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Evaluate the integral by changing to spherical coordinates. The outer boundaries are from 0 to 1. The middle one goes from -sqr...Asked by Mackenzie
Evaluate the integral by changing to spherical coordinates.
The outer boundaries are from 0 to 1.
The middle one goes from -sqrt(1-x^2) to sqrt(1-x^2)
The inner one goes from -sqrt(1-x^2-z^) to sqrt(1-x^2-z^)
for 1/sqrt(x^2+y^2+z^2) dydzdx
I don't understand how to get the limits of integration. I know for rho it will be from 0 to 1. I want to know the process to get the boundaries for phi and theta since I have a few other similar problems to do.
The outer boundaries are from 0 to 1.
The middle one goes from -sqrt(1-x^2) to sqrt(1-x^2)
The inner one goes from -sqrt(1-x^2-z^) to sqrt(1-x^2-z^)
for 1/sqrt(x^2+y^2+z^2) dydzdx
I don't understand how to get the limits of integration. I know for rho it will be from 0 to 1. I want to know the process to get the boundaries for phi and theta since I have a few other similar problems to do.
Answers
Answered by
Steve
you are integrating over the whole sphere, so
0 <= p <= 1 (inside-outside)
0 <= φ <= 2π (whole x-y plane)
0 <= θ <= π (top-to-bottom of sphere)
There must be some examples in your text. And there are surely some online.
0 <= p <= 1 (inside-outside)
0 <= φ <= 2π (whole x-y plane)
0 <= θ <= π (top-to-bottom of sphere)
There must be some examples in your text. And there are surely some online.
Answered by
Mackenzie
The solution says the the boundary for phi is from 0 to pi as well as the one for theta.
Answered by
Steve
I'd have to think about it, but you obviously have both halves of the circle and both halves of the sphere.
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