# transforming from spherical coordinates

• Aug 12th 2009, 04:35 PM
bigdoggy
transforming from spherical coordinates
This question is born from a question relating to Stokes' theorem, where I'm evaluating the surface integral.
The area I'm confused with is where I need to transform the follwing into speherical polar coordinates:

$\displaystyle \int\int_{S} (\nabla \times F)\cdot \bold{\hat{n}}\ dS \ = \frac{1}{2}\int\int_S(-yz-3z)\ dS$

where

$\displaystyle x=2\cos\phi\sin\theta\ y=2\sin\phi\sin\theta\ z=2\cos\theta$

how would I transform the above integral into one wrt $\displaystyle d\theta\ d\phi$?
• Aug 12th 2009, 04:47 PM
arbolis
I think we must know what surface $\displaystyle S$ is.
Out of curiosity, your "$\displaystyle \frac{1}{2}\int\int_S(-yz-3z)\ dS$" shouldn't be $\displaystyle dA$ instead of $\displaystyle dS$?
• Aug 12th 2009, 05:40 PM
luobo
Quote:

Originally Posted by bigdoggy
This question is born from a question relating to Stokes' theorem, where I'm evaluating the surface integral.
The area I'm confused with is where I need to transform the follwing into speherical polar coordinates:

$\displaystyle \int\int_{S} (\nabla \times F)\cdot \bold{\hat{n}}\ dS \ = \frac{1}{2}\int\int_S(-yz-3z)\ dS$

where

$\displaystyle x=2\cos\phi\sin\theta\ y=2\sin\phi\sin\theta\ z=2\cos\theta$

how would I transform the above integral into one wrt $\displaystyle d\theta\ d\phi$?

Use Jacobian matrix.
• Aug 12th 2009, 05:59 PM
bigdoggy
Quote:

I think we must know what surface S is.
Out of curiosity, your "$\displaystyle \frac{1}{2}\int\int_S(-yz-3z)\ dS$" shouldn't be dA instead of dS?
The surface is the hemisphere $\displaystyle x^2 + y^2 + z^2 = 4$ and the boundary curve is the intersection of the hemisphere with the plane $\displaystyle z=0.$

Quote:

Use Jacobian matrix.
I've looked on Wikipedia, Jacobian matrix and determinant - Wikipedia, the free encyclopedia, but how would I go from the Jacobian to $\displaystyle dS = 4\sin\theta d\phi d\phi$ ??

I'm trying to understand how to convert the integral from the surface integral to one which can be evaluated using spherical polar's.[incidentally I know the answer is $\displaystyle dS = 4\sin\theta d\phi d\phi$ but cant figure out how to get to it!(Wondering)]
• Aug 13th 2009, 04:25 AM
bigdoggy
$\displaystyle dS=r^2\sin\theta d\phi d\theta$