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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #661 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(\frac{48}{(\frac{2}{7})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})}{5}- 32cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.28\text{ and }1.58\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.750,0.661)\text{ and }\overrightarrow{u_2}=(-0.509,-0.861)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

38.7

-3.23

19.35

-19.35

Correct answer:

19.35

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(\frac{48}{(\frac{2}{7})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})}{5}- 32cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(48\cdot r)}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})})}- 32\cdot rcos(\frac{r^{2}}{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.661}{-0.750})=0.77\pi;\theta_2=arctan(\frac{-0.861}{-0.509})=1.33\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.77\pi}^{1.33\pi}\int_{0.28}^{1.58}(\frac{(48\cdot r)}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})})}- 32\cdot rcos(\frac{r^{2}}{2}))drd\theta=(- 32sin(\frac{r^{2}}{2}) -\frac{ 36}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})}ln(\frac{2}{7}))})d\theta|_{0.28}^{1.58}\\&\int_{0.77\pi}^{1.33\pi}(11.0)d\theta=(11.0\theta)|_{0.77\pi}^{1.33\pi}=19.35\end{align*}$

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Example Question #662 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})} +\frac{ (2\cdot (\frac{7}{2})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{25})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.34\text{ and }1.91\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.998,0.063)\text{ and }\overrightarrow{u_2}=(-0.094,0.996)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

511.96

-255.98

-42.66

127.99

Correct answer:

127.99

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})} +\frac{ (2\cdot (\frac{7}{2})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{25})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot (\frac{7}{2})^{(\frac{(2\cdot r^{2})}{3})}\cdot r)}{25}+ re^{(\frac{(3\cdot r^{2})}{2})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.063}{0.998})=0.02\pi;\theta_2=arctan(\frac{0.996}{-0.094})=0.53\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.02\pi}^{0.53\pi}\int_{0.34}^{1.91}(\frac{(2\cdot (\frac{7}{2})^{(\frac{(2\cdot r^{2})}{3})}\cdot r)}{25}+ re^{(\frac{(3\cdot r^{2})}{2})})drd\theta=(\frac{e^{(\frac{(3\cdot r^{2})}{2})}}{3}+\frac{ (3\cdot (\frac{7}{2})^{(\frac{(2\cdot r^{2})}{3})})}{(50ln(\frac{7}{2}))})d\theta|_{0.34}^{1.91}\\&\int_{0.02\pi}^{0.53\pi}(79.88)d\theta=(79.88\theta)|_{0.02\pi}^{0.53\pi}=127.99\end{align*}$

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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 : Line Integrals

Evaluate $\int_{C} \nabla f \cdot d\vec{r}$ , where $f(x,y,z)=\cos(2\pi x)+\sin(3\pi y)+xyz$ , and is any path that starts at (1,1,-2) , and ends at (3,4,5) .

Possible Answers:

Correct answer:

Explanation:

Since there isn't a specific path we need to take, we just evaluate f(x,y,z) at the end points.

$\int_{C} \nabla f \cdot d\vec{r}=f(1,1,-2)-f(3,4,5)$

$=\cos(2\pi (1))+\sin(3\pi (1))+(1)(1)(-2)-(\cos(2\pi (3))+\sin(4\pi (4))+(3)(4)(5))$

$=\cos(2\pi)+\sin(3\pi)-2-(\cos(6\pi)+\sin(12\pi)+60)$

=1+0-2-(1+0+60)=-62

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Example Question #2 : Line Integrals

Use Green's Theorem to evaluate $\oint_C 5xy \ dx +x^3y^2 \ dy$ , where is a triangle with vertices (0,0) , (1,0) , (1,2) with positive orientation.

Possible Answers:

Correct answer:

Explanation:

First we need to make sure that the conditions for Green's Theorem are met.

The conditions are met because it is positively oriented, piecewise smooth, simple, and closed under the region (see below).

$0\leq x \leq 1$

$0 \leq y \leq 2x$

In this particular case P=5xy , and Q=x^3y^2 , where , and refer to $\oint_C P \ dx +Q \ dy$ .

We know from Green's Theorem that

$\oint_C P \ dx +Q \ dy=\int \int_D \Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big) dA$

So lets find the partial derivatives.

$\frac{\partial Q}{\partial x}=3x^2y^2$

$\frac{\partial P}{\partial y}=5x$

$\oint_C 5xy \ dx +x^3y^2 \ dy=\int \int_D 3x^2y^2-5x \ dA$

$=\int_{0}^{1} \int_{0}^{2x} 3x^2y^2-5x \ dy\ dx$

$=\int_{0}^{1} \Big(\frac{1}{3}\cdot3x^2y^3-5xy\Big)\Big|_{0}^{2x}\ dx$

$=\int_{0}^{1} \Big(x^2(2x)^3-5x(2x)\Big)\ dx$

$=\int_{0}^{1} 8x^5-10x^2\ dx$

$= \frac{4}{3}x^6-\frac{10}{3}x^3\Big|_{0}^{1}$

$= \frac{4}{3}(1)^6-\frac{10}{3}(1)^3-0=\frac{4}{3}-\frac{10}{3}=-\frac{6}{3}=-2$

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Example Question #2 : Green's Theorem

Use Green's Theorem to evaluate the line integral

$\int_{c}3ydx-4xdy$

over the region R, described by connecting the points (0,0),(2,0),(2,4) , orientated clockwise.

Possible Answers:

Correct answer:

Explanation:

Using Green's theorem

$\int_C{F}dr=\int \int\left[{\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}}\right]dA$

$\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}=-4-3=-7$

$Area=\frac{1}{2}bh=\frac{1}{2}\times2\times4=4$

since the region is oriented clockwise, we would have

$\int_C{F}dr=-\int\int\left[\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right]dA$

which gives us

$\int_C{F} dr=-(-7)\times4=28$

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Example Question #3 : Green's Theorem

Use Greens Theorem to evaluate the line integral

$\int_C 5y dx - 4x dy$

over the region connecting the points (0,0),(2,0),(2,4) oriented clockwise

Possible Answers:

Correct answer:

Explanation:

$\int_C 5y dx - 4x dy$

Using Green's theorem

$\int_C F dr=\int\int \left[\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right]dA$

$Q=-4x\\ \frac{\partial{Q}}{\partial{x}}=-4$ $P=5y\\ \frac{\partial{P}}{\partial{y}}=5$

$Area=\frac{1}{2}bh=\frac{1}{2}\times2\times4=4$

Since the region is oriented clockwise

$\int_C F dr=-\int\int (-4-5)dA$

$\int_C F dr=\int\int 9dA=(9)(4)=36$

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Example Question #1551 : Calculus 3

Evaluate $\int \int \int_F 20z \ dV$ , where is the upper half of the sphere x^2+y^2+z^2=1 .

Possible Answers:

$2\pi$

$5\pi$

$4\pi$

$\pi$

Correct answer:

$5\pi$

Explanation:

Since we are only dealing with the upper half of a sphere, we can determine the boundaries easily, and remember to convert to spherical coordinates.

$0 \leq \rho \leq 1$

$0 \leq \theta \leq2\pi$

$0 \leq\phi\leq \frac{\pi}{2}$

$\int \int \int_F 20z \ dV=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} \int_{0}^{1} \rho^2\sin(\phi)(20\rho\cos(\phi)) d\rho \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} \int_{0}^{1} 20\rho^3\sin(\phi)\cos(\phi) d\rho \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5\rho^4\sin(\phi)\cos(\phi) \Big|_{0}^{1} \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5(1)^4\sin(\phi)\cos(\phi)-0 \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5\sin(\phi)\cos(\phi) \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 5\theta \sin(\phi)\cos(\phi) \Big|_{0}^{2\pi} \ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 10\pi \sin(\phi)\cos(\phi)-0 \ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 10\pi \sin(\phi)\cos(\phi) \ d\phi$

$= -5\pi \cos^2(\phi)\Big|_{0}^{\frac{\pi}{2}}$

$= -5\pi \cos^2(\frac{\pi}{2})+5\pi \cos^2(0)=5\pi$

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Example Question #1 : Gradient Vector, Tangent Planes, And Normal Lines

Find the equation of the tangent plane to $z=\ln(4x^3+10y^2)$ at z=(0,5) .

Possible Answers:

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

$z=-\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}-\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y+\ln(254)$

$z=\frac{6}{127}x-\frac{50}{127}y+\frac{256}{127}+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}$

Correct answer:

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

Explanation:

First, we need to find the partial derivatives in respect to , and , and plug in z=(1,5) .

$f(x,y)=\ln(4x^3+10y^2)$ , $f(0,5)=\ln(4(1)^3+10(5)^2)=\ln(254)$

$f_x(x,y)=\frac{12x^2}{4x^3+10y^2}$ , $f_x(0,5)=\frac{12(1)^2}{4(1)^3+10(5)^2}=\frac{12}{254}=\frac{6}{127}$

$f_y(x,y)=\frac{20y}{4x^3+10y^2}$ , $f_y(0,5)=\frac{20(5)}{4(1)^3+10(5)^2}=\frac{100}{254}=\frac{50}{127}$

Remember that the general equation for a tangent plane is as follows:

z-f(x_0,y_0)=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)

Now lets apply this to our problem

$z-\ln(254)=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)$

$z=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)+\ln(254)$

$z=\frac{6}{127}x-\frac{6}{127}+\frac{50}{127}y-\frac{250}{127}+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #661 : Multiple Integration

Example Question #662 : Multiple Integration

Example Question #663 : Multiple Integration

Example Question #661 : Multiple Integration

Example Question #1 : Line Integrals

Example Question #2 : Line Integrals

Example Question #2 : Green's Theorem

Example Question #3 : Green's Theorem

Example Question #1551 : Calculus 3

Example Question #1 : Gradient Vector, Tangent Planes, And Normal Lines

All Calculus 3 Resources

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