# Surface integral problem - don't need to use Jacobian for polar?

1. Dec 6, 2011

### ishanz

1. The problem statement, all variables and given/known data

Evaluate the surface integral.
∫∫S x^2*z^2 dS
S is the part of the cone z^2 = x^2 + y^2 that lies between the planes z = 1 and z = 3.

2. Relevant equations

$\int \int _{S}F dS = \int \int _D F(r(u,v))|r_u\times r_v|dA$
$x=rcos(\theta)$
$y=rsin(\theta)$

3. The attempt at a solution
First, I parametrized the cone.
$z^2=x^2+y^2\Rightarrow z=\sqrt{x^2+y^2} \Rightarrow z=r$
Therefore, the cone's vector equation should be
${\bf R}(r,\theta)=rcos(\theta){\bf i}+rsin(\theta){\bf j}+r{\bf k}$
${\bf R}_r=cos(\theta){\bf i}+sin(\theta){\bf j}+{\bf k}$
${\bf R}_\theta=-rsin(\theta){\bf i}+rcos(\theta){\bf j}$
$|{\bf R}_r \times {\bf R}_\theta| = r\sqrt{2}$
$\int_0^{2\pi}\int_1 ^3 {(rcos(\theta))^2(r)^2(r\sqrt{2})}drd\theta=\frac{364\sqrt{2}\pi}{3}$

Now, this is the right answer as per the book. My question is, when we go from $dA$ to $drd\theta$, why don't we use the Jacobian for polar coordinates, r? Had we included the Jacobian, the degree of r in the final double integral would have been six instead of five, giving us a completely different answer. My confusion here: Why is $dA$ not equal to $rdrd\theta$?

Last edited by a moderator: Dec 7, 2011
2. Dec 6, 2011

### ishanz

Sorry, the final step of the integration process didn't come out right. Here it is:

$\int_0^{2\pi}\int_1 ^3 {(rcos(\theta))^2(r)^2(r\sqrt{2})}drd\theta=\frac{364\sqrt{2}\pi}{3}$

3. Dec 6, 2011

### Dick

The |R_r x R_theta| factor already includes the r in dA. If the surface were just the x-y plane and they wanted you to find area by integrating 1. Think about what that factor would be.

Last edited: Dec 6, 2011
4. Dec 7, 2011

### ishanz

I don't quite understand. I don't know if it's because of the totally methodical and unintuitive way that our professor has taught us (i.e., simply equating $dA$ to $rdrd\theta$ in all scenarios) or because of my own negligence. Does the $|{\bf R}_r\times{\bf R}_\theta|$ factor always take care of the Jacobian for surface integrals, then? In something akin to spherical coordinate integrals (e.g., if my vector ${/bf R}$ wer

5. Dec 7, 2011

### ishanz

God, I'm bad at this whole Latex + forum thing. I'm so sorry about the double posts... I think I accidentally hit submit or something.

I don't quite understand. I don't know if it's because of the totally methodical and unintuitive way that our professor has taught us (i.e., simply equating $dA$ to $rdrd\theta$ in all scenarios) or because of my own negligence. Does the $|{\bf R}_r\times{\bf R}_\theta|$ factor always take care of the Jacobian for surface integrals, then? In something akin to spherical coordinate integrals, would that factor take care of the whole $\rho^2sin(\phi)$ factor for me?

6. Dec 7, 2011

### Dick

r and theta here are really just a convenient parametrization of the surface. $dS=|r_u\times r_v| du dv$. I wouldn't substitute dA for du dv, if it's going to confuse you.

7. Dec 7, 2011

### ishanz

I see, that makes sense. Are there any situations in which I would ever have to actually worry about a Jacobian factor when doing a surface integral similar to the one I've described above? Or is it something I should only worry about when explicitly executing coordinate transforms?

8. Dec 7, 2011

### Dick

You don't have to worry about it if you use that formula. Like I said, the 'jacobian' part is the $|r_u\times r_v|$ factor.

9. Dec 7, 2011

### ishanz

Got it. Thanks very much, Dick.