Line Integrals

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AP Calculus BC › Line Integrals

Questions 1 - 10

Given the vector field

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

find the divergence of the vector field:

$div \vec{F}$ .

$div \vec{F}=0$

$div \vec{F}=x+y+z$

$div \vec{F}=xy+yz+xz$

$div \vec{F}=x+y$

$div \vec{F}=x+z$

Explanation

Given a vector field

$\vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$

we find its divergence by taking the dot product with the gradient operator:

$div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$

We know that , so we have

$div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}=0+0+0=0$

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

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

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

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

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.

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

Find the curl of the vector field:

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

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

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

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

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

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.

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.

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

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

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

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

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

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.

Find the curl of the vector field:

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

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

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

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

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

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.

Find the curl of the following vector field:

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

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

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

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

$\vec{0}$

Explanation

The curl of the vector field is given by

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

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

Find the curl of the following vector field:

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

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

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

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

$\vec{0}$

Explanation

The curl of the vector field is given by

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

Given the vector field

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

find the divergence of the vector field:

$div \vec{F}$ .

$div \vec{F}=0$

$div \vec{F}=x+y+z$

$div \vec{F}=xy+yz+xz$

$div \vec{F}=x+y$

$div \vec{F}=x+z$

Explanation

Given a vector field

$\vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}$

we find its divergence by taking the dot product with the gradient operator:

$div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$

We know that , so we have

$div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}=0+0+0=0$

Find the divergence of the vector $\left \langle x^2y^2z^2,3y^5,17z\right \rangle$

Explanation

To find the divergence a vector $a=\left \langle A,B,C\right \rangle$ , you use the following definition: $div(a)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ . Applying this to the vector from the problem statement, we get $\frac{\partial A}{\partial x}=2xy^2z^2, \frac{\partial B}{\partial y}=15y^4,\frac{\partial C}{\partial z}=17$ . Adding all of these up, according to the definition, will produce the correct answer.

Find the divergence of the vector $\left \langle x^2y^2z^2,3y^5,17z\right \rangle$

Explanation

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