# surface integrals

• Feb 28th 2009, 06:42 AM
James0502
surface integrals
evaluate double int over S (e^z) dS
where S is the surface of the sphere x^2 + y^2 + z^2 = a^2

I know dS = a^2(sintheta)(d.theta)(d.phi) using the parametrization

r = (a.sintheta.cosphi,a.sintheta.sinphi,acosphi)

so from there do I just calculate

int int e^(acosphi) a^2(sintheta)(d.theta)(d.phi)

?

many thanks
• Feb 28th 2009, 12:35 PM
ThePerfectHacker
Quote:

Originally Posted by James0502
evaluate double int over S (e^z) dS
where S is the surface of the sphere x^2 + y^2 + z^2 = a^2

I know dS = a^2(sintheta)(d.theta)(d.phi) using the parametrization

r = (a.sintheta.cosphi,a.sintheta.sinphi,acosphi)

so from there do I just calculate

int int e^(acosphi) a^2(sintheta)(d.theta)(d.phi)

?

many thanks

The surface can be parametrized as $\bold{g} : [0,2\pi]\times [0,\pi] \to \mathbb{R}^3$ as $\bold{g}(\theta,\phi) = (a^2\cos \theta \sin \phi, a^2 \sin \theta \sin \phi, a^2 \cos \phi)$.

The function that you have is $f(x,y,z) = e^z$.

Therefore, $\iint_S f dS = \int_0^{\pi} \int_0^{2\pi} f(\bold{g}(\theta,\phi)) \cdot \left| \frac{\partial \bold{g}}{\partial \theta} \times \frac{\partial \bold{g}}{\partial \phi} \right| d\theta ~ d\phi$

Compute that for your surface integral.