Hello I'm trying to calculate the following without using Divergence theorem.
Given the vector field $F(x,y,z)=(xy+2xz,x^2+y^2,xy-z)$ and the surface of the cylinder $x^2+y^2=4$ bound by the planes $z=y-2$ and $z=0$ calculate the outward flux.
So I tried to set up the surface integral by parameterizing the surface into two pieces $f(x,y,z)=(\sqrt{4-y^2},y,z)$ and $g(x,y,z)=(-\sqrt{4-y^2},y,z)$ and the bounds of integration y-2<z<0 and -2<y<2. I calculate the normal of the surface to be $(\pm 1, \frac{y}{\sqrt{4-y^2}},0)$ so then I calculate the unit normal to be $(\pm\frac{\sqrt{4-y^2}}{2},\frac{y}{2},0)$ and set up the integral and get $\int^2_{-2}\int^0_{y-2}4ydzdy$. But the correct answer is $-12\pi$. So my question is what should the correct integral be, and where am I going wrong?
Best Answer
First of all, it is always good practice to choose a parametrization that represents the whole surface.
Secondly, $-12 \pi$ is incorrect answer unless there are typo in the question.
Thirdly, there must be mistakes how you arrived at the final integral. It does not simplify to what you wrote. Also note that if you do the dot product with unit normal vector, you will have to use a factor to map surface area element to area of the projection in yz plane. Here is the correct integral using your approach,
The unit normal vector is $ \hat n = \left(\cfrac{x}{2}, \cfrac{y}{2}, 0 \right)$
So, $\vec F \cdot \hat n = x^2 (y + z) + \cfrac{y^3}{2}$
Now using $x = \sqrt{4-y^2}$ and $ds = \sqrt{1 + (x')^2} \ dy \ dz$, the integral is,
$\displaystyle 2 \int_{-2}^2 \int_{y-2}^0 \left((4-y^2) (y + z) + \frac{y^3}{2} \right) \frac{2}{\sqrt{4-y^2}} \ dz \ dy$
$ = - 40 \pi$
In fact, you can confirm this result using divergence theorem.
Here is what would be a better parametrization given it is a cylinder,
$r(\theta, z) = (2\cos\theta, 2 \sin \theta, z), 0 \leq \theta \leq 2\pi, 2 \sin\theta - 2 \leq z \leq 0$
$r_{\theta} \times r_z = (2 \cos\theta, 2\sin\theta, 0)$
$ \vec F(x,y,z)=(xy+2xz,x^2+y^2,xy-z)$
So, $\vec F (r(\theta, z)) = (2 \sin2\theta + 4z \cos\theta, 4, 2 \sin2\theta - z)$
$\vec F \cdot (r_{\theta} \times r_z) = 4 \sin2\theta \cos\theta + 8 z \cos^2\theta + 8 \sin\theta$
So integral is,
$\displaystyle \int_0^{2\pi} \int_{2 \sin\theta - 2}^0 (4 \sin2\theta \cos\theta + 8 z \cos^2\theta + 8 \sin\theta) \ dz \ d\theta = - 40 \pi$
After you integrate wrt $z$ and while integrating wrt $\theta$, you can avoid integrating functions whose integral is zero over $(0, 2\pi)$. That should simplify working to some extent.