I am trying to evaluate this Flux integral and I want to check if my work is correct.
I am given that
$\vec F = <x,y,5>$ and $S...\text{region enclosed by }x^2+z^2=1, y=0,x+y=2.$
So, I am imagining that the region looks like a bamboo cut diagonally.
So, there are three surfaces that I want to integrate and I would like to separate them into three parts $I_1, I_2$ and $I_3$.
$I_1$... $y=0$
The normal vector is $<0,-1,0>$, so $\vec F\cdot\vec n=-y$ but because $y=0$ the flux is $0$.
$I_2$...$x+y=2$
The normal vector is $<1,1,0>$, so $\vec F \cdot \vec n = x+y$.
To evaluate $\int \int_S x+y \space dA$ I used polar coordinates $x=r\cos(\theta), z=r\sin(\theta),y=r^2$
where I get $$\int_0^{2\pi}\int_0^1r\cos (\theta) +r^2 \space rdrd\theta = \pi/2$$
$I_3$...$x^2+z^2=1$
I use the parameterization $x=\cos(\theta), z=\sin(\theta), y=y$ so
$$\vec r (\theta,y) = <\cos(\theta), y, \sin(\theta)>$$
This leads to
$$\vec F \cdot \vec n = \cos^2(\theta) + 5\sin(\theta)$$
where I evaluate the integral
$$\int_0^{2\pi} \int_0^{2-\cos(\theta)} \cos^2(\theta)+5\sin(\theta) \space dy d\theta=2\pi$$
Summing up I get $\frac{5\pi}{2}$ but it's supposedly $4\pi$.
I really am struggling with these kind of problems and I can't seem to really see what the issue is. . .
Please help.
By the Gauss's theorem we should integrate $\mathop{div} F = 2$ over the volume of the cut bamboo. So the flux is twice the volume. But if we join together along the cut plane two copies of our bamboo, we will get a cylinder of radius $1$ and height $4$. Its volume is $4\pi$. So here's the answer.
It's pretty harder to calculate the flux straightforwardly, because the limits of integration are quite nontrivial. And I think you have a mistake in the second integral. You've got $(F,n) = x+y$, but $x+y = 2$ on the plain. And $y \neq r^2$, $y = 2 - x$. Maybe there are some other mistakes (the third integral is the most difficult).