Surface integrals - parametrizing a part of a sphere

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Homework Help Overview

The problem involves finding the area of a specific portion of a sphere defined by the equation x² + y² + z² = 4z, which is constrained by the paraboloid x² + y² = z. The context is within the subject area of surface integrals and spherical coordinates.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to switch to spherical coordinates to simplify the integration process and calculates the limits for the angles. They express confusion regarding the correctness of their area calculation, suspecting an error in the portion of the curve they are considering.

Discussion Status

Some participants have engaged with the original poster's attempts, with one noting a personal issue with formatting in LaTeX. The original poster later indicates they have resolved their confusion regarding the portion of the curve being analyzed, suggesting a shift in understanding.

Contextual Notes

The discussion includes a mention of specific limits for integration and the need to correctly identify the relevant portion of the surface area being calculated. There is also a reference to formatting issues in the mathematical expressions used.

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
\int^{2∏}_{0} \int^{∏}_{∏/3} 4SinΦ dΦdΘ
<br /> \int^{2∏}_{0} \int_{∏/3}_{∏} 4SinΦ dΦdΘ<br />

I get 12∏, which is incorrect. The correct answer is 4∏. Where am I going wrong?
 
Last edited:
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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:
<br /> \int^{2∏}_{0} \int_{∏/3}_{∏} 4SinΦ dΦdΘ<br />

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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