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Calculus 3 : Double Integrals

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Example Questions

Example Question #663 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{50}{(3\cdot (x^{2} + y^{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.22\text{ and }1.31\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.613,0.790)\text{ and }\overrightarrow{u_2}=(-0.454,-0.891)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

29.89

119.58

-59.79

-358.73

Correct answer:

-59.79

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{50}{(3\cdot (x^{2} + y^{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{50}{(3\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.790}{-0.613})=0.71\pi;\theta_2=arctan(\frac{-0.891}{-0.454})=1.35\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.71\pi}^{1.35\pi}\int_{0.22}^{1.31}(-\frac{50}{(3\cdot r)})drd\theta=(-\frac{(50ln(r))}{3})d\theta|_{0.22}^{1.31}\\&\int_{0.71\pi}^{1.35\pi}(-29.74)d\theta=(-29.74\theta)|_{0.71\pi}^{1.35\pi}=-59.79\end{align*}$

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Example Question #661 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(19sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{2}-\frac{ (\frac{47}{(\frac{2}{5})^{(x^{2} + y^{2})}})}{4})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.41\text{ and }1.71\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.790,0.613)\text{ and }\overrightarrow{u_2}=(1.000,-0.031)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

58.46

-292.31

-58.46

1753.86

Correct answer:

-292.31

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(19sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{2}-\frac{ (\frac{47}{(\frac{2}{5})^{(x^{2} + y^{2})}})}{4})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(19\cdot rsin(\frac{r^{2}}{2}))}{2}-\frac{ (47\cdot r)}{(4\cdot (\frac{2}{5})^{(r^{2})})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.613}{-0.790})=0.79\pi;\theta_2=arctan(\frac{-0.031}{1.000})=1.99\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.79\pi}^{1.99\pi}\int_{0.41}^{1.71}(\frac{(19\cdot rsin(\frac{r^{2}}{2}))}{2}-\frac{ (47\cdot r)}{(4\cdot (\frac{2}{5})^{(r^{2})})})drd\theta=(\frac{47}{(8\cdot (\frac{2}{5})^{(r^{2})}ln(\frac{2}{5}))}-\frac{ (19cos(\frac{r^{2}}{2}))}{2})d\theta|_{0.41}^{1.71}\\&\int_{0.79\pi}^{1.99\pi}(-77.54)d\theta=(-77.54\theta)|_{0.79\pi}^{1.99\pi}=-292.31\end{align*}$

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Example Question #1 : Double Integration Over General Regions

Calculate the following Integral.

$\int_{0}^{1}\int_{x}^{x^2} xy^2 dydx$

Possible Answers:

$-\frac{3}{120}$

120

$-\frac{1}{120}$

$\frac{1}{120}$

Correct answer:

$-\frac{3}{120}$

Explanation:

$\int_{0}^{1}\int_{x}^{x^2} xy^2 dydx=\int_{0}^{1}\Big(\int_{x}^{x^2} xy^2 dy\Big)dx$

Lets deal with the inner integral first.

$\int_{x}^{x^2}xy^2 dy$

$\int_{x}^{x^2}xy^2 dy=\frac{1}{3}y^3x\Big|_{x}^{x^2}$

$=(\frac{1}{3}(x^2)^3\cdot x)-(\frac{1}{3}x^3\cdot x)$

$=\frac{1}{3}x^7-\frac{1}{3}x^4$

Now we evaluate this expression in the outer integral.

$\int_{0}^{1} \frac{1}{3}x^7-\frac{1}{3}x^4dx$

$=\frac{1}{3}\int_{0}^{1} x^7-x^4dx$

$=\frac{1}{3}\Big( \frac{1}{8}x^8-\frac{1}{5}x^5\Big)\Big|_{0}^{1}$

$=\frac{1}{3}\Big(\frac{1}{8}1^8-\frac{1}{5}1^5\Big)-\frac{1}{3}\Big(\frac{1}{8}0^8-\frac{1}{5}0^5\Big)$

$=\frac{1}{3}\Big(\frac{1}{8}-\frac{1}{5}\Big)-\frac{1}{3}\Big(0-0\Big)$

$=\frac{1}{3}\Big(\frac{5}{40}-\frac{8}{40}\Big)$

$=\frac{1}{3}\Big(-\frac{3}{40}\Big)=-\frac{3}{120}$

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Example Question #1 : Double Integration Over General Regions

Calculate the definite integral of the function f(x,y) , given below as

$f(x,y)= 3x^3\sqrt{y}+ \sin(2x)\cos(4y)$

Possible Answers:

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{8}\cos(2x)\sin(4y)$

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}+\frac{1}{8}\cos(2x)\sin(4y)+C$

Cannot be solved.

$\iint f(x,y)dxdy=x^4y^{3/2}-\cos(2x)\sin(4y)+C$

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{8}\cos(2x)\sin(4y)+C$

Correct answer:

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{8}\cos(2x)\sin(4y)+C$

Explanation:

Because there are no nested terms containing both and , we can rewrite the integral as

$\iint f(x,y)dxdy=\int3x^3dx \int \sqrt{y}dy+\int\sin(2x)dx \int \cos(4y)dy$

This enables us to evaluate the double integral and the product of two independent single integrals. From the integration rules from single-variable calculus, we should arrive at the result

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{12}\cos(2x)\sin(4y)+C$ .

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Example Question #1 : Double Integration Over General Regions

Evaluate the following integral on the region specified:

$\int\int_R(x^3y)\:dxdy$

Where R is the region defined by the conditions:

$0\leq x \leq 2, \; 1\leq y \leq2$

Possible Answers:

$2\sqrt{2}$

$4\sqrt{2}$

7.5

Correct answer:

Explanation:

$\int^2_1\int^2_0(x^3y)\:dxdy=\int^2_1(\frac{x^4y}{4}|^2_0)dy\int^2_14y\:dy =2y^2|^2_1=2(4-1)=2(3)=6$

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Example Question #2 : Double Integration Over General Regions

Evaluate:

$\int \int x^2dxdy$

Possible Answers:

$\frac{x^2y}{3}+C$

$\frac{x^3}{3}+C$

$\frac{x^3}{3}+y+C$

$\frac{x^3y}{3}+C$

Correct answer:

$\frac{x^3y}{3}+C$

Explanation:

Because the x and y terms in the integrand are independent of one another, we can move them to their respective integrals:

$\int dy \cdot \int x^2dx= \frac{x^3y}{3}+C$

We used the following rules for integration:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$ , $\int dx=x+C$

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Example Question #3 : Double Integration Over General Regions

Evaluate the following integral.

$\int_{0}^{1}\int_{2}^{3}\left (5x^4y \right )dydx$

Possible Answers:

$\frac{5}{2}$

Correct answer:

$\frac{5}{2}$

Explanation:

First, you must evaluate the integral with respect to y (because of the notation dydx ).

Using the rules of integration, this gets us

$5x^4\frac{y^2}{2}dx$ .

Evaluated from y=2 to y=3, we get

$5x^4(\frac{9}{2}-\frac{4}{2})dx=\frac{25}{2}x^4dx$ .

Integrating this with respect to x gets us $\frac{5}{2}x^5$ , and evaluating from x=0 to x=1, you get $\frac{5}{2}$ .

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Example Question #4 : Double Integration Over General Regions

Compute the following integral:

$\int_{1}^{2}\int_{0}^{2}(x^3+y^2)dydx$

Possible Answers:

$\frac{120}{15}$

$\frac{130}{12}$

$\frac{122}{12}$

Correct answer:

$\frac{122}{12}$

Explanation:

First, you must evaluate the integral with respect to y and solving within the bounds.

In doing so, you get $x^3y+\frac{y^3}{3}$ and you evaluate for y from 0 to 2.

This gets you

$\int_{1}^{2}(2x^3+\frac{8}{3})dx$ .

This time evaluating the integral with respect to x gets you

$\frac{2x^4}{4}+\frac{8}{3}x$ .

Evaluating for x from 1 to 2 gets you

$\frac{122}{12}$ .

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Example Question #7 : Double Integration Over General Regions

Evaluate the double integral.

$\int_{0}^{7} \int_{0}^{5} 2x+7\,dxdy$

Possible Answers:

710

125

365

420

Correct answer:

420

Explanation:

When solving double integrals, we compute the integral on the inside first.

$\int_{0}^{7} \int_{0}^{5} 2x+7\,dxdy=\int_0^7x^2+7x|_{x=0}^{x=5}\,dy$

$=\int_0^75^2+7(5)-0^2+7(0)\,dy$

$=\int_0^760\,dy$

$=60y|_{y=0}^{y=7}$

=420-0

=420

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Example Question #1 : Double Integration Over General Regions

Evaluate the double integral.

$\int_{0}^{6} \int_{0}^{4} 6y+8\,dxdy$

Possible Answers:

624

315

540

312

Correct answer:

624

Explanation:

When solving double integrals, we compute the integral on the inside first.

$\int_{0}^{6} \int_{0}^{4} 6y+8\,dxdy=\int_0^66xy+8x|_{x=0}^{x=4}\,dy$

$=\int_0^6 6(4)y+8(4)-6(0)y-8(0)\,dy$

$=\int_0^624y+32\,dy$

$=12y^2+32y|_{y=0}^{y=6}$

=12(6)^2+32(6)-[12(0)^2+32(0)]

=624

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Calculus 3 : Double Integrals

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #663 : Multiple Integration

Example Question #661 : Multiple Integration

Example Question #1 : Double Integration Over General Regions

Example Question #1 : Double Integration Over General Regions

Example Question #1 : Double Integration Over General Regions

Example Question #2 : Double Integration Over General Regions

Example Question #3 : Double Integration Over General Regions

Example Question #4 : Double Integration Over General Regions

Example Question #7 : Double Integration Over General Regions

Example Question #1 : Double Integration Over General Regions

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