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

1. anonymous

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.

2. anonymous

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

3. anonymous

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

4. anonymous

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.

5. anonymous

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)$

6. anonymous

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

7. anonymous

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

8. anonymous

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,$\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}e^{a \cos \theta}a^2 \sin \theta d \theta d \phi$

9. anonymous

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

10. anonymous

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

11. anonymous

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})$

12. anonymous

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?

13. anonymous

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?

14. anonymous