Surface Integral - 2 methods give different answer

[destroyer130]

Thanks for checking this out. Here's the problem:

I attempted to do it by using parametrize it into spherical coordinate.

r(r,t) = (x= cost, y= sint, z=r)
dS=|r_{u} x r_{v}| dA = r\sqrt{2} dA
dA = rdrdt

\int\intx^{2}z^{2}dS = \int\int\sqrt{2} cos^{2} r^{6} drdt

I check my solution manual and this is how they do it. My integral has r^{6} factor. However, solution's only has r^{5} instead. I am very confused because these two are supposed to be from the same source...

 

[clamtrox]

r(r,t) = (x= cost, y= sint, z=r)
dS=|r_{u} x r_{v}| dA = r\sqrt{2} dA
dA = rdrdt

There is one r too many here. dS = |r_u x r_v| dr dθ

 

[destroyer130]

There is one r too many here. dS = |r_u x r_v| dr dθ

Oh, so why dA not equal to rdrdt in this case?

 

[clamtrox]

Oh, so why dA not equal to rdrdt in this case?

Why would it be? What is dA anyway?

 

[destroyer130]

Based on parametric equation, if x = rcosθ, y=rsinθ, then dA = r dr dθ
In my homework case: dS = |ru x rv| dA = |ru x rv| r dr dθ
so that why somehow there is r^1 excess? but idk what r is excess?

 

[Dick]

Based on parametric equation, if x = rcosθ, y=rsinθ, then dA = r dr dθ
In my homework case: dS = |ru x rv| dA = |ru x rv| r dr dθ
so that why somehow there is r^1 excess? but idk what r is excess?

When you are using the formula |ru x rv| dA you don't add the 'volume' part (r) to dr dθ in dA. You'll automatically get that factor from the |ru x rv|.

 

[destroyer130]

Okay, thanks so much for your help :)