Calc III Double Intergral (In Polar Coordinates)

[Seda]

Hey, I'm trying to find the volume bounded by two cylinders:

X^2 + Y^2 = R^2

and Y^2 + Z^2 = R^2


Here's what I have so far, I don't know if this is needed.



So, the first cylinder sets up my limits of integration to be 0<theta<2pi and 0<r<R

Correct?

And the integrand will be the second cylinder in terms of r and theta correct?

Z^2 = R^2- Y^2

Z= sqrt(R^2 - Y^2)
Z= sqrt(R^2 - (RSinTheta)^2)

However, when I try to intergrate this over dA (rdrdtheta), I end up getting zero[/b[/b]

 

[Seda]

BTW i'm multplying whatever volume I get by 2 to account for the volume above and below the XY plane.

 

[tkhunny]

I didn't check your geometry, but what else did you expect besides zero? Take a close look at your argument.

\(\displaystyle \sqrt{R^{2} - (R\sin(\theta))^{2}}\;=\;R\sqrt{1 - (\sin(\theta))^{2}}\;=\;R\sqrt{(\cos(\theta))^{2}}\)

Unless you were VERY careful with that cosine, you managed one entire period of the cosine function. That includes enough positive and negative values to cancel out everything.

If you changed this to simply \(\displaystyle \cos{\theta}\), that's where you wandered off. It is a common error. This is the non-error version.

\(\displaystyle \sqrt{(\cos(\theta))^{2}}\;=\;|\cos(\theta)|\)

The question, then, is how do you deal with that aboslute value?

One way is to point out that you almost tripped over it. That is, Good Call on the symmetry above and below the x-y plane. Why stop there? Isn't there also symmetry when comparing the four octants above the x-y plane? Why not change the limits to [0,pi/2] and multiply again by 4? This simplifies your life substantially, since on [0,pi/2] the cosine is non-negative and you can discard the absolute values!

 

[Seda]

Thanks, the funny thing is i just thought of doing that, and then i came here so see if anyone responded.

It's always the simple things i miss :/