
Surface Area of Surface
http://img408.imageshack.us/img408/4...10examhelp.jpg
I got part a) b) c) however I am just unsure on part d)
For a) I got $\displaystyle r(\theta, \phi) = \sin(\phi)\cos(\theta)i + \sin(\phi)\sin(\theta)j+\cos(\phi)k$
b) $\displaystyle r(x,y) = xi+yj+zk$ but$\displaystyle z^2+y^2+x^2 = 1$ so $\displaystyle z = \sqrt{1(x^2+y^2)}$
So $\displaystyle r(x,y) = xi+yj+\sqrt{1(x^2+y^2)}k$
c) Not required for d)
d) So area of surface $\displaystyle = \int \int_D r_x \times r_y dA$
$\displaystyle r_x \times r_y = \frac{1}{\sqrt{1x^2y^2}}$ (I've double checked this, it's correct :))
So $\displaystyle \int \int_D \frac{1}{\sqrt{1x^2y^2}} dA$
But what are the bounds for the double integral?
Thanks!

You are told that $\displaystyle \pi/6\le \phi\le \pi/3$.
The bounds for the entire sphere would be $\displaystyle 0\le \theta\le 2\pi$ and $\displaystyle 0\le \phi\le \pi$.
With the restriction given, $\displaystyle 0\le \theta\le 2\pi$ and $\displaystyle \pi/6\le \phi\le \pi/3$, of course.