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
<br /> x = \frac 12 u \cos v \\<br /> z = u \sin v <br />
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 \pi/2[/tex] to keep z positive). That way, when x is squared, the &quot;1/2&quot; squared will cancel the &quot;4&quot;. Of course, we can just take y= t, say, from -1 to 1 (or just keep &quot;y&quot; as the parameter).<br /> <br /> With x= (1/2)cos(\theta), y= t, z= sin(\theta) then we can write a &quot;position vector&quot; 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.<br /> <br /> The derivatives, with respect to t and \theta<br /> \vec{r}_t= \vec{j}<br /> \vec{r}_\theta= -(1/2)sin(\theta)\vec{i}+ cos(\theta)\vec{k}<br /> are tangent to the surface.<br /> <br /> The cross product of those vectors, cos(\theta)\vec{i}- (1/2)sin(\theta)\vec{k}, is perpendicular to the surface and gives the &quot;vector differential of surface area&quot;, d\vec{A}= (cos(\theta)\vec{i}- (1/2)sin(\theta)\vec{k})d\theta dt<br /> <br /> 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<br /> \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<br /> = \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)