# Surface Integral of a sphere (or part there of :P!)

• Aug 26th 2006, 05:01 PM
sarahisme
Surface Integral of a sphere (or part there of :P!)
Hi all,

would anyone be able to give me a little bit of help with this question.

http://img222.imageshack.us/img222/4540/picture1qy8.png

for part (a) i did this....

our sphere is given by $\displaystyle x^2 + y^2 +z^2 = 25$

putting in z = 3 to find out part of the sphere gives:
$\displaystyle x^2 + y^2 = 16$

so we change to the parametric representation of the part of the sphere we are interested in sphere which is

$\displaystyle \vec{r}(\theta,\psi) = (4sin\psi cos\theta) \vec{i} + (4sin\psi sin\theta) \vec{j} +(4cos\psi) \vec{k}$

then

$\displaystyle \vec{r_\theta}(\theta,\psi) = (-4sin\psi sin\theta) \vec{i} + (4sin\psi cos\theta) \vec{j}$

$\displaystyle \vec{r_\psi}(\theta,\psi) = (4cos\psi cos\theta) \vec{i} + (4cos\psi sin\theta) \vec{j} +(-4sin\psi) \vec{k}$

http://img222.imageshack.us/img222/526/picture2gf9.png

does this look like i am heading in the right direction for part (a)?

cheers! :)

-Sarah
• Aug 26th 2006, 09:53 PM
sarahisme
nm, got it! :)