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Calculus 3 : Line Integrals

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Example Questions

Example Question #91 : Line Integrals

Find the curl of the following vector field, in vector form:

$\vec{F}=\left \langle \cos(xy), xz, z^2\right \rangle$

Possible Answers:

$\left \langle -x, 0, z+y\sin(xy)\right \rangle$

$\left \langle 0, -x, z+x\sin(xy)\right \rangle$

$\left \langle -x, 0, z-x\sin(xy)\right \rangle$

$\left \langle -x, 0, z+x\sin(xy)\right \rangle$

Correct answer:

$\left \langle -x, 0, z+x\sin(xy)\right \rangle$

Explanation:

The curl of the vector field is given by:

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ \cos(xy)&xz &z^2 \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

Next, we take the cross product. One can do this by multiplying across from the top left to the lower right, and continuing downward, and then subtracting the terms multiplied from top right to the bottom left. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

$\frac{\partial }{\partial y}z^2\hat{i}+\frac{\partial }{\partial x}xz\hat{k}+\frac{\partial }{\partial z}\cos(xy)\hat{j}-(\frac{\partial }{\partial y}\cos(xy)\hat{k}+\frac{\partial }{\partial z}xz\hat{i}+\frac{\partial }{\partial x}z^2\hat{j})=0=z\hat{k}+0+x\sin(xy)\hat{k}-x\hat{i}+0=\left \langle -x, 0, z+x\sin(xy) \right \rangle$

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Example Question #17 : Curl

Find the curl of the following vector field:

$\vec{F}=\left \langle xy, yz, \cos(z)\right \rangle$

Possible Answers:

$\left \langle y, 0, -x\right \rangle$

$\vec{0}$

$\left \langle y, 0, x\right \rangle$

$\left \langle -y, 0, -x\right \rangle$

Correct answer:

$\left \langle -y, 0, -x\right \rangle$

Explanation:

The curl of the vector field is given by

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ xy& yz& \cos(z) \end{vmatrix}$

$\frac{\partial }{\partial y}\cos(z)\hat{i}+\frac{\partial }{\partial x}yz\hat{k}+\frac{\partial }{\partial z}xy\hat{j}-(\frac{\partial }{\partial y}xy\hat{k}+\frac{\partial }{\partial z}yz\hat{i}+\frac{\partial }{\partial x}\cos(z)\hat{j})=-x\hat{k}-y\hat{i}=\left \langle -y, 0, -x\right \rangle$

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Example Question #1741 : Calculus 3

Determine the curl of the following vector field:

$\vec{F}=6x\hat{i}+(e^z+y)\hat{j}+z\hat{k}$

Possible Answers:

$-e^z\hat{i}$

$-(e^z+y)\hat{i}$

$e^z\hat{i}$

$\vec{0}$

Correct answer:

$-e^z\hat{i}$

Explanation:

The curl of the vector field is given by

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ 6x&e^z+y & z \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

$\frac{\partial }{\partial y}z\hat{i}+\frac{\partial }{\partial x}(e^z+j)\hat{k}+\frac{\partial }{\partial z}6x\hat{j}-(\frac{\partial }{\partial y}6x\hat{k}+\frac{\partial }{\partial z}(e^z+j)\hat{i}+\frac{\partial }{\partial x}z\hat{j})=-e^z\hat{i}$

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Example Question #1742 : Calculus 3

Determine the curl of the following vector field:

$\vec{F}=\ln(y)\hat{i}+xz^2\hat{j}+y\hat{k}$

Possible Answers:

$(1-2xz)\hat{i}+(z^2-\frac{1}{y})\hat{k}$

$(1+2xz)\hat{i}+(z^2+\frac{1}{y})\hat{k}$

$(1-xz^2)\hat{i}+(z^2-\frac{1}{y})\hat{k}$

$(1-2xz)\hat{i}+(z^2-\ln(y))\hat{k}$

Correct answer:

$(1-2xz)\hat{i}+(z^2-\frac{1}{y})\hat{k}$

Explanation:

The curl of the vector field is given by

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ \ln(y)&xz^2 & y \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

$\frac{\partial }{\partial y}y\hat{i}+\frac{\partial }{\partial x}xz^2\hat{k}+\frac{\partial }{\partial z}\ln(y)\hat{j}-(\frac{\partial }{\partial y}\ln(y)\hat{k}+\frac{\partial }{\partial z}xz^2\hat{i}+\frac{\partial }{\partial x}y\hat{j})=(1-2xz)\hat{i}+(z^2-\frac{1}{y})\hat{k}$

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Example Question #20 : Curl

Find the curl of the following vector field:

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

Possible Answers:

$(2x-2xy)\hat{j}$

$(2x+2xy)\hat{k}$

2x-2xy

$(2x-2xy)\hat{k}$

$\vec{0}$

Correct answer:

$(2x-2xy)\hat{k}$

Explanation:

The curl of the vector field is given by

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ xy^2&x^2 & \cos(z) \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

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

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Example Question #21 : Curl

Find the curl of the vector field:

$\vec{F}=\sec(y)\hat{i}+xy\hat{j}+y^2\hat{k}$

Possible Answers:

$\left \langle 2y, 0, y-\sec(y)\tan(y)\right \rangle$

$\left \langle y, 0, y-\sec(y)\tan(y)\right \rangle$

$\left \langle 2y, 0, y+\sec(y)\tan(y)\right \rangle$

$\left \langle 2y, 0, y-\sec(y)\tan(y)\right \rangle$

Correct answer:

$\left \langle 2y, 0, y-\sec(y)\tan(y)\right \rangle$

Explanation:

The curl of the vector field is given by

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ \sec(y)&xy &y^2 \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

$\frac{\partial }{\partial y}y^2\hat{i}+\frac{\partial }{\partial x}xy\hat{k}+\frac{\partial }{\partial z}\sec(y)\hat{j}-(\frac{\partial }{\partial y}\sec(y)\hat{k}+\frac{\partial }{\partial z}xy\hat{i}+\frac{\partial }{\partial x}y^2\hat{j})=2y\hat{i}+(y-\sec(y)\tan(y))\hat{k}=\left \langle 2y, 0, y-\sec(y)\tan(y)\right \rangle$

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Example Question #91 : Line Integrals

Find the curl of the vector field:

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

Possible Answers:

$\left \langle 0, \cos(x)-2y, 0\right \rangle$

$\left \langle 0, 0, \cos(x)-2y\right \rangle$

$\left \langle 0, 0, \cos(x)+2y\right \rangle$

$\left \langle 0, 0, -\cos(x)+2y\right \rangle$

Correct answer:

$\left \langle 0, 0, \cos(x)-2y\right \rangle$

Explanation:

The curl of the vector field is given by

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ y^2&\sin(x) &z \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

$\frac{\partial }{\partial y}z\hat{i}+\frac{\partial }{\partial x}\sin(x)\hat{k}+\frac{\partial }{\partial z}y^2\hat{j}-(\frac{\partial }{\partial y}y^2\hat{k}+\frac{\partial }{\partial z}\sin(x)\hat{i}+\frac{\partial }{\partial x}z\hat{j})=(\cos(x)-2y)\hat{k}=\left \langle 0, 0, \cos(x)-2y\right \rangle$

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Example Question #92 : Line Integrals

Determine whether the following vector field is conservative or not, and why:

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

Possible Answers:

The vector field is not conservative because the curl does not equal zero.

The vector field is conservative because the curl equals zero.

The vector field is not conservative because the curl equals zero.

The vector field is conservative because the curl does not equal zero.

Correct answer:

The vector field is not conservative because the curl does not equal zero.

Explanation:

The curl of the vector field is given by

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ x\cos(y)&xz &y^2 \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

$\frac{\partial }{\partial y}y^2\hat{i}+\frac{\partial }{\partial x}xz\hat{k}+\frac{\partial }{\partial z}x\cos(y)\hat{j}-(\frac{\partial }{\partial y}x\cos(y)\hat{k}+\frac{\partial }{\partial z}xz\hat{i}+\frac{\partial }{\partial x}y^2\hat{j})=2y\hat{i}+(z+x\sin(y))\hat{k}\neq \vec{0}$

So, the vector field is not conservative because the curl did not result in the zero vector.

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Example Question #24 : Curl

Determine whether the vector field is conservative or not, and why:

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

Possible Answers:

The vector field is conservative because the curl is zero.

The vector field is not conservative because the curl is zero.

The vector field is conservative because the curl is not zero.

The vector field is not conservative because the curl is not zero.

Correct answer:

The vector field is not conservative because the curl is not zero.

Explanation:

A vector field is conservative if its curl is equal to zero (the zero vector).

The curl of the vector field is given by

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ x^2z& xy\cos(y)& y^2 \end{vmatrix}$

$\frac{\partial }{\partial y}y^2\hat{i}+\frac{\partial }{\partial x}xy\cos(y)\hat{k}+\frac{\partial }{\partial z}x^2z\hat{j}-(\frac{\partial }{\partial y}x^2z\hat{k}+\frac{\partial }{\partial z}xy\cos(y)\hat{i}+\frac{\partial }{\partial x}y^2\hat{j})= 2y \hat{i}+y\cos(y)\hat{k}+x^2\hat{j}= \left \langle 2y, y\cos(y), x^2\right \rangle\neq \vec{0}$

The curl is not equal to the zero vector, so the vector field is not conservative.

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Example Question #22 : Curl

Determine whether the vector field is conservative or not, and why:

$\vec{F}=\left \langle x, y^4e^y, 3z^2\right \rangle$

Possible Answers:

The vector field is not conservative because the curl is zero.

The vector field is conservative because the curl is not zero.

The vector field is conservative because the curl is zero.

The vector field is not conservative because the curl is not zero.

Correct answer:

The vector field is conservative because the curl is zero.

Explanation:

A vector field is conservative if its curl is equal to zero (the zero vector).

The curl of the vector field is given by

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}&\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ x& y^4e^y\cos(y)& 3z^2 \end{vmatrix}$

$\frac{\partial }{\partial y}3z^2\hat{i}+\frac{\partial }{\partial x}y^4e^y\hat{k}+\frac{\partial }{\partial z}x\hat{j}-(\frac{\partial }{\partial y}x\hat{k}+\frac{\partial }{\partial z}y^4e^y\hat{i}+\frac{\partial }{\partial x}3z^2\hat{j})= \vec{0}$

All of the partial derivatives returned zero values for the unit vectors, so the curl indeed is equal to zero. The vector field is conservative.

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Calculus 3 : Line Integrals

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #91 : Line Integrals

Example Question #17 : Curl

Example Question #1741 : Calculus 3

Example Question #1742 : Calculus 3

Example Question #20 : Curl

Example Question #21 : Curl

Example Question #91 : Line Integrals

Example Question #92 : Line Integrals

Example Question #24 : Curl

Example Question #22 : Curl

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