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Calculus 3 : Triple Integrals

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Example Questions

Example Question #461 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{8}^{9.5}\int_{-4}^{0.5}\int_{10}^{12.5}(\frac{(3cos(3z)e^{(-2y)})}{(62x)})dxdydz\end{align*}$

Possible Answers:

3.66

-0.61

7.31

-10.97

Correct answer:

3.66

Explanation:

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\end{align*}$

$\begin{align*}&\int_{8}^{9.5}\int_{-4}^{0.5}\int_{10}^{12.5}(\frac{(3cos(3z)e^{(-2y)})}{(62x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{9.5}\int_{-4}^{0.5}\int_{10}^{12.5}(\frac{(3cos(3z)e^{(-2y)})}{(62x)})dxdydz=\int_{8}^{9.5}\int_{-4}^{0.5}(\frac{(3cos(3z)e^{(-2y)}ln(x))}{62})dydz|_{10}^{12.5}\\&\int_{8}^{9.5}\int_{-4}^{0.5}(\frac{(3cos(3z)e^{(-2y)}ln(\frac{5}{4}))}{62})dydz=\int_{8}^{9.5}(-\frac{(3cos(3z)e^{(-2y)}ln(\frac{5}{4}))}{124})dz|_{-4}^{0.5}\\&\int_{8}^{9.5}(\frac{(3cos(3z)e^{(-1)}ln(\frac{5}{4})\cdot (e^{(9)} - 1))}{124})dz=\frac{(sin(3z)e^{(-1)}ln(\frac{5}{4})\cdot (e^{(9)} - 1))}{124}|_{8}^{9.5}=3.66\end{align*}$

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Example Question #462 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{3.5}^{4.5}\int_{-8}^{-3}\int_{-4.5}^{-2}(\frac{(37\cdot 2^{(\frac{x}{4})}cos(4y)e^{(-z)})}{4})dxdydz\end{align*}$

Possible Answers:

-0.14

$3.45\cdot10^{-2}$

$-1.15\cdot10^{-2}$

$6.89\cdot10^{-2}$

Correct answer:

$6.89\cdot10^{-2}$

Explanation:

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\end{align*}$

$\begin{align*}&\int_{3.5}^{4.5}\int_{-8}^{-3}\int_{-4.5}^{-2}(\frac{(37\cdot 2^{(\frac{x}{4})}cos(4y)e^{(-z)})}{4})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3.5}^{4.5}\int_{-8}^{-3}\int_{-4.5}^{-2}(\frac{(37\cdot 2^{(\frac{x}{4})}cos(4y)e^{(-z)})}{4})dxdydz=\int_{3.5}^{4.5}\int_{-8}^{-3}(\frac{(37\cdot 2^{(\frac{x}{4})}cos(4y)e^{(-z)})}{ln(2)})dydz|_{-4.5}^{-2}\\&\int_{3.5}^{4.5}\int_{-8}^{-3}(\frac{(37cos(4y)e^{(-z)}\cdot (2\cdot 2^{(\frac{1}{2})} - 2^{(\frac{7}{8})}))}{(4ln(2))})dydz=\int_{3.5}^{4.5}(\frac{(37sin(4y)e^{(-z)}\cdot (2\cdot 2^{(\frac{1}{2})} - 2^{(\frac{7}{8})}))}{(16ln(2))})dz|_{-8}^{-3}\\&\int_{3.5}^{4.5}(\frac{(37\cdot 2^{(\frac{1}{2})}e^{(-z)}\cdot (sin(12) - sin(32))\cdot (2^{(\frac{3}{8})} - 2))}{(16ln(2))})dz=-\frac{(37\cdot 2^{(\frac{1}{2})}e^{(-z)}\cdot (sin(12) - sin(32))\cdot (2^{(\frac{3}{8})} - 2))}{(16ln(2))}|_{3.5}^{4.5}=6.89\cdot10^{-2}\end{align*}$

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Example Question #463 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-9}^{-4}\int_{-10}^{-8}\int_{4}^{5.5}(\frac{(10cos(z + 2)sin(y + 1)e^{(-2x)})}{13})dxdydz\end{align*}$

Possible Answers:

$-2.58\cdot10^{-5}$

$2.58\cdot10^{-5}$

$-2.58\cdot10^{-4}$

$5.15\cdot10^{-5}$

Correct answer:

$5.15\cdot10^{-5}$

Explanation:

$\begin{align*}&\int_{-9}^{-4}\int_{-10}^{-8}\int_{4}^{5.5}(\frac{(10cos(z + 2)sin(y + 1)e^{(-2x)})}{13})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-9}^{-4}\int_{-10}^{-8}\int_{4}^{5.5}(\frac{(10cos(z + 2)sin(y + 1)e^{(-2x)})}{13})dxdydz=\int_{-9}^{-4}\int_{-10}^{-8}(-\frac{(5cos(z + 2)sin(y + 1)e^{(-2x)})}{13})dydz|_{4}^{5.5}\\&\int_{-9}^{-4}\int_{-10}^{-8}(\frac{(5cos(z + 2)sin(y + 1)e^{(-11)}\cdot (e^{(3)} - 1))}{13})dydz=\int_{-9}^{-4}(-\frac{(5cos(y + 1)cos(z + 2)e^{(-11)}\cdot (e^{(3)} - 1))}{13})dz|_{-10}^{-8}\\&\int_{-9}^{-4}(-\frac{(5cos(z + 2)e^{(-11)}\cdot (cos(7) - cos(9))\cdot (e^{(3)} - 1))}{13})dz=-\frac{(5sin(z + 2)e^{(-11)}\cdot (cos(7) - cos(9))\cdot (e^{(3)} - 1))}{13}|_{-9}^{-4}=5.15\cdot10^{-5}\end{align*}$

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Example Question #941 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{5}^{6.5}\int_{6}^{7}\int_{-8}^{-7}(\frac{(12\cdot 3^zcos(3x))}{(47y^{2})})dxdydz\end{align*}$

Possible Answers:

-1.64

16.38

-3.28

1.64

Correct answer:

-3.28

Explanation:

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\end{align*}$

$\begin{align*}&\int_{5}^{6.5}\int_{6}^{7}\int_{-8}^{-7}(\frac{(12\cdot 3^zcos(3x))}{(47y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{5}^{6.5}\int_{6}^{7}\int_{-8}^{-7}(\frac{(12\cdot 3^zcos(3x))}{(47y^{2})})dxdydz=\int_{5}^{6.5}\int_{6}^{7}(\frac{(4\cdot 3^zsin(3x))}{(47y^{2})})dydz|_{-8}^{-7}\\&\int_{5}^{6.5}\int_{6}^{7}(-\frac{(12\cdot 3^z\cdot (\frac{sin(21)}{3}-\frac{ sin(24)}{3}))}{(47y^{2})})dydz=\int_{5}^{6.5}(\frac{(\frac{(4\cdot 3^zsin(21))}{47}-\frac{ (4\cdot 3^zsin(24))}{47})}{y})dz|_{6}^{7}\\&\int_{5}^{6.5}(-\frac{(2\cdot 3^z\cdot (sin(21) - sin(24)))}{987})dz=-\frac{(2\cdot 3^z\cdot (sin(21) - sin(24)))}{(987ln(3))}|_{5}^{6.5}=-3.28\end{align*}$

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Example Question #465 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{7}^{10}\int_{8}^{9.5}\int_{-7}^{-5}(\frac{(71cos(z + 1))}{(33y)})dxdydz\end{align*}$

Possible Answers:

-1.47

4.41

0.49

-0.49

Correct answer:

-1.47

Explanation:

$\begin{align*}&\int_{7}^{10}\int_{8}^{9.5}\int_{-7}^{-5}(\frac{(71cos(z + 1))}{(33y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{10}\int_{8}^{9.5}\int_{-7}^{-5}(\frac{(71cos(z + 1))}{(33y)})dxdydz=\int_{7}^{10}\int_{8}^{9.5}(\frac{(71xcos(z + 1))}{(33y)})dydz|_{-7}^{-5}\\&\int_{7}^{10}\int_{8}^{9.5}(\frac{(142cos(z + 1))}{(33y)})dydz=\int_{7}^{10}(\frac{(142cos(z + 1)ln(y))}{33})dz|_{8}^{9.5}\\&\int_{7}^{10}(\frac{(142cos(z + 1)ln(\frac{19}{16}))}{33})dz=\frac{(142sin(z + 1)ln(\frac{19}{16}))}{33}|_{7}^{10}=-1.47\end{align*}$

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Example Question #461 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3.5}^{-1}\int_{-3.5}^{-2.5}\int_{9}^{10.5}(\frac{(3sin(x + 1))}{(35\cdot 2^y\cdot 3^z)})dxdydz\end{align*}$

Possible Answers:

-36.85

7.37

-110.55

147.4

Correct answer:

-36.85

Explanation:

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\end{align*}$

$\begin{align*}&\int_{-3.5}^{-1}\int_{-3.5}^{-2.5}\int_{9}^{10.5}(\frac{(3sin(x + 1))}{(35\cdot 2^y\cdot 3^z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3.5}^{-1}\int_{-3.5}^{-2.5}\int_{9}^{10.5}(\frac{(3sin(x + 1))}{(35\cdot 2^y\cdot 3^z)})dxdydz=\int_{-3.5}^{-1}\int_{-3.5}^{-2.5}(-\frac{(3cos(x + 1))}{(35\cdot 2^y\cdot 3^z)})dydz|_{9}^{10.5}\\&\int_{-3.5}^{-1}\int_{-3.5}^{-2.5}(\frac{(3\cdot (cos(10) - cos(\frac{23}{2})))}{(35\cdot 2^y\cdot 3^z)})dydz=\int_{-3.5}^{-1}(-\frac{(3\cdot (\frac{1}{3})^z\cdot (cos(10) - cos(\frac{23}{2})))}{(35\cdot 2^yln(2))})dz|_{-3.5}^{-2.5}\\&\int_{-3.5}^{-1}(\frac{(12\cdot 2^{(\frac{1}{2})}cos(10) - 12\cdot 2^{(\frac{1}{2})}cos(\frac{23}{2}))}{(35\cdot 3^zln(2))})dz=-\frac{(12\cdot (2^{(\frac{1}{2})}cos(10) - 2^{(\frac{1}{2})}cos(\frac{23}{2})))}{(35\cdot 3^zln(2)ln(3))}|_{-3.5}^{-1}=-36.85\end{align*}$

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Example Question #467 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3}^{-2}\int_{3}^{5}\int_{10}^{11.5}(\frac{(3e^{(2z)})}{(140\cdot 2^yx)})dxdydz\end{align*}$

Possible Answers:

$-3.21\cdot10^{-6}$

$1.60\cdot10^{-6}$

$-6.42\cdot10^{-7}$

$3.21\cdot10^{-6}$

Correct answer:

$3.21\cdot10^{-6}$

Explanation:

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\end{align*}$

$\begin{align*}&\int_{-3}^{-2}\int_{3}^{5}\int_{10}^{11.5}(\frac{(3e^{(2z)})}{(140\cdot 2^yx)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3}^{-2}\int_{3}^{5}\int_{10}^{11.5}(\frac{(3e^{(2z)})}{(140\cdot 2^yx)})dxdydz=\int_{-3}^{-2}\int_{3}^{5}(\frac{(3e^{(2z)}ln(x))}{(140\cdot 2^y)})dydz|_{10}^{11.5}\\&\int_{-3}^{-2}\int_{3}^{5}(\frac{(3e^{(2z)}ln(\frac{23}{20}))}{(140\cdot 2^y)})dydz=\int_{-3}^{-2}(-\frac{(3e^{(2z)}ln(\frac{23}{20}))}{(140\cdot 2^yln(2))})dz|_{3}^{5}\\&\int_{-3}^{-2}(\frac{(9e^{(2z)}ln(\frac{23}{20}))}{(4480ln(2))})dz=\frac{(9e^{(2z)}ln(\frac{23}{20}))}{(8960ln(2))}|_{-3}^{-2}=3.21\cdot10^{-6}\end{align*}$

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Example Question #942 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{-7}^{-3.5}\int_{-8}^{-5}\int_{10}^{12.5}(\frac{(7cos(y + 2)sin(x + 1)sin(3z))}{67})dxdydz\end{align*}$

Possible Answers:

$6.24\cdot10^{-4}$

$1.25\cdot10^{-4}$

$-6.24\cdot10^{-4}$

$-3.12\cdot10^{-3}$

Correct answer:

$-6.24\cdot10^{-4}$

Explanation:

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\end{align*}$

$\begin{align*}&\int_{-7}^{-3.5}\int_{-8}^{-5}\int_{10}^{12.5}(\frac{(7cos(y + 2)sin(x + 1)sin(3z))}{67})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-3.5}\int_{-8}^{-5}\int_{10}^{12.5}(\frac{(7cos(y + 2)sin(x + 1)sin(3z))}{67})dxdydz=\int_{-7}^{-3.5}\int_{-8}^{-5}(-\frac{(7cos(x + 1)cos(y + 2)sin(3z))}{67})dydz|_{10}^{12.5}\\&\int_{-7}^{-3.5}\int_{-8}^{-5}(\frac{(7cos(y + 2)sin(3z)\cdot (cos(11) - cos(\frac{27}{2})))}{67})dydz=\int_{-7}^{-3.5}(\frac{(7sin(y + 2)sin(3z)\cdot (cos(11) - cos(\frac{27}{2})))}{67})dz|_{-8}^{-5}\\&\int_{-7}^{-3.5}(-\frac{(7sin(3z)\cdot (cos(11) - cos(\frac{27}{2}))\cdot (sin(3) - sin(6)))}{67})dz=\frac{(7cos(3z)\cdot (cos(11) - cos(\frac{27}{2}))\cdot (sin(3) - sin(6)))}{201}|_{-7}^{-3.5}=-6.24\cdot10^{-4}\end{align*}$

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Example Question #469 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{10}^{14}\int_{-8}^{-3}\int_{-7}^{-2}(\frac{sin(y + 2)}{(21x^{3}z)})dxdydz\end{align*}$

Possible Answers:

$1.93\cdot10^{-4}$

$-4.63\cdot10^{-3}$

$-7.72\cdot10^{-4}$

$1.54\cdot10^{-3}$

Correct answer:

$-7.72\cdot10^{-4}$

Explanation:

$\begin{align*}&\int_{10}^{14}\int_{-8}^{-3}\int_{-7}^{-2}(\frac{sin(y + 2)}{(21x^{3}z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{14}\int_{-8}^{-3}\int_{-7}^{-2}(\frac{sin(y + 2)}{(21x^{3}z)})dxdydz=\int_{10}^{14}\int_{-8}^{-3}(-\frac{sin(y + 2)}{(42x^{2}z)})dydz|_{-7}^{-2}\\&\int_{10}^{14}\int_{-8}^{-3}(-\frac{(15sin(y + 2))}{(2744z)})dydz=\int_{10}^{14}(\frac{(15cos(y + 2))}{(2744z)})dz|_{-8}^{-3}\\&\int_{10}^{14}(\frac{(15\cdot (cos(1) - cos(6)))}{(2744z)})dz=ln(z)\cdot (\frac{(15cos(1))}{2744}-\frac{ (15cos(6))}{2744})|_{10}^{14}=-7.72\cdot10^{-4}\end{align*}$

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Example Question #470 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{4.5}^{6}\int_{10}^{11}\int_{-7}^{-6}(\frac{(3^{(\frac{z}{2})}cos(4x))}{(11y)})dxdydz\end{align*}$

Possible Answers:

-0.28

0.21

$7.03\cdot10^{-2}$

$-1.17\cdot10^{-2}$

Correct answer:

$7.03\cdot10^{-2}$

Explanation:

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\end{align*}$

$\begin{align*}&\int_{4.5}^{6}\int_{10}^{11}\int_{-7}^{-6}(\frac{(3^{(\frac{z}{2})}cos(4x))}{(11y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{4.5}^{6}\int_{10}^{11}\int_{-7}^{-6}(\frac{(3^{(\frac{z}{2})}cos(4x))}{(11y)})dxdydz=\int_{4.5}^{6}\int_{10}^{11}(\frac{(3^{(\frac{z}{2})}sin(4x))}{(44y)})dydz|_{-7}^{-6}\\&\int_{4.5}^{6}\int_{10}^{11}(-\frac{(3^{(\frac{z}{2})}\cdot (\frac{sin(24)}{4}-\frac{ sin(28)}{4}))}{(11y)})dydz=\int_{4.5}^{6}(-\frac{(3^{(\frac{z}{2})}ln(y)\cdot (sin(24) - sin(28)))}{44})dz|_{10}^{11}\\&\int_{4.5}^{6}(-\frac{(3^{(\frac{z}{2})}ln(\frac{11}{10})\cdot (sin(24) - sin(28)))}{44})dz=-\frac{(3^{(\frac{z}{2})}\cdot (ln(\frac{11}{10})sin(24) - ln(\frac{11}{10})sin(28)))}{(22ln(3))}|_{4.5}^{6}=7.03\cdot10^{-2}\end{align*}$

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Calculus 3 : Triple Integrals

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #461 : Multiple Integration

Example Question #462 : Multiple Integration

Example Question #463 : Multiple Integration

Example Question #941 : Calculus 3

Example Question #465 : Multiple Integration

Example Question #461 : Triple Integrals

Example Question #467 : Multiple Integration

Example Question #942 : Calculus 3

Example Question #469 : Multiple Integration

Example Question #470 : Multiple Integration

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