Area of a cylinder inside a sphere (surface integral)

1. Jun 22, 2014

bawbag

1. The problem statement, all variables and given/known data
Find the area of the cylinder $x^2 + y^2 -y = 0$ inside the sphere $x^2 + y^2 +z^2 =1$

2. Relevant equations

$dA = sec \gamma dydz$ where $$sec \gamma = \frac{|\nabla \phi|}{|\partial \phi/ \partial x|}$$

3. The attempt at a solution

The method shown in this section is to calculate the angle of the area element to the plane ($\gamma$) and integrate over the area of the projection onto the plane, in this case the y-z plane. The final answer will need to be multiplied by 2 to get the total area for both sides of the plane.

In the above equation for $sec \gamma$, phi is the given expression for the cylinder. Evaluating gives $$sec \gamma = \frac {\sqrt{(2x)^2 + (2y-1)^2}}{2x}$$ $$= \frac{\sqrt{4x^2 + 4y^2 -4y +1}}{2x} = \frac{\sqrt{4(x^2 + y^2 - y) +1}}{2x} = \frac{1}{2x}$$$$= \frac{1}{2\sqrt{y-y^2}}$$
Completing the square
$= \frac{1}{2\sqrt{(y-\frac{1}{2})^2 -\frac{1}{4}}}$

So the integral becomes $\int^{1}_{z=-1} \int^{\sqrt{1-z^2}}_{y=0}\frac{1}{2\sqrt{(y-\frac{1}{2})^2 -\frac{1}{4}}}$

Now to integrate this expression I used a trig substitution $(y-\frac{1}{2}) = \frac{1}{2}sec \theta$ but this leads to a big dirty expression inside a natural logarithm that I have no idea how to integrate for the second part. I figure I'm missing something fundamental since the answer is simply 4.

Any suggestions?

P.S. I noticed a thread on a topic very similar to this that was resolved using parameterization of the cylinder, but since I haven't covered this in the textbook, and the given method makes no mention of it, I'd like to solve it without using parameterizations if possible.

2. Jun 23, 2014

bawbag

It's just occurred to me that I completed the square incorrectly. So please ignore that part :D.

Also, I feel like cylindrical coordinates might be helpful here, but when I change the integral, it seems to get even worse.

Any assistance would be appreciated!