1
$\begingroup$

I have seen two different cases where a surface Integral uses a jacobian and does not in spherical coordinates. Here is an example that I saw uses a jacobian

$$\iint_R 2\sin\phi \cos\theta*4\sin\phi\,\mathrm dA = \int_0^{2\pi}\int_0^\pi 16\sin^3\phi \cos^2\theta \,\mathrm d\phi \mathrm d\theta$$

where $4\sin\phi$ on the left is not the jacobian but the norm ($||\mathbf r_u \times \mathbf r_v||)$. and on the right the jacobian is added, thus another factor of $4\sin\phi$ is included in the equation.

OK... here is where I am confused...

In the above case, we used the norm and the jacobian (which are the same in spherical). But in my next example, the jacobian is not used and it is correct! (according to pauls outline).

$$\iint_R 9\sin\phi \cos\theta \cos\phi *9\sin\phi\,\mathrm dA = 81\int_{\pi/2}^{\pi}\int_0^{\pi/2} \sin^2\phi \cos\theta \cos\phi\,\mathrm d\phi\mathrm d\theta$$

What's up with that? $9\sin\phi$ is the norm on the left and if using the same logic as my first example there should be another factor included but there is not!

Is Pauls Outline Incorrect?

Please help me understand this, I am very confused why the jacobian was not used in my second example.

$\endgroup$
2
$\begingroup$

Pauls notes was not incorrect.

What happen was in one problem the normal was found with respect to $x,y,z$ and then converted to spherical coordinates. If you choose to do it this way, you must add another factor of the jacobian when you convert.

Otherwise, if you convert to spherical coordinates beforehand and find the normal, the jacobian will be included in your answer and you do not need to include another factor of the jacobian.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.