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Calculus 3 : Double Integrals

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Example Questions

Example Question #11 : Double Integration Over General Regions

Let f(x) be any continuous, one-to-one function over the interval I=[a,b] , with f(x)>0 on and 0<a<b . Select all integrals which correctly define the indicated area under the curve.

Question 5 calc 3

$1)\: \: \: \: \: \: \: \: A=\int_a^bf(x)dx$

$2)\: \: \: \: \:\, \, \: A=\int_0^{f(x)}\int_{a}^bdydx$

$3)\: \: \: \: \: \: \: \: A = \int_a^b\int_0^{f(x)}dydx$

$4) \: \: \: \: \: \: \: A=\int_{f(a)}^{f(b)}\int_{a}^{b}f(x)dxdy$

Possible Answers:

1 3 and 4

3 and 4

1 and 3

1 and 4

Correct answer:

1 and 3

Explanation:

$1)\: \: \: \: \: \: \: \: A=\int_a^bf(x)dx$

No explanation necessary. This is the familiar integral for computing the area under the curve y=f(x) over [a,b] .

_____________________________________________________________

$2)\: \: \: \: \:\, \, \: A=\int_0^{f(x)}\int_{a}^bdydx$

This integral does not give the area, in fact, it will result in a function of which certainly cannot be interpreted as the area of corresponding to a specific interval I=[a,b] .

$A = \int _0^{f(x)}\int_a^bdydx=\int_0^{f(x)}[y]_{y=a}^{y=b}dx=\int_0^{f(x)}(b-a)dx$

Now carry out the integration with respect to we obtain,

$\int_0^{f(x)}(b-a)dx=\left[ (a-b)x\right ]_{x=0}^{x=f(x)}=(a-b)f(x)$

Certainly this is not the area of the area under f(x) . One obvious problem is that the integral gave back a function of and no real number.

_____________________________________________________________

$3)\: \: \: \: \: \: \: \: A = \int_a^b\int_0^{f(x)}dydx$

This equation will reduce down to the first option, option , after we carry out the integration with respect to .

$A = \int_a^b\int_0^{f(x)}dydx=\int_a^b \left[ \: \: y \, \, \right]_{y=o}^{y=f(x)} dx =\int_a^bf(x)dx$

Therefore, option is a valid representation for the area.

___________________________________________________________

$4) \: \: \: \: \: \: \: A=\int_{f(a)}^{f(b)}\int_{a}^{b}f(x)dxdy$

The problem with this choice can be seen immediately. Clearly the first integration in will simply be the area under the curve on , but the second will now multiply the area by f(b)-f(a) .

$A=\int_{f(a)}^{f(b)}\int_{a}^{b}f(x)dxdy = \int_{f(a)}^{f(b)}Ady = [Ay]_{y=f(a)}^{y=f(b)}=A[f(b)-f(a)]$

A = A[f(b)-f(a)] could only be true if f(b)-f(a)=1 . But because none of the listed characteristics of our function imply this, we cannot assume it. In fact, the problem stated that f(x) is any function that satisfies the stated conditions. Certainly is not true for just any function.

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Example Question #12 : Double Integration Over General Regions

Evaluate the following iterated integrals:

$\int_-_1^3 \int_0^\frac{\pi}{2} y^2 \cos(x) \ dxdy$

Possible Answers:

$\frac{28}{3}$

Correct answer:

$\frac{28}{3}$

Explanation:

When evaluating double integrals, work from the inside out: that is, evaluate the integrand with respect to the first variable listed by the differential operators, and then evaluate the result with respect to the second variable listed by the differential operators.

Here, we have the order of integration specified by dxdy ; hence, we evaluate the double integrals with respect to first, and then integrate the result with respect to , as shown:

$\int_-_1^3 \int_0^\frac{\pi}{2} y^2 \cos(x) \ dxdy$

$=\int_-_1^3 \left [\int_0^\frac{\pi}{2} y^2 \cos(x) \ dx \right ] dy$

$=\int_-_1^3 \left [y^2 \sin(x) \Big|_0^\frac{\pi}{2} \right ] dy$

$=\int_-_1^3 \left [y^2 (1)-y^2(0) \right ] dy$

$=\int_-_1^3 y^2 \ dy$

$= \frac{y^3}{3} \Big|_-_1^3$

$=\frac{27}{3}-\frac{-1}{3}$

$=\frac{28}{3}$

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Example Question #21 : Double Integration Over General Regions

Calculate the following Integral.

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

Possible Answers:

$-\frac{3}{120}$

$-\frac{1}{120}$

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 #21 : Double Integration Over General Regions

Calculate the following Integral.

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

Possible Answers:

$-\frac{1}{120}$

120

$\frac{1}{120}$

$-\frac{3}{120}$

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 With Change Of Variables

Evaluate $\int \int_R 3x+2y \ dA$ , where is the trapezoidal region with vertices given by (0,0) , (5,0) , $(\frac{5}{2}, \frac{5}{2})$ , and $(\frac{5}{2}, -\frac{5}{2})$ ,

using the transformation x=2u+3v , and y=2u-3v .

Possible Answers:

$\frac{375}{2}$

$\frac{35}{4}$

$\frac{375}{4}$

$\frac{37}{4}$

Correct answer:

$\frac{375}{4}$

Explanation:

The first thing we have to do is figure out the general equations for the lines that create the trapezoid.

y=x

y=-x

y=-x+5

y=x-5

Now we have the general equations for out trapezoid, now we need to plug in our transformations into these equations.

y=x

2u+3v=2u-3v

$-6v=0\rightarrow v=0$

y=-x

2u+3v=-2u+3v

$4u=0\rightarrow u=0$

y=-x+5

2u+3v=-2u+3v+5

$4u=5\rightarrow u=\frac{5}{4}$

y=x-5

2u+3v=2u-3v-5

$-6v=-5\rightarrow v=\frac{5}{6}$

So our region is a rectangle given by $0 \leq u \leq \frac{5}{4}$ , $0 \leq v \leq \frac{5}{6}$

Next we need to calculate the Jacobian.

$\begin{vmatrix} 2 & 3\\ 2 & -3 \end{vmatrix} =-6-6=-12$

Now we can put the integral together.

$\int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} (3(2u+3v)+2(2u-3v))) \cdot \left | -12\right | \ du\ dv$

$=\int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} 120u+36v \ du\ dv$

$=\int_{0}^{\frac{5}{6}} 60u^2+36uv \Big|_{0}^{\frac{5}{4}}\ dv$

$=\int_{0}^{\frac{5}{6}} 60( \frac{5}{4})^2+36(\frac{5}{4})v-0 \ dv$

$=\int_{0}^{\frac{5}{6}} \frac{375}{4}+45v \ dv$

$=\frac{375}{4}v+\frac{45}{2}v^2\Big|_{0}^{\frac{5}{6}} \ dv$

$=\frac{375}{4}\cdot\frac{5}{6}+\frac{45}{2}\cdot\Big(\frac{5}{6}\Big)^2 -0=\frac{375}{4}$

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Example Question #3831 : Calculus 3

Find the surface area of the part of the plane 4x+4y+2z=6 in the first octant.

Possible Answers:

$\frac{13}{8}$

$\frac{17}{8}$

$\frac{27}{8}$

Correct answer:

$\frac{27}{8}$

Explanation:

Lets recall the equation of surface area.

$S=\int \int_D \sqrt{[f_x]^2+[f_y]^2+1}\ dA$

Now we need to find all the neccessary equations to be able to evaluate the integral.

z=3-2x-2y

$\frac{\partial z}{\partial x}=-2$

$\frac{\partial z}{\partial y}=-2$

We will plug in z=0 , into the plane equation in order to get a line that intersects with the z axis.

$4x+4y+2(0)=6\rightarrow y=\frac{3}{2}-x$

Now we are going to set y=0 , in the previous equation and solve for .

$0=\frac{3}{2}-x\rightarrow x=\frac{3}{2}$

We now have all the bounds for our double integral

$0\leq x \leq \frac{3}{2}$

$0\leq y \leq \frac{3}{2}-x$

$S=\int_{0}^{\frac{3}{2}} \int_{0}^{\frac{3}{2}-x} \sqrt{[-2]^2+[-2]^2+1} \ dy \ dx$

$S=\int_{0}^{\frac{3}{2}} \int_{0}^{\frac{3}{2}-x} \sqrt{9} \ dy \ dx$

$S=\int_{0}^{\frac{3}{2}} \int_{0}^{\frac{3}{2}-x} 3 \ dy \ dx$

$S=\int_{0}^{\frac{3}{2}} 3y \Big|_{0}^{\frac{3}{2}-x} \ dx$

$S=\int_{0}^{\frac{3}{2}} 3(\frac{3}{2}-x)-0 \ dx$

$S=\int_{0}^{\frac{3}{2}} \frac{9}{2}-3x \ dx$

$S= \frac{9}{2}x-\frac{3}{2}x^2 \Big|_{0}^{\frac{3}{2}}$

$S= \frac{9}{2}(\frac{3}{2})-\frac{3}{2}(\frac{3}{2})^2-0=\frac{27}{8}$

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Calculus 3 : Double Integrals

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #11 : Double Integration Over General Regions

Example Question #12 : Double Integration Over General Regions

Example Question #21 : Double Integration Over General Regions

Example Question #21 : Double Integration Over General Regions

Example Question #1 : Double Integration With Change Of Variables

Example Question #3831 : Calculus 3

All Calculus 3 Resources

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