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

**sarahisme** · August 26th 2006, 05:01 PM

Hi all,

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

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

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

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

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

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

then

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

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

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

cheers!

-Sarah

**sarahisme** · August 26th 2006, 09:53 PM

nm, got it!

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