In order to calculate the curl, we need to recall the formula.

\(\displaystyle \\curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{vmatrix}\\ \\ \\ =\Big(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\ \Big)\vec{i}+\Big(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\ \Big)\vec{j}+\Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\ \Big)\vec{k}\)

where \(\displaystyle P\), \(\displaystyle Q\), and \(\displaystyle R\) correspond to the components of a given vector field: \(\displaystyle \vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}\)

Now lets apply this to out situation.

\(\displaystyle curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\\\ x^3y^2 & x^2y^3z^4 & x^2z^2 \end{vmatrix}\)

\(\displaystyle \\=\Big(\frac{\partial }{\partial y}\Big(x^2z^2\Big)-\frac{\partial}{\partial z}\Big(x^2y^3z^4\Big)\ \Big)\vec{i}+\Big(\frac{\partial}{\partial z}\Big(x^3y^2\Big)-\frac{\partial}{\partial x}\Big(x^2z^2\Big)\ \Big)\vec{j}+\Big(\frac{\partial}{\partial x}\Big(x^2y^3z^4\Big)-\frac{\partial}{\partial y}\Big(x^3y^2\Big)\ \Big)\vec{k}\)

\(\displaystyle =\Big(0-4x^2y^3z^3\Big)\vec{i}+\Big(0-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}\)

Thus the curl is

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

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

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

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

In order to calculate the curl, we need to recall the formula.

\(\displaystyle \\curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{vmatrix}\\ \\ \\ =\Big(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\ \Big)\vec{i}+\Big(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\ \Big)\vec{j}+\Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\ \Big)\vec{k}\)

where \(\displaystyle P\), \(\displaystyle Q\), and \(\displaystyle R\) correspond to the components of a given vector field: \(\displaystyle \vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}\)

Now lets apply this to out situation.

\(\displaystyle curl\ \vec{F}=\triangledown \times \vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\\\ x^3y^2 & x^2y^3z^4 & x^2z^2 \end{vmatrix}\)

\(\displaystyle \\=\Big(\frac{\partial }{\partial y}\Big(x^2z^2\Big)-\frac{\partial}{\partial z}\Big(x^2y^3z^4\Big)\ \Big)\vec{i}+\Big(\frac{\partial}{\partial z}\Big(x^3y^2\Big)-\frac{\partial}{\partial x}\Big(x^2z^2\Big)\ \Big)\vec{j}+\Big(\frac{\partial}{\partial x}\Big(x^2y^3z^4\Big)-\frac{\partial}{\partial y}\Big(x^3y^2\Big)\ \Big)\vec{k}\)

\(\displaystyle =\Big(0-4x^2y^3z^3\Big)\vec{i}+\Big(0-2xz^2\Big)\vec{j}+\Big(2xy^3z^4-2x^3y\Big)\vec{k}\)

Thus the curl is

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

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

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

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

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the \(\displaystyle xy\) plane, this means we are above \(\displaystyle z=0\).

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

The region \(\displaystyle D\) is between two circles \(\displaystyle x^2+y^2=1\), and \(\displaystyle x^2+y^2=9\).

This means that 

\(\displaystyle 0 \leq \theta \leq 2\pi\)

\(\displaystyle 1\leq r \leq 3\)

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

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

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

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

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

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

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

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

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

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

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

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

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

We need to figure out our boundaries for our integral.

We need to convert everything into cylindrical coordinates. Remeber we are above the \(\displaystyle xy\) plane, this means we are above \(\displaystyle z=0\).

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

The region \(\displaystyle D\) is between two circles \(\displaystyle x^2+y^2=1\), and \(\displaystyle x^2+y^2=9\).

This means that 

\(\displaystyle 0 \leq \theta \leq 2\pi\)

\(\displaystyle 1\leq r \leq 3\)

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

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

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

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

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

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

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

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

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

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

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

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

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