Here goes try number two with help from Galactus:
Hemisphere cut by cylinder... Find the area of the Hemisphere x^2+y^2+z^2=4 cut by the cylinder x^2+y^2=2x
\(\displaystyle {
f(x,y,z) = x^2 + y^2 + z^2 = 4 \cr
\nabla f = 2x + 2y + 2z \cr
\left| {\nabla f} \right| = \sqrt {4x^2 + 4y^2 + 4z^2 } \Rightarrow 4\sqrt {x^2 + y^2 + z^2 } \Rightarrow 4\sqrt 4 \cr}\)
See above, x^2+y^2+z^2=4 so I substituted;
\(\displaystyle \int\!\!\!\int\limits_R {{{\left| {\nabla f} \right|} \over {\left| {f.p} \right|}}dA}\) ;where p is equal to the k component of the gradient vector.
\(\displaystyle r = 2\cos \theta \cr
\int\limits_0^{2\pi } {\int\limits_0^{2Cos\theta } {{{4\sqrt 4 } \over {2z}}dA \Rightarrow } } \int\limits_0^{2\pi } {\int\limits_0^{2Cos\theta } {{{2\sqrt 4 } \over {\sqrt {4 - r^2 } }}*r dr d\theta = 16( - 2 + \pi ) = 18.2655} \cr}\)
PHEW!!! MUCH SHORTER....I had to redeem myself. Hopefully I didn't make another mistake. Let me know if it looks good!
Hemisphere cut by cylinder... Find the area of the Hemisphere x^2+y^2+z^2=4 cut by the cylinder x^2+y^2=2x
\(\displaystyle {
f(x,y,z) = x^2 + y^2 + z^2 = 4 \cr
\nabla f = 2x + 2y + 2z \cr
\left| {\nabla f} \right| = \sqrt {4x^2 + 4y^2 + 4z^2 } \Rightarrow 4\sqrt {x^2 + y^2 + z^2 } \Rightarrow 4\sqrt 4 \cr}\)
See above, x^2+y^2+z^2=4 so I substituted;
\(\displaystyle \int\!\!\!\int\limits_R {{{\left| {\nabla f} \right|} \over {\left| {f.p} \right|}}dA}\) ;where p is equal to the k component of the gradient vector.
\(\displaystyle r = 2\cos \theta \cr
\int\limits_0^{2\pi } {\int\limits_0^{2Cos\theta } {{{4\sqrt 4 } \over {2z}}dA \Rightarrow } } \int\limits_0^{2\pi } {\int\limits_0^{2Cos\theta } {{{2\sqrt 4 } \over {\sqrt {4 - r^2 } }}*r dr d\theta = 16( - 2 + \pi ) = 18.2655} \cr}\)
PHEW!!! MUCH SHORTER....I had to redeem myself. Hopefully I didn't make another mistake. Let me know if it looks good!