1. ## Surface Integrals Implicit

I have to evaluate the given surface integrals $\displaystyle \iint_ s 3{y}^{2 }z d\sigma$ where S is the portion of the cylinder $\displaystyle {x}^{2 } +{y}^{2 } =1$ between the planes z = 0 and z = 1, and on the positive x side of the yz-plane

First of all i think i need to find n

Which is $\displaystyle \frac{2xi + 2yj}{4 }$ I think.

Now i'm not sure on my next step.

I have to evaluate the given surface integrals $\displaystyle \iint_ s 3{y}^{2 }z d\sigma$ where S is the portion of the cylinder $\displaystyle {x}^{2 } +{y}^{2 } =1$ between the planes z = 0 and z = 1, and on the positive x side of the yz-plane

First of all i think i need to find n

Which is $\displaystyle \frac{2xi + 2yj}{4 }$ I think.

Now i'm not sure on my next step.
I would parameterize the surface of the cylindar using cylindrical polar coordinates.

$\displaystyle x=1\cdot \cos(\phi) \quad y=1\cdot =sin(\phi) \quad z=z$

This gives

$\displaystyle \mathbf{r}(\phi,z)=\cos(\phi)\mathbf{i}+\sin(\phi) \mathbf{j}+z\mathbf{k}$

This gives

$\displaystyle \mathbf{r}_\phi \times \mathbf{r}_z=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -\sin(\phi) & \cos(\phi) & 0 \\ 0 & 0 & 1 \end{vmatrix}= \cos(\phi)\mathbf{i}+\sin(\phi)\mathbf{j}$

Now if we find its magnitude we get

$\displaystyle |\mathbf{r}_\phi \times \mathbf{r}_z|=\sqrt{\cos^2(\phi)+\sin^2(\phi)}=1$

So this gives the integral

$\displaystyle \iint 3y^2zd\sigma=3\int_{0}^{1} \int_{0}^{2\pi}sin^2(\phi)zd\phi dz =3\left(\int_{0}^{1}zdz \right)\left(\int_{0}^{2\pi}\sin^2(\phi)d\phi \right)$

3. Originally Posted by TheEmptySet
$\displaystyle \iint 3y^2zd\sigma=3\int_{0}^{1} \int_{0}^{2\pi}sin^2(\phi)zd\phi dz =3\left(\int_{0}^{1}zdz \right)\left(\int_{0}^{2\pi}\sin^2(\phi)d\phi \right)$
I just integrated this and got 3pi/2 but the answer is 3pi/4. Am i right or have i done poor integration.

$\displaystyle \phi \in \left[-\frac{\pi}{2}, \frac{\pi}{2} \right]$