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Calculus 3 : Line Integrals

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Example Questions

Example Question #65 : Divergence

Find the divergence of the vector $\left \langle 2xy^4,y^5,\tan(z)\right \rangle$

Possible Answers:

$7y^3+\cos^2(z)$

$7y^4+\sec^2(z)$

$y^4-\sec^2(z)$

$7y^4+\sec^2(x)$

Correct answer:

$7y^4+\sec^2(z)$

Explanation:

To find the divergence of a vector $v=\left \langle P,Q,R\right \rangle$ , we use the formula:

$div(v)=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$

Applying to the vector from the problem statement, we get

$\frac{\partial }{\partial x}(2xy^4)+\frac{\partial }{\partial y}(y^5)+\frac{\partial }{\partial z}(\tan(z))=7y^4+\sec^2(z)$

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Example Question #66 : Divergence

Find the divergence of the vector $\left \langle \sin(x),5yz^3,\cos(z)\right \rangle$

Possible Answers:

$\cos(x)+4z^3-\sin(z)$

$-\cos(x)+5z^3-\sin(z)$

$\cos(x)+5z^3-\sin(z)$

$\cos(x)+5z^3+\sin(z)$

Correct answer:

$\cos(x)+5z^3-\sin(z)$

Explanation:

To find the divergence of a vector $v=\left \langle P,Q,R\right \rangle$ , we use the formula:

$div(v)=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$

Applying to the vector from the problem statement, we get

$\frac{\partial }{\partial x}(\sin(x))+\frac{\partial }{\partial y}(5yz^3)+\frac{\partial }{\partial z}(\cos(z))=\cos(x)+5y^3-\sin(z)$

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Example Question #67 : Divergence

Find the divergence of the function:

$\vec{F}=\left \langle \cos(x)\sin(y), x^2y+y^2, z\right \rangle$

Possible Answers:

$-\sin(x)\sin(y)+x^2+2y$

$-\sin(x)\sin(y)+x^2+2y+1$

$\sin(x)\sin(y)+x^2+2y+1$

$-\sin(x)\sin(y)+2xy+1$

Correct answer:

$-\sin(x)\sin(y)+x^2+2y+1$

Explanation:

The divergence of a vector field is given by

$div \vec{F}=\bigtriangledown \cdot \vec{F}$ , where $\bigtriangledown = \left \langle \frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z} \right \rangle$

When we take the dot product of the gradient and the vector field, we get the sum of the respective partial derivatives. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=-\sin(x)\sin(y)$ , f_y=x^2+2y , f_z=1

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Example Question #71 : Line Integrals

Find the divergence of the vector field:

$\vec{F}=\left \langle 6x+x^2, xy^2z, x+y \right \rangle$

Possible Answers:

8+2xyz

6+3x+2xyz+y

3+x+xyz

6+2x+2xyz

Correct answer:

6+2x+2xyz

Explanation:

The divergence of a vector field is given by

$div \vec{F}=\bigtriangledown \cdot \vec{F}$ , where $\bigtriangledown = \left \langle \frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z} \right \rangle$

The partial derivatives are

f_x=6+2x , f_y=2xyz , f_z=0

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Example Question #69 : Divergence

Find the divergence of the vector field:

$\vec{F}=\left \langle y^2, xz^3, xe^{xy} \right \rangle$

Possible Answers:

$\vec{0}$

$e^{xy}+xye^{xy}$

2y+3xz^2

Correct answer:

Explanation:

The divergence of a vector field is given by

$div \vec{F}=\bigtriangledown \cdot \vec{F}$ , where $\bigtriangledown = \left \langle \frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z} \right \rangle$

The partial derivatives are

f_x=0 , f_y=0 , f_z=0

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Example Question #1 : Curl

Calculate the curl for the following vector field.

$\vec{F}=x^3y^2\ \vec{i}+x^2y^3z^4\ \vec{j}+x^2z^2\ \vec{k}$

Possible Answers:

$curl\ \vec{F}=\Big(-4x^2y^3z^3\Big)\vec{i}+\Big(-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}$

$curl\ \vec{F}=\Big(4x^2y^3z^3\Big)\vec{i}+\Big(2xz^2\Big)\vec{j}+\Big(-2xy^3z^4+2x^3y\Big)\vec{k}$

$curl\ \vec{F}=0$

$curl\ \vec{F}=\Big(2x^2z^3\Big)\vec{i}+\Big(2yz^2\Big)\vec{j}+\Big(-2x^3y\Big)\vec{k}$

Correct answer:

$curl\ \vec{F}=\Big(-4x^2y^3z^3\Big)\vec{i}+\Big(-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}$

Explanation:

In order to calculate the curl, we need to recall the formula.

$\\curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{vmatrix}\\ \\ \\ =\Big(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\ \Big)\vec{i}+\Big(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\ \Big)\vec{j}+\Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\ \Big)\vec{k}$

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 out situation.

$curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\\\ x^3y^2 & x^2y^3z^4 & x^2z^2 \end{vmatrix}$

$\\=\Big(\frac{\partial }{\partial y}\Big(x^2z^2\Big)-\frac{\partial}{\partial z}\Big(x^2y^3z^4\Big)\ \Big)\vec{i}+\Big(\frac{\partial}{\partial z}\Big(x^3y^2\Big)-\frac{\partial}{\partial x}\Big(x^2z^2\Big)\ \Big)\vec{j}+\Big(\frac{\partial}{\partial x}\Big(x^2y^3z^4\Big)-\frac{\partial}{\partial y}\Big(x^3y^2\Big)\ \Big)\vec{k}$

$=\Big(0-4x^2y^3z^3\Big)\vec{i}+\Big(0-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}$

Thus the curl is

$=\Big(-4x^2y^3z^3\Big)\vec{i}+\Big(-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}$

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Example Question #2 : Curl

Calculate the curl for the following vector field.

$\vec{F}=xz\ \vec{i}+x^2yz\ \vec{j}+y^2z^2\ \vec{k}$

Possible Answers:

$curl \ \vec{F}=\Big(-2yz^2+x^2y\Big)\vec{i}+\Big(-x\Big)\vec{j}+\Big(-2xyz\Big)\vec{k}$

$curl \ \vec{F}=\Big(2yz^2\Big)\vec{i}+\Big(z\Big)\vec{j}+\Big(2xz\Big)\vec{k}$

$curl \ \vec{F}=0$

$curl \ \vec{F}=\Big(-x^2y\Big)\vec{i}+\Big(-x\Big)\vec{j}+\Big(xyz\Big)\vec{k}$

$curl \ \vec{F}=\Big(2yz^2-x^2y\Big)\vec{i}+\Big(x\Big)\vec{j}+\Big(2xyz\Big)\vec{k}$

Correct answer:

$curl \ \vec{F}=\Big(2yz^2-x^2y\Big)\vec{i}+\Big(x\Big)\vec{j}+\Big(2xyz\Big)\vec{k}$

Explanation:

In order to calculate the curl, we need to recall the formula.

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 out situation.

$\\=\Big(\frac{\partial }{\partial y}\Big(y^2z^2\Big)-\frac{\partial}{\partial z}\Big(x^2yz\Big)\ \Big)\vec{i}+\Big(\frac{\partial}{\partial z}\Big(xz\Big)-\frac{\partial}{\partial x}\Big(y^2z^2\Big)\ \Big)\vec{j}+\Big(\frac{\partial}{\partial x}\Big(x^2yz\Big)-\frac{\partial}{\partial y}\Big(xz\Big)\ \Big)\vec{k}$

$=\Big(2yz^2-x^2y\Big)\vec{i}+\Big(x-0\Big)\vec{j}+\Big(2xyz-0\Big)\vec{k}$

Thus the curl is

$=\Big(2yz^2-x^2y\Big)\vec{i}+\Big(x\Big)\vec{j}+\Big(2xyz\Big)\vec{k}$

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Example Question #3 : Curl

Find the curl of the force field $\mathbf{F}(x,y,z) = \mathbf{i} +xy\mathbf{j} +z^2\mathbf{k}$

Possible Answers:

None of the other answers

All of the other answers

$\mathbf{i} +xy\mathbf{j} +z^2\mathbf{k}$

1+xy+z^2

$y\mathbf{k}$

Correct answer:

$y\mathbf{k}$

Explanation:

Curl is probably best remembered by the determinant formula

$\bigtriangledown \times \mathbf{F}(x,y,z)$ , which is used here as follows.

$\bigtriangledown \times \mathbf{F}(x,y,z) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} &\frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 1& xy & z^2 \end{vmatrix}$

$=\begin{vmatrix} \frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ xy & z^2 \end{vmatrix}\mathbf{i}- \begin{vmatrix} \frac{\partial}{\partial x} &\frac{\partial}{\partial z} \\ 1 & z^2 \end{vmatrix} \mathbf{j}+ \begin{vmatrix} \frac{\partial}{\partial x} &\frac{\partial}{\partial y} \\ 1 & xy \end{vmatrix}\mathbf{k}$

$=(\frac{\partial}{\partial y}z^2 -\frac{\partial}{\partial z}(xy))\mathbf{i} -(\frac{\partial}{\partial x}z^2 - \frac{\partial}{\partial z}(1))\mathbf{j} + (\frac{\partial}{\partial x}(xy)-\frac{\partial}{\partial y}(1))\mathbf{k}$ $=0\mathbf{i} - 0\mathbf{j}+ (y-0)\mathbf{k}$

$=y\mathbf{k}$

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Example Question #1 : Curl

Let be any arbitrary real valued vector field. Find the div(curl F)

Possible Answers:

$\pm\frac{1}{2}$

Correct answer:

Explanation:

Take any field, the curl gives us the amount of rotation in the vector field. The purpose of the divergence is to tell us how much the vectors move in a linear motion.

When vectors are moving in circular motion only, there are no possible linear motion. Thus the divergence of the curl of any arbitrary vector field is zero.

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Example Question #1 : Curl

Evaluate the curl of the force field $F = (x^3+1)\mathbf{i}+xy\mathbf{j}+\ln(z)\mathbf{k}$ .

Possible Answers:

$\mathbf{0}$

$z\mathbf{k}$

$y\mathbf{i}$

$y\mathbf{k}$

$x\mathbf{i}$

Correct answer:

$y\mathbf{k}$

Explanation:

To evaluate the curl of a force field, we use Curl $(F)=\bigtriangledown \times F.$

$\bigtriangledown \times F = \begin{vmatrix} \mathbf{i}&\mathbf{j} &\mathbf{k} \\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ (x^3 +1)& xy & \ln(z) \end{vmatrix}$ . Start

$=(\frac{\partial}{\partial y} \ln(z) - \frac{\partial}{\partial z}(xy))\mathbf{i} -(\frac{\partial}{\partial x}\ln(z)-\frac{\partial}{\partial z}(x^3+1))\mathbf{j} +(\frac{\partial}{\partial x}(xy)-\frac{\partial}{\partial y}(x^3+1))\mathbf{k}$ Evaluate along the first row using cofactor expansion.

$= y\mathbf{k}$ . Evaluate partial derivatives. All terms except the 2nd to last one are .

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Calculus 3 : Line Integrals

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #65 : Divergence

Example Question #66 : Divergence

Example Question #67 : Divergence

Example Question #71 : Line Integrals

Example Question #69 : Divergence

Example Question #1 : Curl

Example Question #2 : Curl

Example Question #3 : Curl

Example Question #1 : Curl

Example Question #1 : Curl

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