- #1
Saketh
- 261
- 2
Okay - I thought that I figured this stuff out, but I didn't.
The Problem
When [tex]G(x, y, z) = (1-x^2-y^2)^{3/2}[/tex], and [tex]z = \sqrt{1-x^2-y^2}[/tex], evaluate the surface integral.
My Work
I keep trying this but I end up with the following integral that I cannot evaluate:
[tex]
\int_{-1}^{1} \!\!\! \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (1-x^2-y^2)^{3/2}\sqrt{1+4x^2+4y^2} \,dx \,dy
[/tex].
Conversion to polar coordinates doesn't help much, either. How can I find this surface integral? (Ans: [itex]\frac{\pi}{2}[/itex])
Thanks!
The Problem
When [tex]G(x, y, z) = (1-x^2-y^2)^{3/2}[/tex], and [tex]z = \sqrt{1-x^2-y^2}[/tex], evaluate the surface integral.
My Work
I keep trying this but I end up with the following integral that I cannot evaluate:
[tex]
\int_{-1}^{1} \!\!\! \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (1-x^2-y^2)^{3/2}\sqrt{1+4x^2+4y^2} \,dx \,dy
[/tex].
Conversion to polar coordinates doesn't help much, either. How can I find this surface integral? (Ans: [itex]\frac{\pi}{2}[/itex])
Thanks!