Triple Integration of Surface - Multivariable Calculus

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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

$=\Big(-4x^2y^3z^3\Big)\vec{i}+\Big(-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}$

Compare your answer with the correct one above

Question

Compute $div\vec{F}$ , where $\vec{F}=10x^2\ln(yz)\vec{i}+xyz\vec{j}+xz\ln(y)\vec{k}$ .

Answer

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

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

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

Compare your answer with the correct one above

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

$=\Big(-4x^2y^3z^3\Big)\vec{i}+\Big(-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}$

Compare your answer with the correct one above

Question

Compute $div\vec{F}$ , where $\vec{F}=10x^2\ln(yz)\vec{i}+xyz\vec{j}+xz\ln(y)\vec{k}$ .

Answer

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

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

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

Compare your answer with the correct one above

Question

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

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

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

$1\leq r \leq 3$

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

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

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

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

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

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

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

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

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

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

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

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

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

Compare your answer with the correct one above

Question

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

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

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

$1\leq r \leq 3$

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

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

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

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

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

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

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

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

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

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

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

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

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

Compare your answer with the correct one above

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

$=\Big(-4x^2y^3z^3\Big)\vec{i}+\Big(-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}$

Compare your answer with the correct one above

Question

Compute $div\vec{F}$ , where $\vec{F}=10x^2\ln(yz)\vec{i}+xyz\vec{j}+xz\ln(y)\vec{k}$ .

Answer

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

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

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

Compare your answer with the correct one above

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

$=\Big(-4x^2y^3z^3\Big)\vec{i}+\Big(-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}$

Compare your answer with the correct one above

Question

Compute $div\vec{F}$ , where $\vec{F}=10x^2\ln(yz)\vec{i}+xyz\vec{j}+xz\ln(y)\vec{k}$ .

Answer

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

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

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

Compare your answer with the correct one above

Question

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

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

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

$1\leq r \leq 3$

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

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

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

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

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

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

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

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

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

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

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

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

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

Compare your answer with the correct one above

Question

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

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

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

$1\leq r \leq 3$

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

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

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

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

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

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

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

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

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

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

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

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

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

Compare your answer with the correct one above

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

$=\Big(-4x^2y^3z^3\Big)\vec{i}+\Big(-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}$

Compare your answer with the correct one above

Question

Compute $div\vec{F}$ , where $\vec{F}=10x^2\ln(yz)\vec{i}+xyz\vec{j}+xz\ln(y)\vec{k}$ .

Answer

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

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

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

Compare your answer with the correct one above

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

$=\Big(-4x^2y^3z^3\Big)\vec{i}+\Big(-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}$

Compare your answer with the correct one above

Question

Compute $div\vec{F}$ , where $\vec{F}=10x^2\ln(yz)\vec{i}+xyz\vec{j}+xz\ln(y)\vec{k}$ .

Answer

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

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

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

Compare your answer with the correct one above

Question

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

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 .

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.

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

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

Compute $div\vec{F}$ , where $\vec{F}=10x^2\ln(yz)\vec{i}+xyz\vec{j}+xz\ln(y)\vec{k}$ .

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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