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Calculus 3 : Triple Integration in Cartesian Coordinates

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Example Questions

Example Question #851 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{4}^{5.5}\int_{-5}^{-3.5}\int_{-7}^{-5.5}(\frac{(cos(3x)e^{(-2y)})}{(6z)})dxdydz\end{align*}$

Possible Answers:

-143.3

286.7

47.78

Correct answer:

286.7

Explanation:

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

$\begin{align*}&\int_{4}^{5.5}\int_{-5}^{-3.5}\int_{-7}^{-5.5}(\frac{(cos(3x)e^{(-2y)})}{(6z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{4}^{5.5}\int_{-5}^{-3.5}\int_{-7}^{-5.5}(\frac{(cos(3x)e^{(-2y)})}{(6z)})dxdydz=\int_{4}^{5.5}\int_{-5}^{-3.5}(\frac{(sin(3x)e^{(-2y)})}{(18z)})dydz|_{-7}^{-5.5}\\&\int_{4}^{5.5}\int_{-5}^{-3.5}(\frac{(e^{(-2y)}\cdot (sin(21) - sin(\frac{33}{2})))}{(18z)})dydz=\int_{4}^{5.5}(-\frac{(e^{(-2y)}\cdot (sin(21) - sin(\frac{33}{2})))}{(36z)})dz|_{-5}^{-3.5}\\&\int_{4}^{5.5}(\frac{(e^{(7)}\cdot (sin(21) - sin(\frac{33}{2}))\cdot (e^{(3)} - 1))}{(36z)})dz=-ln(z)\cdot (\frac{(e^{(7)}sin(21))}{36}-\frac{ (e^{(10)}sin(21))}{36}-\frac{ (e^{(7)}sin(\frac{33}{2}))}{36}+\frac{ (e^{(10)}sin(\frac{33}{2}))}{36})|_{4}^{5.5}=286.7\end{align*}$

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Example Question #371 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{8}^{10.5}\int_{-10}^{-7.5}\int_{10}^{15}(\frac{(17cos(3x)cos(z + 1))}{y})dxdydz\end{align*}$

Possible Answers:

-0.77

3.86

-3.86

7.72

Correct answer:

3.86

Explanation:

$\begin{align*}&\int_{8}^{10.5}\int_{-10}^{-7.5}\int_{10}^{15}(\frac{(17cos(3x)cos(z + 1))}{y})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{10.5}\int_{-10}^{-7.5}\int_{10}^{15}(\frac{(17cos(3x)cos(z + 1))}{y})dxdydz=\int_{8}^{10.5}\int_{-10}^{-7.5}(\frac{(17cos(z + 1)sin(3x))}{(3y)})dydz|_{10}^{15}\\&\int_{8}^{10.5}\int_{-10}^{-7.5}(-\frac{(17cos(z + 1)\cdot (sin(30) - sin(45)))}{(3y)})dydz=\int_{8}^{10.5}(-ln(y)\cdot (\frac{(17cos(z + 1)sin(30))}{3}-\frac{ (17cos(z + 1)sin(45))}{3}))dz|_{-10}^{-7.5}\\&\int_{8}^{10.5}(-\frac{(17cos(z + 1)ln(\frac{3}{4})\cdot (sin(30) - sin(45)))}{3})dz=-\frac{(17sin(z + 1)ln(\frac{3}{4})\cdot (sin(30) - sin(45)))}{3}|_{8}^{10.5}=3.86\end{align*}$

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Example Question #371 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{-3.5}^{-1}\int_{-4}^{-0.5}\int_{-9}^{-6}(14\cdot 3^{(\frac{z}{2})}sin(x + 2)e^{(-2y)})dxdydz\end{align*}$

Possible Answers:

23033

46066

Correct answer:

23033

Explanation:

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

$\begin{align*}&\int_{-3.5}^{-1}\int_{-4}^{-0.5}\int_{-9}^{-6}(14\cdot 3^{(\frac{z}{2})}sin(x + 2)e^{(-2y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3.5}^{-1}\int_{-4}^{-0.5}\int_{-9}^{-6}(14\cdot 3^{(\frac{z}{2})}sin(x + 2)e^{(-2y)})dxdydz=\int_{-3.5}^{-1}\int_{-4}^{-0.5}(-14\cdot 3^{(\frac{z}{2})}cos(x + 2)e^{(-2y)})dydz|_{-9}^{-6}\\&\int_{-3.5}^{-1}\int_{-4}^{-0.5}(-14\cdot 3^{(\frac{z}{2})}e^{(-2y)}\cdot (cos(4) - cos(7)))dydz=\int_{-3.5}^{-1}(7\cdot 3^{(\frac{z}{2})}e^{(-2y)}\cdot (cos(4) - cos(7)))dz|_{-4}^{-0.5}\\&\int_{-3.5}^{-1}(-7\cdot 3^{(\frac{z}{2})}e^{(1)}\cdot (cos(4) - cos(7))\cdot (e^{(7)} - 1))dz=\frac{(14\cdot 3^{(\frac{z}{2})}\cdot (cos(4)e^{(1)} - cos(7)e^{(1)} - cos(4)e^{(8)} + cos(7)e^{(8)}))}{ln(3)}|_{-3.5}^{-1}=23033\end{align*}$

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Example Question #374 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{9}^{11}\int_{7}^{9.5}\int_{10}^{14}(\frac{(7z^{2}cos(4y)sin(x + 1))}{64})dxdydz\end{align*}$

Possible Answers:

0.11

-0.43

0.02

-0.02

Correct answer:

0.11

Explanation:

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

$\begin{align*}&\int_{9}^{11}\int_{7}^{9.5}\int_{10}^{14}(\frac{(7z^{2}cos(4y)sin(x + 1))}{64})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{11}\int_{7}^{9.5}\int_{10}^{14}(\frac{(7z^{2}cos(4y)sin(x + 1))}{64})dxdydz=\int_{9}^{11}\int_{7}^{9.5}(-\frac{(7z^{2}cos(x + 1)cos(4y))}{64})dydz|_{10}^{14}\\&\int_{9}^{11}\int_{7}^{9.5}(\frac{(7z^{2}cos(4y)\cdot (cos(11) - cos(15)))}{64})dydz=\int_{9}^{11}(\frac{(7z^{2}sin(4y)\cdot (cos(11) - cos(15)))}{256})dz|_{7}^{9.5}\\&\int_{9}^{11}(-\frac{(7z^{2}\cdot (cos(11) - cos(15))\cdot (sin(28) - sin(38)))}{256})dz=-z^{3}\cdot (\frac{(7cos(11)sin(28))}{768}-\frac{ (7cos(15)sin(28))}{768}-\frac{ (7cos(11)sin(38))}{768}+\frac{ (7cos(15)sin(38))}{768})|_{9}^{11}=0.11\end{align*}$

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Example Question #371 : Multiple Integration

$\begin{align*}&\text{Evaluate the triple integral}\int_{6}^{10.5}\int_{9}^{12.5}\int_{10}^{11.5}(\frac{(4y^{2}cos(z + 2))}{(15x)})dxdydz\end{align*}$

Possible Answers:

5.35

64.22

-16.05

-3.21

Correct answer:

-16.05

Explanation:

$\begin{align*}&\int_{6}^{10.5}\int_{9}^{12.5}\int_{10}^{11.5}(\frac{(4y^{2}cos(z + 2))}{(15x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{10.5}\int_{9}^{12.5}\int_{10}^{11.5}(\frac{(4y^{2}cos(z + 2))}{(15x)})dxdydz=\int_{6}^{10.5}\int_{9}^{12.5}(\frac{(4y^{2}cos(z + 2)ln(x))}{15})dydz|_{10}^{11.5}\\&\int_{6}^{10.5}\int_{9}^{12.5}(\frac{(4y^{2}cos(z + 2)ln(\frac{23}{20}))}{15})dydz=\int_{6}^{10.5}(\frac{(4y^{3}cos(z + 2)ln(\frac{23}{20}))}{45})dz|_{9}^{12.5}\\&\int_{6}^{10.5}(\frac{(9793cos(z + 2)ln(\frac{23}{20}))}{90})dz=\frac{(9793sin(z + 2)ln(\frac{23}{20}))}{90}|_{6}^{10.5}=-16.05\end{align*}$

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Example Question #852 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{-4.5}^{-0.5}\int_{-7}^{-3.5}\int_{6}^{11}(\frac{(2cos(y + 2))}{(3\cdot 3^zx)})dxdydz\end{align*}$

Possible Answers:

-24.93

498.6

-99.71

24.93

Correct answer:

-99.71

Explanation:

$\begin{align*}&\int_{-4.5}^{-0.5}\int_{-7}^{-3.5}\int_{6}^{11}(\frac{(2cos(y + 2))}{(3\cdot 3^zx)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-4.5}^{-0.5}\int_{-7}^{-3.5}\int_{6}^{11}(\frac{(2cos(y + 2))}{(3\cdot 3^zx)})dxdydz=\int_{-4.5}^{-0.5}\int_{-7}^{-3.5}(\frac{(2\cdot (\frac{1}{3})^zcos(y + 2)ln(x))}{3})dydz|_{6}^{11}\\&\int_{-4.5}^{-0.5}\int_{-7}^{-3.5}(\frac{(2cos(y + 2)ln(\frac{11}{6}))}{(3\cdot 3^z)})dydz=\int_{-4.5}^{-0.5}(\frac{(2sin(y + 2)ln(\frac{11}{6}))}{(3\cdot 3^z)})dz|_{-7}^{-3.5}\\&\int_{-4.5}^{-0.5}(-\frac{(2ln(\frac{11}{6})\cdot (sin(\frac{3}{2}) - sin(5)))}{(3\cdot 3^z)})dz=\frac{(2\cdot (ln(\frac{11}{6})sin(\frac{3}{2}) - ln(\frac{11}{6})sin(5)))}{(3\cdot 3^zln(3))}|_{-4.5}^{-0.5}=-99.71\end{align*}$

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Example Question #853 : Calculus 3

$\begin{align*}&\text{Evaluate the triple integral}\int_{-10}^{-6}\int_{-8}^{-6.5}\int_{4}^{8.5}(\frac{(45cos(3y))}{(32xz)})dxdydz\end{align*}$

Possible Answers:

0.27

1.36

-1.64

-0.05

Correct answer:

0.27

Explanation:

$\begin{align*}&\int_{-10}^{-6}\int_{-8}^{-6.5}\int_{4}^{8.5}(\frac{(45cos(3y))}{(32xz)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-10}^{-6}\int_{-8}^{-6.5}\int_{4}^{8.5}(\frac{(45cos(3y))}{(32xz)})dxdydz=\int_{-10}^{-6}\int_{-8}^{-6.5}(\frac{(45cos(3y)ln(x))}{(32z)})dydz|_{4}^{8.5}\\&\int_{-10}^{-6}\int_{-8}^{-6.5}(\frac{(45cos(3y)ln(\frac{17}{8}))}{(32z)})dydz=\int_{-10}^{-6}(\frac{(15sin(3y)ln(\frac{17}{8}))}{(32z)})dz|_{-8}^{-6.5}\\&\int_{-10}^{-6}(\frac{(15ln(\frac{17}{8})\cdot (sin(24) - sin(\frac{39}{2})))}{(32z)})dz=ln(z)\cdot (\frac{(15ln(\frac{17}{8})sin(24))}{32}-\frac{ (15ln(\frac{17}{8})sin(\frac{39}{2}))}{32})|_{-10}^{-6}=0.27\end{align*}$

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Example Question #381 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{6}^{11}\int_{-7}^{-5.5}\int_{9}^{12}(\frac{(z^{2}cos(3x)cos(3y))}{62})dxdydz\end{align*}$

Possible Answers:

-2.01

-4.02

6.03

Correct answer:

-2.01

Explanation:

$\begin{align*}&\int_{6}^{11}\int_{-7}^{-5.5}\int_{9}^{12}(\frac{(z^{2}cos(3x)cos(3y))}{62})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{11}\int_{-7}^{-5.5}\int_{9}^{12}(\frac{(z^{2}cos(3x)cos(3y))}{62})dxdydz=\int_{6}^{11}\int_{-7}^{-5.5}(\frac{(z^{2}cos(3y)sin(3x))}{186})dydz|_{9}^{12}\\&\int_{6}^{11}\int_{-7}^{-5.5}(-\frac{(z^{2}cos(3y)\cdot (\frac{sin(27)}{3}-\frac{ sin(36)}{3}))}{62})dydz=\int_{6}^{11}(-\frac{(z^{2}sin(3y)\cdot (sin(27) - sin(36)))}{558})dz|_{-7}^{-5.5}\\&\int_{6}^{11}(-\frac{(z^{2}\cdot (sin(21) - sin(\frac{33}{2}))\cdot (sin(27) - sin(36)))}{558})dz=-z^{3}\cdot (\frac{(sin(21)sin(27))}{1674}-\frac{ (sin(21)sin(36))}{1674}-\frac{ (sin(27)sin(\frac{33}{2}))}{1674}+\frac{ (sin(\frac{33}{2})sin(36))}{1674})|_{6}^{11}=-2.01\end{align*}$

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Example Question #381 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{3.5}^{7.5}\int_{-6}^{-4.5}\int_{9}^{13}(34xcos(3y)e^{(-2z)})dxdydz\end{align*}$

Possible Answers:

0.07

-0.09

2.12

-0.35

Correct answer:

-0.35

Explanation:

$\begin{align*}&\int_{3.5}^{7.5}\int_{-6}^{-4.5}\int_{9}^{13}(34xcos(3y)e^{(-2z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3.5}^{7.5}\int_{-6}^{-4.5}\int_{9}^{13}(34xcos(3y)e^{(-2z)})dxdydz=\int_{3.5}^{7.5}\int_{-6}^{-4.5}(17x^{2}cos(3y)e^{(-2z)})dydz|_{9}^{13}\\&\int_{3.5}^{7.5}\int_{-6}^{-4.5}(1496cos(3y)e^{(-2z)})dydz=\int_{3.5}^{7.5}(\frac{(1496sin(3y)e^{(-2z)})}{3})dz|_{-6}^{-4.5}\\&\int_{3.5}^{7.5}(\frac{(1496e^{(-2z)}\cdot (sin(18) - sin(\frac{27}{2})))}{3})dz=-\frac{(748e^{(-2z)}\cdot (sin(18) - sin(\frac{27}{2})))}{3}|_{3.5}^{7.5}=-0.35\end{align*}$

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Example Question #383 : Triple Integrals

$\begin{align*}&\text{Evaluate the triple integral}\int_{-5}^{-2}\int_{-4}^{1}\int_{9}^{12.5}(\frac{(11e^{(-2y)}e^{(z)})}{28})dxdydz\end{align*}$

Possible Answers:

263.5

-52.71

-790.6

1581

Correct answer:

263.5

Explanation:

$\begin{align*}&\int_{-5}^{-2}\int_{-4}^{1}\int_{9}^{12.5}(\frac{(11e^{(-2y)}e^{(z)})}{28})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-5}^{-2}\int_{-4}^{1}\int_{9}^{12.5}(\frac{(11e^{(-2y)}e^{(z)})}{28})dxdydz=\int_{-5}^{-2}\int_{-4}^{1}(\frac{(11xe^{(-2y)}e^{(z)})}{28})dydz|_{9}^{12.5}\\&\int_{-5}^{-2}\int_{-4}^{1}(\frac{(11e^{(z - 2y)})}{8})dydz=\int_{-5}^{-2}(-\frac{(11e^{(z - 2y)})}{16})dz|_{-4}^{1}\\&\int_{-5}^{-2}(\frac{(11e^{(-2)}e^{(z)}\cdot (e^{(10)} - 1))}{16})dz=e^{(z - 2)}\cdot (\frac{(11e^{(10)})}{16}-\frac{ 11}{16})|_{-5}^{-2}=263.5\end{align*}$

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Calculus 3 : Triple Integration in Cartesian Coordinates

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #851 : Calculus 3

Example Question #371 : Triple Integrals

Example Question #371 : Multiple Integration

Example Question #374 : Triple Integrals

Example Question #371 : Multiple Integration

Example Question #852 : Calculus 3

Example Question #853 : Calculus 3

Example Question #381 : Triple Integrals

Example Question #381 : Triple Integrals

Example Question #383 : Triple Integrals

All Calculus 3 Resources

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