Surface Area of Surface

**usagi_killer** · November 1st 2010, 01:11 AM

I got part a) b) c) however I am just unsure on part d)

For a) I got $r(\theta, \phi) = \sin(\phi)\cos(\theta)i + \sin(\phi)\sin(\theta)j+\cos(\phi)k$

b) $r(x,y) = xi+yj+zk$ but $z^2+y^2+x^2 = 1$ so $z = \sqrt{1-(x^2+y^2)}$

So $r(x,y) = xi+yj+\sqrt{1-(x^2+y^2)}k$

c) Not required for d)

d) So area of surface $= \int \int_D |r_x \times r_y| dA$

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

)

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

But what are the bounds for the double integral?

Thanks!

**HallsofIvy** · November 1st 2010, 09:53 AM

You are told that $\pi/6\le \phi\le \pi/3$ .

The bounds for the entire sphere would be $0\le \theta\le 2\pi$ and $0\le \phi\le \pi$ .

With the restriction given, $0\le \theta\le 2\pi$ and $\pi/6\le \phi\le \pi/3$ , of course.

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