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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I can help, but I need to know if theta is your zenith or azimuthal angle. Is it the angle that sweeps down from the zaxis, or the angle that sweeps across the xy plane? The convention is different between pure and applied mathematics.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you for responding lokisan. Here theta is the angle coming down from the positive zaxis, which I believe is the zenith?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hi StandardCarpet. I can help you, just have to wait till I'm at a computer (on iPhone).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Note 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:\[\frac{\partial x}{\partial \theta}=a(\cos \theta \cos \phi , \cos \theta \sin \phi, \sin \theta)\]\[\frac{\partial x}{\partial \phi}=a(\sin \theta \sin \phi, \sin \theta \cos \phi, 0)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The cross product gives, \[\frac{\partial x}{\partial \theta} \times \frac{\partial x}{\partial \phi}=a^2(\sin^2 \theta \cos \phi, \sin^2 \theta \sin \phi, \cos \theta \sin \theta)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0with magnitude,\[\frac{\partial x}{\partial \theta} \times \frac{\partial x}{\partial \phi}=a^2 \sin \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, 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,\[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}e^{a \cos \theta}a^2 \sin \theta d \theta d \phi\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Move the factor a^2 through and integrate phi out (since the integrand does not depend on it) to give,\[2\pi a^2\int\limits_{0}^{\pi}e^{a \cos \theta} d \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, should be\[2 \pi a^2 \int\limits_{0}^{\pi}e^{a \cos \theta}\sin \theta d \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Recognizing 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, \[2\pi a^2 \int\limits_{u_1}^{u_2}e^u \left( \frac{du}{a} \right)=2\pi a \int\limits_{u_1}^{u_2}e^udu=2\pi a \left[ e^u \right]^{u_1}_{u_2}\]\[=2\pi a \left[ e^{a \cos \theta} \right]^{\pi}_{0}=2\pi a (e^{a}1)=2 \pi a (1e^{a})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yep. 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0'sphere *of* radius a' ... it should read.
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