# Surface Area of Surface

• Nov 1st 2010, 01:11 AM
usagi_killer
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 $\displaystyle = \int \int_D |r_x \times r_y| dA$

$\displaystyle |r_x \times r_y| = \frac{1}{\sqrt{1-x^2-y^2}}$ (I've double checked this, it's correct :))

So $\displaystyle \int \int_D \frac{1}{\sqrt{1-x^2-y^2}} dA$

But what are the bounds for the double integral?

Thanks!
• Nov 1st 2010, 09:53 AM
HallsofIvy
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.