Surface integrals - parametrizing a part of a sphere

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Feodalherren
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Homework Statement



Find the area of the part of the sphere x^2 + y^2 + z^2 = 4z
that lies inside the paraboloid x^2 + y^2 = z

Homework Equations


The Attempt at a Solution


I solved for the intercepts and found that they are z=0 and z=3.
The sphere is centered two units in the z-direction above the origin.

Hence I wanted to switch to spherical coordinates and got:

0≤Θ≤2∏
ρ=2

Now, since we know that the sphere is centered on (0,0,2) we can take find the angle from the Z axis easily.

(∏/3)≤Φ≤∏.

The Surface Area of this portion of the sphere should then become
[itex]\int^{2∏}_{0} \int^{∏}_{∏/3} 4SinΦ dΦdΘ[/itex]
[itex] \int^{2∏}_{0} \int_{∏/3}_{∏} 4SinΦ dΦdΘ[/itex]

I get 12∏, which is incorrect. The correct answer is 4∏. Where am I going wrong?
 
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That itex thing never works for me either :/
 
Nevermind I solved it. I was taking the wrong portion of the curve.
 
Feodalherren said:
[itex] \int^{2∏}_{0} \int_{∏/3}_{∏} 4SinΦ dΦdΘ[/itex]

Feodalherren said:
That itex thing never works for me either :/

In the tex, don't use special characters; instead use the tex code for them. Quote this to see how I changed your code.$$
\int_0^{2\pi}\int_{\pi/3}^{\pi} 4\sin\theta d\phi d\theta$$
 
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