Line Integrals - Calculus 3

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Question

Calculate the curl for the following vector field.

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

Answer

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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Question

Calculate the curl for the following vector field.

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

Answer

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.

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

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

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

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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Question

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

Answer

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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Question

Let be any arbitrary real valued vector field. Find the

Answer

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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Question

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

Answer

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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Question

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

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

Answer

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

$\triangledown \times \mathbf{F}=\textup{det} \begin{vmatrix} \mathbf{i}& \mathbf{j} &\mathbf{k} \ \frac{\partial}{\partial x}&\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \ xz& xy& yz \end{vmatrix}$

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

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Question

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

Answer

For each of the given expressions:

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

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

$abla \cdot( abla \cdot\mathbf{F})$ - 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.

$abla \times( abla f)$ - 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:

$abla \cdot( abla \times \mathbf{F})$ - 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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Question

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

Answer

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

$abla \times ( abla\times f)$

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

For the other solutions:

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

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

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

$abla \times ( abla f)$ - 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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Question

Compute the curl of the following vector function:

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

Answer

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

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

For this function, we calculate the curl as:

$abla \times \mathbf{F}=\textup{det}\begin{vmatrix} \mathbf{i}&\mathbf{j} &\mathbf{k} \ \frac{\partial}{\partial x}&\frac{\partial}{\partial y} &\frac{\partial}{\partial z} \ -y^2& 3y& 0 \end{vmatrix}$

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

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Question

Compute the curl of the following vector function:

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

Answer

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

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

For this function, we calculate the curl as:

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

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

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Question

Find the curl of the vector function:

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

Answer

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.

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:

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

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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Question

Find the curl of the vector function:

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

Answer

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.

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:

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

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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Question

Find the curl of the vector function:

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

Answer

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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Question

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

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

Answer

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.

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:

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

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Question

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

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

Answer

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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Question

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

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

Answer

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.

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.

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

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Question

Find the curl of the following vector field:

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

Answer

The curl of the vector field is given by

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

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.

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

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Question

Determine the curl of the following vector field:

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

Answer

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.

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.

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

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Question

Determine the curl of the following vector field:

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

Answer

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.

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.

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

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Question

Find the curl of the following vector field:

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

Answer

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.

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.

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

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