Question

Hi! [Surface integrals]Determine surface area of upper (z positive) hemisphere x^2+y^2+z^2=a^2 inside the cylinder x^2+y^2=a*x. Thanks to anyone viewing this!

Answer 1

consider switching to spherical coordinates
what does the region projected on the xy-plane look like?

Answer 2

This is the xy plane viewed from the top.

Answer 3

The black region "D" is the region of integration. I'm supposed to solve it without using divergence nor stokes theorem, so I think I can't use spherical cordinates, for the region is in R2. In the attached drawing I represent D in polar coordinates though.

Answer 4

ok admittedly I'm a bit rusty here, but I'm thinking cylindrical coordniates. the formula for surface area is
int int_D sqrt([f_x]^2+[f_y]^2+1)dA

in your case, z=f(x,y)=a^2-x^2-y^2

Answer 5

so the integral you want is
int int sqrt(4x^2+4y^2+1)dA
in cylindrical coordinates this is
int int sqrt(4r^2+1)rdrdt (t=theta)

then you just have to find the bounds of D in cylindrical coordinates. I hope I'm not too far off here.

Answer 6

thing is, I don't think you have the cylindrical symmetry. That;s why I said you may want to be using spherical coordinates with the radius held constant as a

Answer 7

I did all that, but the answer I got does not verify with the one in the book (It says I should get S=a^2 * (pi-2). Can you find any mistakes in what I've done?

Answer 8

I see, I made the mistake of forgetting the sqrt myself. I'm sorry I have to go, but if you don't have an answer when I come back later today I'll look over it again.
Sorry I couldn't help.

Answer 9

Oh, not at all, much thanks for your time!! Have a nice day!

Answer 10

For anyone struggling with this excercise, I just found my mistake. When I use Barrow's rule to evaluate the first integral { sqrt(u) from a^2 to (a*sin(theta))^2 } I didn't apply absolute value after canceling the square roots, so the sin(theta) part of the second integral cancelled out and gave 0 while it should have given 2. I attach a graph to show the difference. See you! (Corrections are shown in green and blue)