Question

Multivariable Calc

Find the area of the part of the sphere
x2 + y2 + z2 = a2
that lies within the cylinder
x2 + y2 = ax
and above the xy-plane.

Answer 1

I know that:
\[Surface\ Area =\int\limits \int\limits \sqrt{f_x^2+f_y^2+1}\]
And I have:
\[z=\sqrt{a^2-x^2-y^2} \\ \frac{\partial z}{\partial x} = \frac{-x}{\sqrt{a^2-x^2-y^2}} \\ \frac{\partial z}{\partial y} = \frac{-y}{\sqrt{a^2-x^2-y^2}}\]

Answer 2

Plugging these into the equation, I have:
\[Surface\ Area = \int\limits\limits \int\limits\limits \sqrt{\frac{x^2 + y^2}{a^2-x^2-y^2}+1}\ dA \\ = \int\limits \int\limits \sqrt{\frac{a^2}{a^2-x^2-y^2}}\ dA\]

Answer 3

I tried converting to polar:
\[=\int\limits \int\limits \frac{a\ r}{\sqrt{a^2-r^2}} dA\]
but I can't find the limits of integration.

Answer 4

In polar, the cylinder would be:
\[r^2 = a r \cos(\theta) \\ r=acos(\theta)\]
So, shouldn't it be:
\[\int\limits_{0}^{2 \pi} \int\limits_{0}^{a \cos(\theta)} \frac{a\ r}{\sqrt{a^2-r^2}} dr\ d\theta ?\]

I've evaluated this several times and I keep getting 2(pi)a², which is apparently wrong.