y\le 2, z≥0z\ge 0, y+z≤3y+z\le 3. The figure on the right is a sketch.

Figure from prob_s3.5, line 1521

Figure from prob_s3.5, line 1521

Figure from prob_s3.5, line 1521

Figure from prob_s3.5, line 1521

So

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

(a) On EE

  • yy runs from 00 to 22.

  • For each fixed yy in that range, (x,z)(x,z) runs over { (x,z) ∣ x2≤z≤3−y }\Set{(x,z)}{x^2\le z\le 3-y}.

  • In particular, the largest x2x^2 is 3−y3-y (when z=3−yz=3-y). So xx runs from −3−y-\sqrt{3-y} to 3−y\sqrt{3-y}.

  • For fixed yy and xx as above, zz runs from x2x^2 to 3−y3-y.

so that

I=∭Ef(x,y,z) dV=∫02∫−3−y3−y∫x23−yf(x,y,z) dz dx dy\begin{align*} I=\tripInt_E f(x,y,z)\ \dee{V} = \int_0^2 \int_{-\sqrt{3-y}}^{\sqrt{3-y}} \int_{x^2}^{3-y} f(x,y,z)\ \dee{z}\,\dee{x}\,\dee{y} \end{align*}

(b) On EE

  • zz runs from 00 to 33.

  • For each fixed zz in that range, (x,y)(x,y) runs over

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

    In particular, yy runs from 00 to the minimum of 22 and 3−z3-z.

  • So if 0≤z≤10\le z\le 1 (so that 3−z≥23-z\ge 2), (x,y)(x,y) runs over { (x,y) ∣ 0≤y≤2, x2≤z }\Set{(x,y)}{0\le y\le 2,\ x^2\le z}, while

  • if 1≤z≤31\le z\le 3, (so that 3−z≤23-z\le 2), (x,y)(x,y) runs over { (x,y) ∣ 0≤y≤3−z, x2≤z }\Set{(x,y)}{0\le y\le 3-z,\ x^2\le z},

so that

I=∫01∫02∫−zzf(x,y,z) dx dy dz+∫13∫03−z∫−zzf(x,y,z) dx dy dz\begin{align*} I= \int_0^1 \int_0^2 \int_{-\sqrt{z}}^{\sqrt{z}} f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z} +\int_1^3 \int_0^{3-z} \int_{-\sqrt{z}}^{\sqrt{z}} f(x,y,z)\ \dee{x}\,\dee{y}\,\dee{z} \end{align*}

(c) On EE

  • zz runs from 00 to 33.

  • For each fixed zz in that range, (x,y)(x,y) runs over

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

    In particular, yy runs from 00 to the minimum of 22 and 3−z3-z.

  • So if 0≤z≤10\le z\le 1 (so that 3−z≥23-z\ge 2), (x,y)(x,y) runs over { (x,y) ∣ 0≤y≤2, x2≤z }\Set{(x,y)}{0\le y\le 2,\ x^2\le z}, while

  • if 1≤z≤31\le z\le 3, (so that 3−z≤23-z\le 2), (x,y)(x,y) runs over { (x,y) ∣ 0≤y≤3−z, x2≤z }\Set{(x,y)}{0\le y\le 3-z,\ x^2\le z},

so that

I=∫01∫−zz∫02f(x,y,z) dy dx dz+∫13∫−zz∫03−zf(x,y,z) dy dx dz\begin{align*} I= \int_0^1 \int_{-\sqrt{z}}^{\sqrt{z}}\int_0^2 f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z} +\int_1^3 \int_{-\sqrt{z}}^{\sqrt{z}} \int_0^{3-z} f(x,y,z)\ \dee{y}\,\dee{x}\,\dee{z} \end{align*}
Q20Stage 2Past exam · M200 2002D

Evaluate ∭Ez dV\tripInt_E z\,\dee{V}, where EE is the region bounded by the planes y=0y=0, z=0z=0 x+y=2x+y=2 and the cylinder y2+z2=1y^2+z^2=1 in the first octant.

Hint

Sketch EE. You can picture EE by thinking of the region bounded by the planes x=0x=0, y=0y=0, z=0z=0 and x+y=2x+y=2 as a large wedge of cheese and thinking of the cylinder y2+z2=1y^2+z^2=1 as a drill hole in the wedge. Then to set up the limits of integration, first sketch a top view of EE.

Answer

1324≈0.5417\frac{13}{24}\approx 0.5417

Full solution

The cylinder y2+z2=1y^2+z^2=1 is centred on the xx axis. The part of the cylinder in the first octant intersects the plane z=0z=0 in the line y=1y= 1, intersects to plane y=0y=0 in the line z=1z=1 and intersects the plane x=0x=0 in the quarter circle y2+z2=1y^2+z^2=1, x=0x=0, y,z≥0y,z\ge 0. Here is a sketch of EE.

Figure from prob_s3.5, line 1642

Figure from prob_s3.5, line 1642

Viewed from above, the region EE is bounded by the lines x=0x=0, y=0y=0, x+y=2x+y=2 and y=1y=1. This base region is pictured below.

Figure from prob_s3.5, line 1642

Figure from prob_s3.5, line 1642

To set up the domain of integration, let's decompose the base region into horizontal strips as in the figure above. On the base region

  • yy runs from 00 to 11 and

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

  • For each fixed (x,y)(x,y) in the base region zz runs from 00 to 1−y2\sqrt{1-y^2}

So

E={ (x,y,z) ∣ 0≤y≤1, 0≤x≤2−y, 0≤z≤1−y2 }\begin{equation*} E=\Set{(x,y,z)}{0\le y\le 1,\ 0\le x\le 2-y,\ 0\le z\le\sqrt{1-y^2} } \end{equation*}

and

∭Ez dV=∫01dy∫02−ydx∫01−y2dz z=∫01dy∫02−ydx 12z2∣01−y2=∫01dy∫02−ydx 12(1−y2)=∫01dy 12(1−y2)(2−y)=12∫01dy (2−y−2y2+y3)=12[2−12−23+14]=1324≈0.5417\begin{align*} \tripInt_E z\,\dee{V}&=\int_0^1 \dee{y}\int_0^{2-y}\dee{x}\int_0^{\sqrt{1-y^2}}\dee{z}\ z\\ &=\int_0^1 \dee{y}\int_0^{2-y}\dee{x}\ \frac{1}{2} z^2\Big|_0^{\sqrt{1-y^2}} \\ &=\int_0^1 \dee{y}\int_0^{2-y}\dee{x}\ \frac{1}{2} (1-y^2) \\ &=\int_0^1 \dee{y}\ \frac{1}{2} (1-y^2)(2-y) =\frac{1}{2}\int_0^1 \dee{y}\ \big(2-y-2y^2+y^3\big) \\ &=\frac{1}{2}\left[2-\frac{1}{2}-\frac{2}{3}+\frac{1}{4}\right] =\frac{13}{24}\approx 0.5417 \end{align*}
Q21Stage 2Past exam · M200 2001D

Find ∭Dx dV\dst\tripInt_D x\,\dee{V} where DD is the tetrahedron bounded by the planes x=1x=1, y=1y=1, z=1z=1, and x+y+z=2x+y+z=2.

Answer

18=0.125\frac{1}{8}=0.125

Full solution

The planes x=1x=1, y=1y=1, z=1z=1, and x+y+z=2x+y+z=2 and the region DD are sketched below.

Figure from prob_s3.5, line 1698

Figure from prob_s3.5, line 1698

And here is a sketch of DD without the planes cluttering up the figure.

Figure from prob_s3.5, line 1698

Figure from prob_s3.5, line 1698

On DD

  • zz runs from 00 to 11 and

  • for each fixed zz, between 0 and 1, (x,y)(x,y) runs over the triangle TzT_z bounded by x=1x=1, y=1y=1 and x+y=2−zx+y=2-z. Observe that when z=0z=0, this triangle is just a point (the bottom vertex of the tetrahedron). As zz increases, the triangle grows, reaching its maximum size when z=1z=1.

Here is a sketch of TzT_z.

Figure from prob_s3.5, line 1698

Figure from prob_s3.5, line 1698

In setting up the domain of integration, we'll decompose, for each 0≤z≤10\le z\le 1, TzT_z into vertical strips as in the figure above. On TzT_z

  • xx runs from 1−z1-z to 11 and

  • for each fixed xx between 1−z1-z and 11, yy runs from 2−x−z2-x-z to 11

so that

∭Dx dV=∫01dz∬Tzdx dy x=∫01dz∫1−z1dx∫2−x−z1dy x=∫01dz∫1−z1dx x(x+z−1)=∫01dz [13x3+12x2(z−1)]1−z1=∫01dz [13+12(z−1)−13(1−z)3−12(1−z)2(z−1)]=∫01dz [13+12(z−1)−16(z−1)3]=[13z+14(z−1)2−124(z−1)4]01=13−14+124=324=18=0.125\begin{align*} \tripInt_D x\,\dee{V} &=\int_0^1\dee{z}\dblInt_{T_z}\dee{x}\,\dee{y}\ x =\int_0^1 \dee{z}\int_{1-z}^1 \dee{x}\int_{2-x-z}^1 \dee{y}\ x \\ &=\int_0^1 \dee{z}\int_{1-z}^1 \dee{x}\ x(x+z-1)\\ &=\int_0^1 \dee{z}\ \left[\frac{1}{3}x^3+\frac{1}{2} x^2(z-1)\right]_{1-z}^1\\ &=\int_0^1 \dee{z}\ \left[\frac{1}{3}+\frac{1}{2} (z-1)-\frac{1}{3}(1-z)^3 -\frac{1}{2}(1-z)^2(z-1)\right]\\ &=\int_0^1 \dee{z}\ \left[\frac{1}{3}+\frac{1}{2} (z-1) -\frac{1}{6}(z-1)^3\right]\\ &=\left[\frac{1}{3}z+\frac{1}{4} (z-1)^2-\frac{1}{24}(z-1)^4\right]_0^1 =\frac{1}{3}-\frac{1}{4}+\frac{1}{24}=\frac{3}{24} \\ &=\frac{1}{8}=0.125 \end{align*}
Q22Stage 2Past exam · M200 2001A

The solid region TT is bounded by the planes x=0x=0, y=0y=0, z=0z=0, and x+y+z=2x+y+z=2 and the surface x2+z=1x^2+z=1.

  1. Draw the region indicating coordinates of all corners.

  2. Calculate ∭Tx dV\tripInt_T x\,\dee{V}.

Answer

(a) Here is a 3d sketch of the region. The coordinates of the labelled corners are

a=(0,0,1)b=(0,0,0)c=(1,0,0)d=(0,1,1)f=(0,2,0)g=(1,1,0)\begin{equation*} a=(0,0,1)\quad b=(0,0,0)\quad c=(1,0,0)\quad d=(0,1,1)\quad f=(0,2,0)\quad g=(1,1,0) \end{equation*}

Figure from prob_s3.5, line 1766

Figure from prob_s3.5, line 1766

(b) 1760\frac{17}{60}

Full solution

(a) Here is a 3d sketch of the region. The coordinates of the labelled corners are

a=(0,0,1)b=(0,0,0)c=(1,0,0)d=(0,1,1)f=(0,2,0)g=(1,1,0)\begin{equation*} a=(0,0,1)\quad b=(0,0,0)\quad c=(1,0,0)\quad d=(0,1,1)\quad f=(0,2,0)\quad g=(1,1,0) \end{equation*}

Figure from prob_s3.5, line 1766

Figure from prob_s3.5, line 1766

(b) Here is a sketch of the side view of TT, looking down the yy axis.

Figure from prob_s3.5, line 1785

Figure from prob_s3.5, line 1785

We'll set up the limits of integration by using it as the base region. We decompose the base region into vertical strips as in the figure above. On the base region

  • xx runs from 00 to 11 and

  • for each fixed xx between 00 and 11, zz runs from 00 to 1−x21-x^2.

  • In TT, for each fixed (x,y)(x,y) in the base region, yy runs from 00 to 2−x−z2-x-z.

So

∭Tx dV=∫01dx∫01−x2dz∫02−x−zdy x=∫01dx∫01−x2dz (2−x−z)x=∫01dx [x(2−x)(1−x2)−12x(1−x2)2]=∫01dx [2x−x2−2x3+x4−12x+x3−12x5]=∫01dx [32x−x2−x3+x4−12x5]=[34x2−13x3−14x4+15x5−112x6]01=34−13−14+15−112=1760\begin{align*} \tripInt_T x\,\dee{V} &=\int_0^1\dee{x}\int_0^{1-x^2}\dee{z}\int_0^{2-x-z}\dee{y}\ x =\int_0^1\dee{x}\int_0^{1-x^2}\dee{z}\ (2-x-z)x\cr &=\int_0^1 \dee{x}\ \left[x(2-x)(1-x^2)-\frac{1}{2} x{(1-x^2)}^2\right] \\ &=\int_0^1 \dee{x}\ \left[2x-x^2-2x^3+x^4-\frac{1}{2} x+x^3 -\frac{1}{2} x^5\right] \\ &=\int_0^1 \dee{x}\ \left[\frac{3}{2}x-x^2-x^3+x^4-\frac{1}{2} x^5\right] \\ &=\left[\frac{3}{4}x^2-\frac{1}{3}x^3-\frac{1}{4}x^4+\frac{1}{5}x^5 -\frac{1}{12}x^6\right]_0^1 \\ &=\frac{3}{4}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{12} =\frac{17}{60} \end{align*}

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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.