Flux through oddly shaped cylinder

[Painguy]

Homework Statement

Homework Equations

The Attempt at a Solution

Here are my attempt.
The top is when i tried using cylindrical coordinates. Instead of using dA=(cosθ i +sinθj)rdzdθ
I used (cosθ i +sinθk)rdydθ since the cylinder is going along the y-axis rather than z.
The bottom is when i tried using dA=(-fx i - fy +k)dx dy

The oddly shaped cylinder is what's throwing me off.

 

[pasmith]

I would suggest setting
$$x = \frac 12 u \cos v \\ z = u \sin v$$
for $u \geq 0$, $v \in [0, 2\pi]$ so that $4x^2 + z^2 = u^2$.

 

[HallsofIvy]

You don't seem to have a firm grasp on "surface integrals". For one thing, there is no "r" here. If, as you appear to be doing, you want to use polar (it's certainly what I would do) then to get "$4x^2+ z^2= 1$", you will need to use $x= (1/2) cos(\theta)$, $z= sin(\theta)$ (with $\theta$ from $-\pi/2$ to [itex]\pi/2[/tex] to keep z positive). That way, when x is squared, the "1/2" squared will cancel the "4". Of course, we can just take y= t, say, from -1 to 1 (or just keep "y" as the parameter).

With $x= (1/2)cos(\theta)$, $y= t$, $z= sin(\theta)$ then we can write a "position vector" for any point on the surface as $\vec{r}= (1/2)cos(\theta)\vec{i}+ t\vec{j}+ sin(\theta)\vec{k}$ in terms of the two parameters t and $\theta$.

The derivatives, with respect to t and $\theta$
$$\vec{r}_t= \vec{j}$$
$$\vec{r}_\theta= -(1/2)sin(\theta)\vec{i}+ cos(\theta)\vec{k}$$
are tangent to the surface.

The cross product of those vectors, $cos(\theta)\vec{i}- (1/2)sin(\theta)\vec{k}$, is perpendicular to the surface and gives the "vector differential of surface area", $d\vec{A}= (cos(\theta)\vec{i}- (1/2)sin(\theta)\vec{k})d\theta dt$

The integrand vector is $2xy^2z\vec{i}+ 2cos(z)\vec{j}+ 3\vec{k}$ which, in terms of t and $\theta$ is $cos(\theta)sin(\theta)t^2\vec{i}+ 2cos(sin(\theta))\vec{j} + 3\vec{k}$ so that the integral becomes
$$\int_{t= -1}^1\int_{\theta= -\pi/2}^{\pi/2} \left(cos^2(\theta)sin(\theta)t^2- \frac{3}{2} sin(\theta)\right)d\theta dt$$
$$= \left(\int_{-1}^1 t^2dt\right)\left(i\int_{-\pi/2}^{\pi/2} cos^2(\theta)sin(\theta)d\theta\right)- \frac{3}{2}\left(\int_{-1}^1 dt\right)\right(\int_{-\pi/2}{\pi/2} sin(\theta) d\theta\right)$$