Calculating Flux Integral for Cylindrical Portion with Outward Normal Vectors

[kthejohn]

Homework Statement

Flux integral:
S is the portion of the cylinder with these cylindrical coordinates:
0<=z<=3 r=1 0<=theta<=pi/2

orient S with outward normal vectors from z-axis and compute the flux of F = <2x,y,-3z> across S

Homework Equations

Flux integral ∫∫_SFdS = ∫∫_S F (ruxrv)dS

The Attempt at a Solution

i know that i should convert double integral but don't know how to get the partial derivs. with respect to r and theta of the portion of the cylinder but i figured out that F should be <2costheta, sintheta, -3z>

 

[HallsofIvy]

The cylinder is given by $x= cos(\theta)$, $y= sin(\theta)$, z= z. Any point on the cylinder is given by the "position vector"
$$\vec{r}(\theta, z)= cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ z\vec{k}$$
The two vectors
$$\vec{r}_\theta= -sin(\theta)\vec{i}+ cos(\theta)\vec{j}$$
$$\vec{r}_z= \vec{k}$$
are in the tangent plane at every point on the cylinder.

Their cross product,
$$cos(\theta)\vec{i}+ sin(\theta)\vec{j}$$
also called the "fundamental vector product" for the surface, gives the "vector differential of surface area:
$$d\vec{S}= \vec{n}dS= (cos(\theta)\vec{i}+ sin(\theta)\vec{j})d\theta dz$$

Integrate the dot product of the flux vector with that.