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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
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

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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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

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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.