Surface Integral involving Paraboloid

[LunaFly]

Homework Statement

Evaluate the surface integral:

∫∫_s y dS

S is the part of the paraboloid y= x² + z² that lies inside the cylinder x² + z² =4.

Homework Equations

∫∫_sf(x,y,z)dS = ∫∫_Df(r(u,v))*|r_u x r_v|dA

The Attempt at a Solution

I've drawn the region D in the xz-plane as a circle with radius 2. My limits of integration are r between 0 and 2, and t between 0 and 2∏.

I've parametrized the surface as:
r(r,t)= <rcos(t), r², rsin(t)>

I've gotten the vectors:
r_r=<cos(t), 2r, sin(t)>
r_t=<-rsin(t), 0, rcos(t)>

Trying to find dS, I did:
r_r X r_t = <2r²cos(t), -r, 2r²sin(t)>,
and the magnitude of that is r√(4r² +1).

Overall, I end up with dS = r²*√(4r²+1)drdt.

The integral becomes:

∫∫r⁴*√(4r²+1)drdt, with limits as mentioned above.

However this is not the correct answer. I am having a difficult time with these surface integrals and I am not really sure why. I keep getting confused about whether or not I am parametrizing
the surfaces correctly, so that may be where my error is. Any help would be much appreciated! Thanks
-Luna

 

[haruspex]

Overall, I end up with dS = r²*√(4r²+1)drdt.

The integral becomes:

∫∫r⁴*√(4r²+1)drdt, with limits as mentioned above.

How did the r² become r⁴? Not saying it's wrong, ... this is not an area I'm expert on.

 

[HallsofIvy]

Homework Statement

Evaluate the surface integral:

∫∫_s y dS

S is the part of the paraboloid y= x² + z² that lies inside the cylinder x² + z² =4.

Homework Equations

∫∫_sf(x,y,z)dS = ∫∫_Df(r(u,v))*|r_u x r_v|dA

The Attempt at a Solution

I've drawn the region D in the xz-plane as a circle with radius 2. My limits of integration are r between 0 and 2, and t between 0 and 2∏.

I've parametrized the surface as:
r(r,t)= <rcos(t), r², rsin(t)>

I've gotten the vectors:
r_r=<cos(t), 2r, sin(t)>
r_t=<-rsin(t), 0, rcos(t)>

Trying to find dS, I did:
r_r X r_t = <2r²cos(t), -r, 2r²sin(t)>,
and the magnitude of that is r√(4r² +1).

Overall, I end up with dS = r²*√(4r²+1)drdt.

Yes, this is what I get.

The integral becomes:

∫∫r⁴*√(4r²+1)drdt, with limits as mentioned above.

What? \int\int y dS= \int \int (r^2)dS= \int \int (r^2)(r\sqrt{4r^2+ 1})drd\theta
You should have r^3 outside the square root, not r^4.

However this is not the correct answer. I am having a difficult time with these surface integrals and I am not really sure why. I keep getting confused about whether or not I am parametrizing
the surfaces correctly, so that may be where my error is. Any help would be much appreciated! Thanks
-Luna

 

[LunaFly]

Yes, this is what I get.


What? \int\int y dS= \int \int (r^2)dS= \int \int (r^2)(r\sqrt{4r^2+ 1})drd\theta
You should have r^3 outside the square root, not r^4.

I believe it is r⁴ because:

My paramatrization has y=r², so f(r(r,t))=r²

dS = |r_r X r_t|dA = r√(4r² + 1) * rdrdt

dA = r drdt (because of the relationship between polar and Cartesian coordinates) which is where the r⁴ comes from.

Altogether, ∫∫_sf(r(r,t))dS = ∫∫_Dr²*r√(4r² + 1) * rdrdt= ∫∫_Dr⁴√(4r²+1) drdt

This may be helpful: According to the book, the integral equals (∏/60)(391√(17) +1).

 

[LunaFly]

So I tried solving the integral that HallsofIvy thought was correct, and it was! Thanks for the response.

But that brings a new question to light: why does dA become drdt rather than r*drdt?

I had the impression that every time we switched from Cartesian to polar, dA became r*drdt, but it seems that in this problem, that's not the case.

Why?

 

[LCKurtz]

So I tried solving the integral that HallsofIvy thought was correct, and it was! Thanks for the response.

But that brings a new question to light: why does dA become drdt rather than r*drdt?

I had the impression that every time we switched from Cartesian to polar, dA became r*drdt, but it seems that in this problem, that's not the case.

Why?

The formula for surface area element for a surface parameterized as $\vec R = \vec R(u,v)$ is $dS = |\vec R_u \times \vec R_v|dudv$. There is no extra $u$ or $v$, and in your case, no extra $r$.

Suppose your surface had been a circle in the xy plane. You might parameterize it as $\vec R(r,\theta) = \langle r\cos\theta,r\sin\theta,0\rangle$. If you calculate $dS = |\vec R_r\times \vec R_\theta|drd\theta$ you will find that that cross product is where the familiar $r$ in $rdrd\theta$ comes from. You don't add an extra $r$ when you parameterize it from the beginning.