Surface of a sphere intersecting two cylinders

[jarondl]

Homework Statement

Find the surface area of the surface $$x^2 + y^2 + z^2 = a^2$$, that is outside the two cylinders $$x^2+y^2=\pm ax$$.
$$(a>0)$$

Homework Equations

$$ds= \sqrt{1+(Z_x')^2+(Z_y')^2}dxdy$$

The Attempt at a Solution

The intersections are clearly four ellipses. I will try to find a area of one of these ellipses in order to multiply it by four and subtract that from the known area of a whole sphere.
I have tried two different attitudes to solve it. First, I used the equation mentioned above, to convert this integral to a double integral.
$$S = \iint_D \sqrt{\frac{a^2}{a^2-x^2-y^2}}dxdy$$ when D is a circle given by $$\left(x-\frac{a}{2}\right)^2 + y^2 \le \left(\frac{a}{2}\right)^2$$
Alas, I haven't succeeded in solving this one (not when staying cartesian, nor with polar co-ordinates).
The second idea was to use spheric coordinates in the first place, but then the limits for theta and phi turn out to be $$0<\phi < \frac{\pi}{2} , -a\cos\phi<\sin^2\theta < a\cos\phi$$ Which ain't nice either.
ANY help would be appreciated, especially before my exam on Sunday
Thanks,
Jaron

 

[tiny-tim]

Welcome to PF!

Hi Jaron! Welcome to PF!

Find the surface area of the surface $$x^2 + y^2 + z^2 = a^2$$, that is outside the two cylinders $$x^2+y^2=\pm ax$$.
$$(a>0)$$
…
The intersections are clearly four ellipses.

Nooo … it's not even a plane figure … an ellipse is a plane figure, and any plane intersects a sphere in a circle, but these cylinders in ellipses.

$$ds= \sqrt{1+(Z_x')^2+(Z_y')^2}dxdy$$
e it.

hmm … clever isn't always the easiest.

have you tried the dumb approach, just using horizontal slices of thickness dy?

(and didn't Archimedes have a theorem about areas on spheres and cylinders?)

 

[jarondl]

Thanks tiny Tim!

Nooo … it's not even a plane figure …

Yep.. You were right.
About Archimedes, I only find his ideas over a cylinder of the same radius as the sphere.

Anyway, thanks for you help