Question

Find the volume of the solid enclosed by the cone z=sqrt(x^2+y^2) between the planes z=1 and z=2

Answer 1

@perl

Answer 2

@tkhunny

Answer 3

I got it all the way through on the attachment...I just want to know if I did it correctly.

Answer 4

You must do z before r if you want it to make more sense. Limits on z are \([2,\sqrt{r}]\). You seem to be doing a cylindrical chunk and ignoring the cone.

Answer 5

Actually, it makes less sense the more I look at it. I think you should do most of it with solid geometry and use the calculus to find the last piece.

\(\pi\cdot(1^{2})\cdot 1 + \int\limits_{0}^{2\pi}\int\limits_{1}^{2}\int\limits_{2}^{\sqrt{r}} r\;dz\;dr\;d\theta\)

Answer 6

Whoops! I have the z limits backwards! Sorry about that.

Answer 7

so z would be going from sqrt(r) to 2 if it was backwards... alright I'll try to evaluate this.

Answer 8

Do you understand what that first part is? The non-integral piece?

Answer 9

I'll be honest. my prof really didn't do a great job at explaining this

Answer 10

@perl

Answer 11

Something funny in there.

\(-\dfrac{8}{5}\sqrt{2} + \dfrac{2}{5} = \dfrac{2-8\sqrt{2}}{5}\) -- You have "2 +"!

You seem to have tripped over the distributive property. Be more careful. I know. It's tedious. :-)

gtg - Good luck.

Answer 12

@UnkleRhaukus

Answer 13

give me a minute

Answer 14

the equation of this cone is phi = pi/4

Answer 15

how did you get pi/4?