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Calculus 3 : Line Integrals

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Example Questions

Example Question #1 : Line Integrals

Write the parametric equations of the line that passes through the points $\left ( 0,1,3\right )$ and $\left ( 2,5,8\right )$ .

Possible Answers:

x=2t+4,y+3t+1,z=t

x=-2t,y=-4t+1,z=-5t+3

x=-2t,y=4t+1,z=5t+3

x=5t+3,y=3t=1,z=-5t+1

Correct answer:

x=-2t,y=-4t+1,z=-5t+3

Explanation:

First, you must find the vector that is parallel to the line.

This vector is

$v=\left \langle x_1-x_2,y_1-y_2,z_1-z_2\right \rangle$ .

From the points we were given, this becomes

$\left \langle -2,-4,-5\right \rangle$ .

To form the parametric equations, we need to pick a point that lies on the line we want.

The point $\left ( 0,1,3\right )$ is used.

The vector form of the line is from the following equation

$r=\left ( x_1,y_1,z_1\right )+t\left \langle v\right \rangle=\left ( 0,1,3\right )+t\left \langle -2,-4,-5\right \rangle$ .

We then rewrite each expression in terms of the variables x, y, and z.

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Example Question #2 : Line Integrals

Evaluate the line integral $\int_C f(x,y)ds$ of the function

f(x,y)=y+cos(x) over the line segment from (0,0) to (4,5)

Possible Answers:

$\frac{11\sqrt{21}}{3}$

$\frac{9\sqrt{21}}{7}$

$\frac{\sqrt{31}}{3}$

$\frac{5\sqrt{41}}{2}$

$\frac{11\sqrt{41}}{4}$

Correct answer:

$\frac{5\sqrt{41}}{2}$

Explanation:

Evaluate the line integral $\int_C f(x,y)ds$ using the function

$f(x,y)=y+\cos(\pi x)$ over the line segment from (0,0) to (4,5)

Define the Parametric Equations to Represent

The points given lie on the line $y = \frac{5}{4}x$ . Define the parameter x = t , then can be written $y =\frac{5}{4}t$ . Therefore, the parametric equations for are:

x(t)=t

$y(t)=\frac{5}{4}t$

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The line integral of a function f(x,y) along the curve with the parametric equation x(t) and y(t) with $a\leq t\leq b$ is defined by:

$\int_C f(x,y)ds = \int_a^bf(x(t),y(t))\left \| \vec{r'}(t)\right \|dt$ (1)

Where $\vec{r'(t)}$ is the vector derivative of the vector $\vec{r}(t)=\left<x(t),y(t)\right>$ , therefore $\left \| \vec{r'(t)}\right \|$ is simply the magnitude of the vector derivative.

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Write the vector $\vec{r(t)}$ :

$\vec{r(t)}=\left<t, \frac{5}{4}t\right>$

Differentiate,

$\vec{r'(t)}=\left<1,\frac{5}{4}\right>$

The absolute value (magnitude) of this vector is:

$\left \| \vec{r'}(t)\right \|=\left \| \left<1,\frac{5}{4 }\right>\right \|=\sqrt{1^2 +\left(\frac{5}{4} \right )^2}=\frac{\sqrt{41}}{4}$

Write the function in terms of the parameter :

$f(x(t),y(t))=f\left(t,\frac{5}{4}t\right) =\frac{5}{4}t+\cos(\pi t)$

Insert everything into Equation (1) noting that the limits of integration will be $0\leq t \leq 4$ due to the fact that the parameter x(t)=t varies from to over the line segment we are integrating over.

$\int_C(y+\cos(\pi x))ds=\int_0^4\left(\frac{5}{4}t+\cos(\pi t)\right)\frac{\sqrt{41}}{4}dt$

$=\frac{\sqrt{41}}{4}\left[\frac{5}{8}t^2+\frac{1}{\pi}\sin(\pi t) \right ]_{t=0}^{t=4}$

$=\frac{\sqrt{41}}{4}\left[\frac{5}{8}(16)+\frac{1}{\pi }\sin(4 \pi) -0-\frac{1}{\pi}\sin(0)\right ]$

$=\frac{5\sqrt{41}}{2}$

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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 #1 : Line Integrals

Compute $div \vec{F}$ for

$\vec{F}=yz\ln(x)\vec{i}+2xyz\vec{j}+z\vec{k}$

Possible Answers:

$div\vec{F}=\frac{yz}{x}$

$div\vec{F}=0$

$div\vec{F}=\frac{yz}{x}+2xz+1$

$div\vec{F}=2xz+1$

$div\vec{F}=-\frac{yz}{x}+2xz$

Correct answer:

$div\vec{F}=\frac{yz}{x}+2xz+1$

Explanation:

In order to find the divergence, we need to remember the formula.

Divergence Formula:

$div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$ , where , , and correspond to the components of a given vector field $\vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$ .

Now lets apply this to our situation.

$div\vec{F}=\frac{\partial}{\partial x}\Big(yz\ln(x)\Big)+\frac{\partial}{\partial y}\Big(2xyz\Big)+\frac{\partial}{\partial z}\Big(z\Big)$

$div\vec{F}=\frac{yz}{x}+2xz+1$

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Example Question #2 : Line Integrals

Compute $div\vec{F}$ for

$\vec{F}=x^2z\ln(y)\vec{i}+x2^{y^2}z\vec{j}+x^x\vec{k}$

Possible Answers:

$div\vec{F}=2xz\ln(y)$

$div\vec{F}=2xz\ln(y)+2\log(2)xyz2^{y^2}$

$div\vec{F}=2\log(2)xyz2^{y^2}$

$div\vec{F}=0$

$div\vec{F}=-2xz\ln(y)-2\log(2)xyz2^{y^2}$

Correct answer:

$div\vec{F}=2xz\ln(y)+2\log(2)xyz2^{y^2}$

Explanation:

In order to find the divergence, we need to remember the formula.

Divergence Formula:

Now lets apply this to our situation.

$div\vec{F}=\frac{\partial}{\partial x}\Big(x^2z\ln(y)\Big)+\frac{\partial}{\partial y}\Big(x2^{y^2}z\Big)+\frac{\partial}{\partial z}\Big(x^x\Big)$

$div\vec{F}=2xz\ln(y)+2\log(2)xyz2^{y^2}+0$

$div\vec{F}=2xz\ln(y)+2\log(2)xyz2^{y^2}$

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Example Question #4 : Line Integrals

Compute $div\vec{F}$ for

$\vec{F}=xy\vec{i}+xz\vec{j}+xyz\vec{k}$

Possible Answers:

$div \vec{F}=(1+x)$

$div \vec{F}=y$

$div \vec{F}=x$

$div \vec{F}=1$

$div \vec{F}=y(1+x)$

Correct answer:

$div \vec{F}=y(1+x)$

Explanation:

In order to find the divergence, we need to remember the formula.

Divergence Formula:

Now lets apply this to our situation.

$div\vec{F}=\frac{\partial}{\partial x}\Big(xy\Big)+\frac{\partial}{\partial y}\Big(xz\Big)+\frac{\partial}{\partial z}\Big(xyz\Big)$

$div \vec{F}=y+0+xy$

$div \vec{F}=y+xy=y(1+x)$

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Example Question #5 : Line Integrals

Find $div\vec{F}$ , where $\vec{F}=xyz\vec{i}+x^2z^{-4}\vec{j}+y^{-10}z^{3}\vec{k}$

Possible Answers:

$div\vec{F}=yz+z^4+\frac{3z^2}{y^{10}}$

$div\vec{F}=\frac{3z^2}{y^{10}}$

$div\vec{F}=yz+\frac{3z^2}{y^{-10}}$

$div\vec{F}=xyz+\frac{3z^2}{y^{10}}$

$div\vec{F}=yz+\frac{3z^2}{y^{10}}$

Correct answer:

$div\vec{F}=yz+\frac{3z^2}{y^{10}}$

Explanation:

In order to find the divergence, we need to remember the formula.

Divergence Formula:

Now lets apply this to our situation.

$div\vec{F}=\frac{\partial}{\partial x}\Big(xyz\Big)+\frac{\partial}{\partial y}\Big(x^2z^{-4}\Big)+\frac{\partial}{\partial z}\Big(y^{-10}z^3\Big)$

$div\vec{F}={yz}+0+3y^{-10}z^2=yz+\frac{3z^2}{y^{10}}$

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Calculus 3 : Line Integrals

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Line Integrals

Example Question #2 : Line Integrals

Example Question #1 : Line Integrals

Example Question #2 : Line Integrals

Example Question #2 : Green's Theorem

Example Question #3 : Green's Theorem

Example Question #1 : Line Integrals

Example Question #2 : Line Integrals

Example Question #4 : Line Integrals

Example Question #5 : Line Integrals

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