What is the integral of this function over a cylinder? What is the $\iint_G x^2 z $ where $G$ is the closed cylinder $x^2 + y^2 = 4$, and $0 \le z \le 3$.
I parametrized the outside of the cylinder as $G(\theta,z)=(2\cos\theta,2\sin\theta,z)$ , with $0 \le \theta \le 2\pi$ and $0 \le z \le 3$ Plugging that into $x^2z$ gives $\int_0^{2\pi} \int_0^3 4\cos^2\theta z \; dz \,d\theta$, which is equal to $18\pi$.
I parametrized the top of the cylinder as $G(r,\theta)=(r\cos\theta,r\sin\theta,3)$ with $0 \le r \le 2$ and $0 \le \theta \le 2\pi$. This gives $3\int_0^{2\pi} \int_0^2 r^2\cos^2\theta \; dr \, d\theta$, which is equal to $8\pi$.
I believe the bottom of the cylinder is 0, because z is zero on the bottom of the cylinder and $0\int_0^{2\pi} \int_0^2 r^2\cos^2\theta \; dr \,d\theta$ is 0.
When I add these together, I get an answer of $26\pi$. According to my math book, the answer is $48\pi$. Where am I going wrong?
 A: In your first integral (over the outer cylindrical shell) you forgot that the integral over the surface doesn't have a mere $d \theta$ but instead an $r d \theta$, where $r = 2$.  This gives that integral:  $36 \pi$.
Likewise for your second integral you made the same mistake, thinking that the unit area included a mere $d \theta$ when instead it is $r d \theta$, where $r$ is a variable.  This gives that integral $12 \pi$.
A: You have to calculate $\iint_Gf(x,y,z)dS$ for $f(x,y,z)=x^2z$. Using your first parametrization $(x,y,z)=(2\cos\theta,2\sin\theta,z)=\vec{r}(\theta,z)$, the integral should be $\iint_Df(\vec{r}(\theta,z))\|\vec{r}_\theta\times\vec{r}_z\|dA$ where $D$ is the rectangle delimited by $0\le\theta\le2\pi$ and $0\le z\le3$.
Now $\vec{r}_\theta=(-2\sin\theta,2\cos\theta,0)$ and $\vec{r}_z=(0,0,1)$, thus $\vec{r}_\theta\times\vec{r}_z=(2\cos\theta,2\sin\theta,0)$ and finally $\|\vec{r}_\theta\times\vec{r}_z\|=2$.
Upon substitution on the integral:
$$\iint_Gx^2z=\int_0^{2\pi}\int_0^34\cos^2\theta z\underline{2}\;dzd\theta$$
Which in fact is equal to $36\pi$ (note the extra $2$ on the integral). A similar problem occur on your second integral.
For the formulas used on this answer you can check here.
