Triple Integration of Surface - Multivariable Calculus

Card 0 of 24

Calculate the curl for the following vector field.

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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$^2\Big)\ \Big)\vec{k}$

Thus the curl is

Compute $div\vec{F}$ , where .

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All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

Calculate the curl for the following vector field.

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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$^2\Big)\ \Big)\vec{k}$

Thus the curl is

Compute $div\vec{F}$ , where .

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All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

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We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ $r^2$\sin(\theta) \ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $zr^2$\sin(\theta) \Big|_${0}^{r\cos(\theta)+1}$ \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $(r\cos(\theta)+1)r^2$\sin(\theta)-0 \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $r^3$$\sin(\theta)\cos(\theta)+r^2$\sin(\theta) \ dr \ d\theta$

$=\int_${0}^{2\pi}$ $$\frac{1}{4}$r^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$r^3$\sin(\theta) \Big|_${1}^{3}$\ d\theta$

$=\int_${0}^{2\pi}$ $($\frac{1}{4}$(3)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(3)^3$$\sin(\theta))-($\frac{1}{4}$(1)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(1)^3$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ ($\frac{81}{4}$\sin(\theta)\cos(\theta)+9\sin(\theta))-($\frac{1}{4}$\sin(\theta)\cos(\theta)+\frac{1}{3}$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ 20\sin(\theta)\cos(\theta)+\frac{26}{3}$\sin(\theta) \ d\theta$

$= $-$\frac{1}{2}$\cdot20\cos^2$(\theta)-$\frac{26}{3}$\cos(\theta)\Big|_${0}^{2\pi}$$

$= $-$\frac{1}{2}$\cdot20\cos^2$$(2\pi)-$\frac{26}{3}$\cos(2\pi)-(-$\frac{1}{2}$\cdot20\cos^2$(0)-$\frac{26}{3}$\cos(0))$

$= -$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1)-(-$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1))$

$= -10-$\frac{26}{3}$-(-10-$\frac{26}{3}$)=0$

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

Tap to see back →

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ $r^2$\sin(\theta) \ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $zr^2$\sin(\theta) \Big|_${0}^{r\cos(\theta)+1}$ \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $(r\cos(\theta)+1)r^2$\sin(\theta)-0 \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $r^3$$\sin(\theta)\cos(\theta)+r^2$\sin(\theta) \ dr \ d\theta$

$=\int_${0}^{2\pi}$ $$\frac{1}{4}$r^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$r^3$\sin(\theta) \Big|_${1}^{3}$\ d\theta$

$=\int_${0}^{2\pi}$ $($\frac{1}{4}$(3)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(3)^3$$\sin(\theta))-($\frac{1}{4}$(1)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(1)^3$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ ($\frac{81}{4}$\sin(\theta)\cos(\theta)+9\sin(\theta))-($\frac{1}{4}$\sin(\theta)\cos(\theta)+\frac{1}{3}$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ 20\sin(\theta)\cos(\theta)+\frac{26}{3}$\sin(\theta) \ d\theta$

$= $-$\frac{1}{2}$\cdot20\cos^2$(\theta)-$\frac{26}{3}$\cos(\theta)\Big|_${0}^{2\pi}$$

$= $-$\frac{1}{2}$\cdot20\cos^2$$(2\pi)-$\frac{26}{3}$\cos(2\pi)-(-$\frac{1}{2}$\cdot20\cos^2$(0)-$\frac{26}{3}$\cos(0))$

$= -$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1)-(-$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1))$

$= -10-$\frac{26}{3}$-(-10-$\frac{26}{3}$)=0$

Calculate the curl for the following vector field.

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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$^2\Big)\ \Big)\vec{k}$

Thus the curl is

Compute $div\vec{F}$ , where .

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All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

Calculate the curl for the following vector field.

Tap to see back →

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$^2\Big)\ \Big)\vec{k}$

Thus the curl is

Compute $div\vec{F}$ , where .

Tap to see back →

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

Tap to see back →

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ $r^2$\sin(\theta) \ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $zr^2$\sin(\theta) \Big|_${0}^{r\cos(\theta)+1}$ \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $(r\cos(\theta)+1)r^2$\sin(\theta)-0 \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $r^3$$\sin(\theta)\cos(\theta)+r^2$\sin(\theta) \ dr \ d\theta$

$=\int_${0}^{2\pi}$ $$\frac{1}{4}$r^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$r^3$\sin(\theta) \Big|_${1}^{3}$\ d\theta$

$=\int_${0}^{2\pi}$ $($\frac{1}{4}$(3)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(3)^3$$\sin(\theta))-($\frac{1}{4}$(1)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(1)^3$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ ($\frac{81}{4}$\sin(\theta)\cos(\theta)+9\sin(\theta))-($\frac{1}{4}$\sin(\theta)\cos(\theta)+\frac{1}{3}$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ 20\sin(\theta)\cos(\theta)+\frac{26}{3}$\sin(\theta) \ d\theta$

$= $-$\frac{1}{2}$\cdot20\cos^2$(\theta)-$\frac{26}{3}$\cos(\theta)\Big|_${0}^{2\pi}$$

$= $-$\frac{1}{2}$\cdot20\cos^2$$(2\pi)-$\frac{26}{3}$\cos(2\pi)-(-$\frac{1}{2}$\cdot20\cos^2$(0)-$\frac{26}{3}$\cos(0))$

$= -$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1)-(-$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1))$

$= -10-$\frac{26}{3}$-(-10-$\frac{26}{3}$)=0$

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

Tap to see back →

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ $r^2$\sin(\theta) \ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $zr^2$\sin(\theta) \Big|_${0}^{r\cos(\theta)+1}$ \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $(r\cos(\theta)+1)r^2$\sin(\theta)-0 \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $r^3$$\sin(\theta)\cos(\theta)+r^2$\sin(\theta) \ dr \ d\theta$

$=\int_${0}^{2\pi}$ $$\frac{1}{4}$r^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$r^3$\sin(\theta) \Big|_${1}^{3}$\ d\theta$

$=\int_${0}^{2\pi}$ $($\frac{1}{4}$(3)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(3)^3$$\sin(\theta))-($\frac{1}{4}$(1)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(1)^3$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ ($\frac{81}{4}$\sin(\theta)\cos(\theta)+9\sin(\theta))-($\frac{1}{4}$\sin(\theta)\cos(\theta)+\frac{1}{3}$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ 20\sin(\theta)\cos(\theta)+\frac{26}{3}$\sin(\theta) \ d\theta$

$= $-$\frac{1}{2}$\cdot20\cos^2$(\theta)-$\frac{26}{3}$\cos(\theta)\Big|_${0}^{2\pi}$$

$= $-$\frac{1}{2}$\cdot20\cos^2$$(2\pi)-$\frac{26}{3}$\cos(2\pi)-(-$\frac{1}{2}$\cdot20\cos^2$(0)-$\frac{26}{3}$\cos(0))$

$= -$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1)-(-$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1))$

$= -10-$\frac{26}{3}$-(-10-$\frac{26}{3}$)=0$

Calculate the curl for the following vector field.

Tap to see back →

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$^2\Big)\ \Big)\vec{k}$

Thus the curl is

Compute $div\vec{F}$ , where .

Tap to see back →

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

Calculate the curl for the following vector field.

Tap to see back →

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$^2\Big)\ \Big)\vec{k}$

Thus the curl is

Compute $div\vec{F}$ , where .

Tap to see back →

All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

Tap to see back →

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ $r^2$\sin(\theta) \ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $zr^2$\sin(\theta) \Big|_${0}^{r\cos(\theta)+1}$ \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $(r\cos(\theta)+1)r^2$\sin(\theta)-0 \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $r^3$$\sin(\theta)\cos(\theta)+r^2$\sin(\theta) \ dr \ d\theta$

$=\int_${0}^{2\pi}$ $$\frac{1}{4}$r^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$r^3$\sin(\theta) \Big|_${1}^{3}$\ d\theta$

$=\int_${0}^{2\pi}$ $($\frac{1}{4}$(3)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(3)^3$$\sin(\theta))-($\frac{1}{4}$(1)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(1)^3$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ ($\frac{81}{4}$\sin(\theta)\cos(\theta)+9\sin(\theta))-($\frac{1}{4}$\sin(\theta)\cos(\theta)+\frac{1}{3}$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ 20\sin(\theta)\cos(\theta)+\frac{26}{3}$\sin(\theta) \ d\theta$

$= $-$\frac{1}{2}$\cdot20\cos^2$(\theta)-$\frac{26}{3}$\cos(\theta)\Big|_${0}^{2\pi}$$

$= $-$\frac{1}{2}$\cdot20\cos^2$$(2\pi)-$\frac{26}{3}$\cos(2\pi)-(-$\frac{1}{2}$\cdot20\cos^2$(0)-$\frac{26}{3}$\cos(0))$

$= -$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1)-(-$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1))$

$= -10-$\frac{26}{3}$-(-10-$\frac{26}{3}$)=0$

Evaluate $\int \int \int_D y \ dV$ , where is the region below the plane , above the plane and between the cylinders , and .

Tap to see back →

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the plane, this means we are above .

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

The region is between two circles , and .

This means that

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ \int_${0}^{r\cos(\theta)+1}$ $r^2$\sin(\theta) \ dz \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $zr^2$\sin(\theta) \Big|_${0}^{r\cos(\theta)+1}$ \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $(r\cos(\theta)+1)r^2$\sin(\theta)-0 \ dr \ d\theta$

$=\int_${0}^{2\pi}$ \int_${1}^{3}$ $r^3$$\sin(\theta)\cos(\theta)+r^2$\sin(\theta) \ dr \ d\theta$

$=\int_${0}^{2\pi}$ $$\frac{1}{4}$r^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$r^3$\sin(\theta) \Big|_${1}^{3}$\ d\theta$

$=\int_${0}^{2\pi}$ $($\frac{1}{4}$(3)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(3)^3$$\sin(\theta))-($\frac{1}{4}$(1)^4$$\sin(\theta)\cos(\theta)+\frac{1}{3}$(1)^3$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ ($\frac{81}{4}$\sin(\theta)\cos(\theta)+9\sin(\theta))-($\frac{1}{4}$\sin(\theta)\cos(\theta)+\frac{1}{3}$\sin(\theta)) \ d\theta$

$=\int_${0}^{2\pi}$ 20\sin(\theta)\cos(\theta)+\frac{26}{3}$\sin(\theta) \ d\theta$

$= $-$\frac{1}{2}$\cdot20\cos^2$(\theta)-$\frac{26}{3}$\cos(\theta)\Big|_${0}^{2\pi}$$

$= $-$\frac{1}{2}$\cdot20\cos^2$$(2\pi)-$\frac{26}{3}$\cos(2\pi)-(-$\frac{1}{2}$\cdot20\cos^2$(0)-$\frac{26}{3}$\cos(0))$

$= -$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1)-(-$\frac{1}{2}$\cdot20(1)-$\frac{26}{3}$(1))$

$= -10-$\frac{26}{3}$-(-10-$\frac{26}{3}$)=0$

Calculate the curl for the following vector field.

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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$^2\Big)\ \Big)\vec{k}$

Thus the curl is

Compute $div\vec{F}$ , where .

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All we need to do is calculate the partial derivatives and add them together.

$div\vec{F}=\frac{\partial}{\partial $x}$(10x^2$\ln(yz))+\frac{\partial}{\partial y}$(xyz)+\frac{\partial}{\partial z}$(xz\ln(y))$

$div\vec{F}=20x\ln(yz)+xz+x\ln(y)$