(Triple integral problem).

Problem/Solution: http://f.imgtmp.com/CZcp1.jpg

I think it should be:
http://www.wolframalpha.com/input/?i=(integrate+(integrate+(integrate+r+dr+from+0+to+3)+d(theta)+from+0+to+2pi)+dz+from+0+to+9)
but it's not so why am I wrong?

Question

(Triple integral problem).

Problem/Solution: http://f.imgtmp.com/CZcp1.jpg

I think it should be:
http://www.wolframalpha.com/input/?i=(integrate+(integrate+(integrate+r+dr+from+0+to+3)+d(theta)+from+0+to+2pi)+dz+from+0+to+9)
but it's not so why am I wrong?

zarkon · Answer

your integral is over a cylinder ... the picture is not a cylinder.

s3a · Answer

Okay, so let's say that I want to keep the order dr d(theta) dz, how would the integral look like?

zarkon · Answer

$$\int\limits_{0}^{9}\int\limits_{0}^{2\pi}\int\limits_{0}^{\sqrt{9-z}}r^2\cdot r\,dr\,d\theta\,dz$$

s3a · Answer

Why is the r^2 part in the innermost integrand (the way it is set up above) there? Also, same question for that of the solution I provided. Are both there for the exact same reason? I get the r = sqrt(9 - z) part. I just don't get the r^2 part.

zarkon · Answer

you are given in the problem that $$f(r,\theta,z)=r^2$$

s3a · Answer

Oh!

s3a · Answer

Two secs, let me make sure I get the solution manual's way as well.

s3a · Answer

Okay, it's all good! Thanks again!