First we need to find the tangent vector, and find its magnitude.

\(\displaystyle \vec{v}(t)=\vec{r}\ '(t)=\left \langle 1, 8\cos(2t), -8\sin(2t)\right \rangle\)

\(\displaystyle \left \| \vec{v}(t)\right \|=\sqrt{1^2+(8\cos(2t))^2+(-8\sin(2t))^2}\)

\(\displaystyle \left \| \vec{v}(t)\right \|=\sqrt{1+64\cos^2(2t)+64\sin^2(2t)}\)

\(\displaystyle \left \| \vec{v}(t)\right \|=\sqrt{1+64(\cos^2(2t)+\sin^2(2t))}\)

\(\displaystyle \left \| \vec{v}(t)\right \|=\sqrt{1+64}=\sqrt{65}\)

Now we can set up our arc length integral

\(\displaystyle L=\int_{a}^{b}\left \| \vec{v}(t)\right \|dt\)

\(\displaystyle L=\int_{0}^{2\pi} \sqrt{65}\ dt=t\sqrt{65}\Big|_{0}^{2\pi}=2\pi\sqrt{65}\)

In order to convert to spherical coordinates , we need to remember the conversion equations.

\(\displaystyle \rho=\sqrt{x^2+y^2+z^2}\)

\(\displaystyle \theta=\tan^{-1}\Big(\frac{y}{x}\Big)\)

\(\displaystyle \phi=\tan^{-1}\Big(\frac{\sqrt{x^2+y^2}}{z}\Big)\)

Now lets apply this to our problem.

\(\displaystyle \rho=\sqrt{(\sin(t))^2+(\cos(t))^2+1^2}=\sqrt{1+1}=\sqrt{2}\)

\(\displaystyle \phi=\tan^{-1}\Big(\frac{\sqrt{(\sin{t})^2+(\cos{t})^2}}{1}\Big)=\tan^{-1}(1)=\frac{\pi}{4}\)

\(\displaystyle \theta=\tan^{-1}\Big(\frac{\cos(t)}{\sin(t)}\Big)=\tan^{-1}(\cot(t))\)

The first thing we have to do is figure out the general equations for the lines that create the trapezoid.

\(\displaystyle y=x\)

\(\displaystyle y=-x\)

\(\displaystyle y=-x+5\)

\(\displaystyle y=x-5\)

Now we have the general equations for out trapezoid, now we need to plug in our transformations into these equations.

\(\displaystyle y=x\)

\(\displaystyle 2u+3v=2u-3v\)

\(\displaystyle -6v=0\rightarrow v=0\)

\(\displaystyle y=-x\)

\(\displaystyle 2u+3v=-2u+3v\)

\(\displaystyle 4u=0\rightarrow u=0\)

\(\displaystyle y=-x+5\)

\(\displaystyle 2u+3v=-2u+3v+5\)

\(\displaystyle 4u=5\rightarrow u=\frac{5}{4}\)

\(\displaystyle y=x-5\)

\(\displaystyle 2u+3v=2u-3v-5\)

\(\displaystyle -6v=-5\rightarrow v=\frac{5}{6}\)

So our region is a rectangle given by \(\displaystyle 0 \leq u \leq \frac{5}{4}\), \(\displaystyle 0 \leq v \leq \frac{5}{6}\)

Next we need to calculate the Jacobian.

\(\displaystyle \begin{vmatrix} 2 & 3\\ 2 & -3 \end{vmatrix} =-6-6=-12\)

Now we can put the integral together.

\(\displaystyle \int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} (3(2u+3v)+2(2u-3v))) \cdot \left | -12\right | \ du\ dv\)

\(\displaystyle =\int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} 120u+36v \ du\ dv\)

\(\displaystyle =\int_{0}^{\frac{5}{6}} 60u^2+36uv \Big|_{0}^{\frac{5}{4}}\ dv\)

\(\displaystyle =\int_{0}^{\frac{5}{6}} 60( \frac{5}{4})^2+36(\frac{5}{4})v-0 \ dv\)

\(\displaystyle =\int_{0}^{\frac{5}{6}} \frac{375}{4}+45v \ dv\)

\(\displaystyle =\frac{375}{4}v+\frac{45}{2}v^2\Big|_{0}^{\frac{5}{6}} \ dv\)

\(\displaystyle =\frac{375}{4}\cdot\frac{5}{6}+\frac{45}{2}\cdot\Big(\frac{5}{6}\Big)^2 -0=\frac{375}{4}\)

In order to convert to spherical coordinates , we need to remember the conversion equations.

\(\displaystyle \rho=\sqrt{x^2+y^2+z^2}\)

\(\displaystyle \theta=\tan^{-1}\Big(\frac{y}{x}\Big)\)

\(\displaystyle \phi=\tan^{-1}\Big(\frac{\sqrt{x^2+y^2}}{z}\Big)\)

Now lets apply this to our problem.

\(\displaystyle \rho=\sqrt{(\sin(t))^2+(\cos(t))^2+1^2}=\sqrt{1+1}=\sqrt{2}\)

\(\displaystyle \phi=\tan^{-1}\Big(\frac{\sqrt{(\sin{t})^2+(\cos{t})^2}}{1}\Big)=\tan^{-1}(1)=\frac{\pi}{4}\)

\(\displaystyle \theta=\tan^{-1}\Big(\frac{\cos(t)}{\sin(t)}\Big)=\tan^{-1}(\cot(t))\)

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 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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