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

[ishanz]

Homework Statement

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.

Homework 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)

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?

 

[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}

 

[Dick]

Homework Statement

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.

Homework 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)

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?

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.

 

[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

 

[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?

 

[Dick]

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?

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.

 

[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?

 

[Dick]

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?

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.

 

[ishanz]

Got it. Thanks very much, Dick.