Volume Integral of f over Sphere: Find Solution

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Homework Help Overview

The problem involves finding the volume integral of the function f = x² + y² + z² over the region inside a sphere of radius R, centered at the origin. The context is within multivariable calculus, specifically focusing on volume integrals in spherical coordinates.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to set up the triple integral in spherical coordinates but expresses uncertainty about the limits of integration. Some participants suggest simplifying the integrand and clarify the correct limits for theta.

Discussion Status

The discussion is ongoing, with participants providing guidance on the limits of integration and suggesting simplifications to the integrand. There is an acknowledgment of a mistake in the title of the thread, which indicates a light-hearted engagement in the discussion.

Contextual Notes

There is a mention of confusion regarding the limits of integration for theta, as well as the original poster's initial oversight in simplifying the integrand.

kidsmoker
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Homework Statement



Find the volume integral of the function [tex]f=x^{2}+y^{2}+z^{2}[/tex] over the region inside a sphere of radius R, centered on the origin.

Homework Equations



Spherical polars [tex]x=rsin(\theta)cos(\phi), y=rsin(\theta)sin(\phi), z=rcos(\theta)[/tex]

Jacobian in spherical polars = [tex]r^2sin(\theta)[/tex]

The Attempt at a Solution



When i work through it I end up with the triple integral

[tex]V=\int^{R}_{0}dr\int^{\pi}_{-\pi}d\phi\int^{\pi}_{-\pi}d\theta (r^{2}sin^{2}\theta cos^{2}\phi+r^{2}sin^{2}\theta sin^{2}\phi + r^{2}cos^{2}\theta)r^2sin\theta[/tex]

but I'm not too sure whether this is right. Mainly I'm not sure about the limits of integration.

Is this correct please?

Thanks.
 
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If you want to integrate over all theta, that's only theta in [-pi/2,pi/2], isn't it? And you can simplify the integrand a LOT. x^2+y^2+z^2=r^2.
 
Ah yeah I didn't bother simplifying the integrand but I can see I should have done cos it would have made it a lot easier to type lol. I thought about it some more and understand why the limits are as you said now. Thanks!
 
Is there are reason why you titled this "surface integral"?
 
Because I was tired :D Oops lol
 

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