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Calculus 3 : Line Integrals

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Example Questions

Example Question #3 : Curl

Calculate the curl $\triangledown \times \mathbf{F}$ of the following vector:

$\mathbf{F}=xz\mathbf{i}+xy\mathbf{j}+yz\mathbf{k}$

Possible Answers:

$-z\mathbf{i}+x\mathbf{j}-y\mathbf{k}$

$-z\mathbf{i}-x\mathbf{j}-y\mathbf{k}$

$z\mathbf{i}+x\mathbf{j}-y\mathbf{k}$

$z\mathbf{i}+x\mathbf{j}+y\mathbf{k}$

$z\mathbf{i}-x\mathbf{j}+y\mathbf{k}$

Correct answer:

$z\mathbf{i}+x\mathbf{j}+y\mathbf{k}$

Explanation:

The curl of a vector $\mathbf{F}=U\mathbf{i}+V\mathbf{j}+W\mathbf{k}$

is defined by the determinant of the following 3x3 matrix:

$\triangledown \times \mathbf{F}=\textup{det} \begin{vmatrix} \mathbf{i}& \mathbf{j} &\mathbf{k} \\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ U& V& W \end{vmatrix}$

For the given vector, we can calculate this determinant

$\triangledown \times \mathbf{F}=(z-0)\mathbf{i} -(0-x)\mathbf{j}+(y-0)\mathbf{k}=z\mathbf{i}+x\mathbf{j}+y\mathbf{k}$

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Example Question #7 : Curl

Given that F is a vector function and f is a scalar function, which of the following operations results in a scalar?

Possible Answers:

$\nabla \cdot(\nabla \times \mathbf{F})$

$\nabla \cdot(\nabla \cdot\mathbf{F})$

$\nabla \times(\nabla f)$

$\nabla(\nabla \cdot\mathbf{F})$

$\nabla(\nabla\cdot f)$

Correct answer:

$\nabla \cdot(\nabla \times \mathbf{F})$

Explanation:

For each of the given expressions:

$\nabla(\nabla \cdot f)$ - The divergence of a scalar function does not exist, so this expression is undefined.

$\nabla(\nabla \cdot\mathbf{F})$ - The dot product of a vector function is a scalar, so the gradient of the term in parenthesis results in a vector.

$\nabla \cdot(\nabla \cdot\mathbf{F})$ - The divergence of a vector function is a scalar. Taking the divergence of the term in parenthesis would be taking the divergence of a scalar, which doesn't exist. This expression is undefined.

$\nabla \times(\nabla f)$ - The gradient of a scalar function is a vector. Thus, the curl of the term in parenthesis is also a vector.

The remaining answer is:

$\nabla \cdot(\nabla \times \mathbf{F})$ - The term in parenthesis is the curl of a vector function, which is also a vector. Taking the divergence of the term in parenthesis, we get the divergence of a vector, which is a scalar.

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Example Question #1 : Curl

Given that F is a vector function and f is a scalar function, which of the following expressions is undefined?

Possible Answers:

$\nabla \times (\nabla f)$

$\nabla\cdot(\nabla f)$

$\nabla(\nabla\cdot\mathbf{F})$

$\nabla \times (\nabla\times f)$

$\nabla\cdot(\nabla\times\mathbf{F})$

Correct answer:

$\nabla \times (\nabla\times f)$

Explanation:

The cross product of a scalar function is undefined. The expression in the parenthesis of:

$\nabla \times (\nabla\times f)$

is the cross product of a scalar function, therefore the entire expression is undefined.

For the other solutions:

$\nabla\cdot(\nabla\times\mathbf{F})$ - The cross product of a vector is also a vector, and the divergence of a vector is defined. This expression is a scalar.

$\nabla\cdot(\nabla f)$ - The gradient of a scalar is a vector, and the divergence of a vector is defined. This expression is also a scalar.

$\nabla(\nabla\cdot\mathbf{F})$ - The divergence of a vector is scalar, and the gradient of a scalar is defined. This expression is a vector.

$\nabla \times (\nabla f)$ - The gradient of a scalar is a vector, and the curl of a vector is defined. This expression is a vector.

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Example Question #1 : Curl

Compute the curl of the following vector function:

$\mathbf{F}=-y^2\mathbf{i}+3y\mathbf{j}$

Possible Answers:

$-2y\mathbf{i}+3\mathbf{k}$

$2y\mathbf{k}$

$3\mathbf{i}-2y\mathbf{k}$

$-2y\mathbf{i}$

$-3\mathbf{k}$

Correct answer:

$2y\mathbf{k}$

Explanation:

For a vector function $\mathbf{F}=U\mathbf{i}+V\mathbf{j}+W\mathbf{k}$ , the curl is given by:

$\nabla \times \mathbf{F}=\textup{det}\begin{vmatrix} \mathbf{i}&\mathbf{j} &\mathbf{k} \\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ U& V& W \end{vmatrix}$

For this function, we calculate the curl as:

$\nabla \times \mathbf{F}=(0-0)\mathbf{i}-(0-0)\mathbf{j}+(0+2y)\mathbf{k}=2y\mathbf{k}$

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Example Question #10 : Curl

Compute the curl of the following vector function:

$\mathbf{F}=(y^2+z^2)\mathbf{i}+y\mathbf{j}+z\mathbf{k}$

Possible Answers:

$2y\mathbf{i}+\mathbf{j}$

$(2y-1)\mathbf{i}-\mathbf{k}$

$(1-2z)\mathbf{i}+\mathbf{j}$

$2z\mathbf{j}-2y\mathbf{k}$

$-2z\mathbf{i}-\mathbf{k}$

Correct answer:

$2z\mathbf{j}-2y\mathbf{k}$

Explanation:

For a vector function $\mathbf{F}=U\mathbf{i}+V\mathbf{j}+W\mathbf{k}$ , the curl is given by:

For this function, we calculate the curl as:

$\nabla \times \mathbf{F} =\textup{det}\begin{vmatrix} \mathbf{i}&\mathbf{j} &\mathbf{k} \\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ (y^2+z^2)& y& z \end{vmatrix}$

$\nabla\times \mathbf{F}=(0-0)\mathbf{i}-(0-2z)\mathbf{j}+(0-2y)\mathbf{k}= 2z\mathbf{j}-2y\mathbf{k}$

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Example Question #1731 : Calculus 3

Find the curl of the vector function:

$\vec{F}=\left \langle 2xy, xyz, z^2\right \rangle$

Possible Answers:

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

$\left \langle -xy, 0, yz\right \rangle$

$\left \langle -xy, 0, yz-2x\right \rangle$

$\left \langle xy, 0, yz+2x\right \rangle$

Correct answer:

$\left \langle -xy, 0, yz-2x\right \rangle$

Explanation:

The curl of the function is given by

$\bigtriangledown X \vec{F}$

First, we must write the determinant in order to take the cross product:

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

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

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

The partial derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #1732 : Calculus 3

Find the curl of the vector function:

$\vec{F}=e^x\hat{i}+x^2y\hat{j}+x\hat{k}$

Possible Answers:

$\left \langle 0, 1, 2xy\right \rangle$

$\left \langle 0, -1, -2xy\right \rangle$

$\left \langle 0, -1, 2xy\right \rangle$

$\left \langle y, -1, 2xy\right \rangle$

Correct answer:

$\left \langle 0, -1, 2xy\right \rangle$

Explanation:

The curl of the function is given by

$\bigtriangledown X \vec{F}$

First, we must write the determinant in order to take the cross product:

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

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

$\\ \frac{\partial }{\partial y}x\hat{i}+\frac{\partial }{\partial x} x^2y\hat{k}+\frac{\partial }{\partial z}e^x\hat{j}-(\frac{\partial }{\partial y}e^x\hat{k}+\frac{\partial }{\partial z}x^2y\hat{i}+\frac{\partial }{\partial x}x\hat{j})\\ \\=0+2xy\hat{k}+0-0-0-1\hat{j}\\ \\=\left \langle 0, -1, 2xy\right \rangle$

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #1733 : Calculus 3

Find the curl of the vector function:

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The curl of a vector function $\vec{F}=\left \langle P, Q, R\right \rangle$ is given by

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

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:

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

The partial derivatives were found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Example Question #1734 : Calculus 3

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

$\left \langle xz, y^2z, z\right \rangle$

Possible Answers:

The vector field is conservative because the curl is not equal to $\vec{0}$ .

The vector field is not conservative because the curl does not equal to $\vec{0}$ .

The vector field is not conservative because the curl is equal to $\vec{0}$ .

The vector field is conservative because the curl is equal to $\vec{0}$ .

Correct answer:

The vector field is not conservative because the curl does not equal to $\vec{0}$ .

Explanation:

The curl of the function is given by the cross product of the gradient and the vector function. If a vector function is conservative if the curl equals zero.

First, we can write the determinant in order to take the cross product of the two vectors:

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

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

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

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Example Question #1735 : Calculus 3

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

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The curl of the vector field is given by:

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

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

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

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Calculus 3 : Line Integrals

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3 : Curl

Example Question #7 : Curl

Example Question #1 : Curl

Example Question #1 : Curl

Example Question #10 : Curl

Example Question #1731 : Calculus 3

$\left \langle -xy, 0, yz-2x\right \rangle$

$\left \langle -xy, 0, yz-2x\right \rangle$

Example Question #1732 : Calculus 3

Example Question #1733 : Calculus 3

Example Question #1734 : Calculus 3

Example Question #1735 : Calculus 3

All Calculus 3 Resources

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