Triple Integration of Surface - Multivariable Calculus

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Question

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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Answer

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}$)=0$

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Question

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

Tap to reveal answer

Answer

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}$)=0$

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Question

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

Tap to reveal answer

Answer

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}$)=0$

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Question

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

Tap to reveal answer

Answer

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}$)=0$

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Question

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

Tap to reveal answer

Answer

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}$)=0$

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Question

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

Tap to reveal answer

Answer

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}$)=0$

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Question

Calculate the curl for the following vector field.

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

Thus the curl is

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Question

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

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Answer

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}$(xz\ln(y))$

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

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Question

Calculate the curl for the following vector field.

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

Thus the curl is

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Question

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

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Answer

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}$(xz\ln(y))$

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

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Question

Calculate the curl for the following vector field.

Tap to reveal answer

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

Thus the curl is

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Question

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

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Answer

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}$(xz\ln(y))$

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

← Didn't Know|Knew It →

Question

Calculate the curl for the following vector field.

Tap to reveal answer

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

Thus the curl is

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Question

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

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Answer

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}$(xz\ln(y))$

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

← Didn't Know|Knew It →

Question

Calculate the curl for the following vector field.

Tap to reveal answer

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

Thus the curl is

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Question

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

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Answer

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}$(xz\ln(y))$

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

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Question

Calculate the curl for the following vector field.

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

Thus the curl is

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Question

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

Tap to reveal answer

Answer

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}$(xz\ln(y))$

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

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Question

Calculate the curl for the following vector field.

Tap to reveal answer

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

Thus the curl is

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Question

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

Tap to reveal answer

Answer

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}$(xz\ln(y))$

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

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