Question

Could anyone show me how to integrate exp(rcos(theta)) over a sphere, radius a?

Answer 1

I can help, but I need to know if theta is your zenith or azimuthal angle. Is it the angle that sweeps down from the z-axis, or the angle that sweeps across the x-y plane? The convention is different between pure and applied mathematics.

Answer 2

Thank you for responding lokisan. Here theta is the angle coming down from the positive z-axis, which I believe is the zenith?

Answer 3

Hi StandardCarpet. I can help you, just have to wait till I'm at a computer (on iPhone).

Answer 4

Okay, your integral is\[\int\limits_{}_{s}\int\limits_{}{}e^{a \cos \theta}dS\]which we need to convert to a standard double integral,\[\int\limits_{}_{D}\int\limits_{}{}e^{a \cos \theta}||\frac{\partial x}{\partial \theta }\times \frac{\partial x}{\partial \phi}||d \theta d \phi\]where \[x(\theta , \phi)\]is a vector parametrization of the surface. The magnitude of the cross product of the partials exists in the integrand as well.

Answer 5

Note that I have put 'a' in place of 'r' in your function. This is because the value for r over this surface will be a constant, namely, 'a'.

The vector parametrization of a spherical surface is\[x=a(\sin \theta \cos \phi , \sin \theta \sin \phi, \cos \theta)\] where I've taken out the common factor of 'a'.

The partial derivatives are:\[\frac{\partial x}{\partial \theta}=a(\cos \theta \cos \phi , \cos \theta \sin \phi, -\sin \theta)\]\[\frac{\partial x}{\partial \phi}=a(-\sin \theta \sin \phi, \sin \theta \cos \phi, 0)\]

Answer 6

The cross product gives, \[\frac{\partial x}{\partial \theta} \times \frac{\partial x}{\partial \phi}=a^2(\sin^2 \theta \cos \phi, -\sin^2 \theta \sin \phi, \cos \theta \sin \theta)\]

Answer 7

with magnitude,\[||\frac{\partial x}{\partial \theta} \times \frac{\partial x}{\partial \phi}||=a^2 \sin \theta\]

Answer 8

Now, for your integral, you're integrating over the whole sphere, so your limits for theta are 0 to pi, and your limits for phi would be 0 to 2pi. Using all this, we have,\[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}e^{a \cos \theta}a^2 \sin \theta d \theta d \phi\]

Answer 9

Move the factor a^2 through and integrate phi out (since the integrand does not depend on it) to give,\[2\pi a^2\int\limits_{0}^{\pi}e^{a \cos \theta} d \theta\]

Answer 10

sorry, should be\[2 \pi a^2 \int\limits_{0}^{\pi}e^{a \cos \theta}\sin \theta d \theta\]

Answer 11

Recognizing that the derivation of cos(theta) is -sin(theta), you can assume the substitution, \[u = a \cos \theta \rightarrow du = -a \sin \theta d \theta \rightarrow \sin \theta d \theta = -\frac{du}{a}\]so that the integral becomes, \[2\pi a^2 \int\limits_{u_1}^{u_2}e^u \left( -\frac{du}{a} \right)=-2\pi a \int\limits_{u_1}^{u_2}e^udu=-2\pi a \left[ e^u \right]^{u_1}_{u_2}\]\[=-2\pi a \left[ e^{a \cos \theta} \right]^{\pi}_{0}=-2\pi a (e^{-a}-1)=2 \pi a (1-e^{-a})\]

Answer 12

Thank you again Lokisan. My only question is regarding treating r as constant a. Normally when I do integrals over volumes, I set them up as triple integrals, so here I would try to integrate between 0 and a for dr. Could you explain to me why it's valid to just treat r as a constant in this case?

Answer 13

Yep. r is constant because you're integrating over the surface of a sphere AND *conveniently* using spherical polar coordinates.

The surface integral is such that you're integrating your function over a surface, meaning you're taking values *from* that surface to plug into you function which you then integrate.

Remember, in school, you integrate over the real line - you're taking values *from* that line to put into your function which you then integrate.

Here, the values you plugged in came from a sphere or radius a AND you were working in a convenient system in which, as you moved theta and phi, keeping a constant allowed you to stay on the surface that would feed your function.

Is that clearer...lol?

Answer 14

'sphere *of* radius a' ... it should read.