# Parameterizing and finding its boundaries (vector calculus)

1. Sep 6, 2008

### pcjang

1. The problem statement, all variables and given/known data

I was studying for my finals, and then the only thing that I got stuck on was parameterizing a surface and finding the area of the surface.

and my problem states that x2 + y2 + z2 = 4
and z $$\geq$$$$\sqrt{2}$$

2. Relevant equations

so when parameterizing a sphere, it comes out to be
x = r sin$$\phi$$cos$$\theta$$
y = r sin$$\phi$$sin$$\theta$$
z = r cos$$\phi$$

3. The attempt at a solution

so i'm sure that $$\theta$$ goes from 0 to 2$$\pi$$. But i was getting confused what boundaries should be for $$\phi$$.

What i tried is that
since z $$\geq$$$$\sqrt{2}$$ and should be less than 2 since 2 is the radius, i put $$\sqrt{2}$$ $$\leq$$ z $$\leq$$ 2.
which z = 2 cos$$\phi$$
so when i solved it, it came out to be $$\pi$$/4 $$\leq$$ $$\phi$$ $$leq$$ 0.

but it just doesn't make sense to me how the boundary can go backwards. can someone explain to me~??

2. Sep 6, 2008

### pcjang

oops, last one, i meant to say

phi is greater than or equal to pi/4
and
phi is less than or equal to 0

3. Sep 6, 2008

### Defennder

When you have $$\cos \phi > \frac{1}{\sqrt{2}}$$, you should note that the graph of cos x decreases continuously from 1 to 0 in the interval 0 to pi/2. Which is why you have to reverse the inequality when taking the arc cos the interval for phi.