# 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!