Question

Calculus 3: Find volume of a solid enclosed by cone \( z= \sqrt{x^2 +y^2}\) and sphere \( x^2 +y^2 +z^2 =2\)

Answer 1

Now thats what i am talking about. :)

Answer 2

but darn i gotta be at work in half hour.

Answer 3

in 10 minutes. DOnt rush me :P

Answer 4

I know that \( z = \sqrt{x^2+y^2} = \sqrt{r^2} = r\)

Answer 5

@perl @dan815

Answer 6

OHHH

Answer 7

\( \color{red}{x^2+y^2} +z^2 = 2\) so \( \color{red}{r^2} +z^2 = 2\)?

Answer 8

\(\implies z^2 = 2-r^2 \implies z= \sqrt{2-r}\)

Answer 9

no wait... we have \(\ z=r\) and \(r^2 + z^2 = 2\)

Answer 10

Is it okay to square z and r as so? \[z^2 =r^2\]

Answer 11

If I do that then I have \(\ r^2 + r^2 = 2 \implies 2r^2=2 \implies r=1\)

Answer 12

I think I have my limits. \[\ 0 \le r\le1\]\[ 0 \le \theta \le 2\pi\]\[\ r \le z \le \sqrt{2-r^2}\]

Answer 13

Omg I was so close.....

Answer 14

Yeah, I'm bad, lol. Sorry.

Answer 15

\[\LARGE \int\limits_0^{2\pi}\int\limits_0^{\sqrt{2}}\int\limits_{\sqrt{2-r^2}}^r rdzdrd \theta\] This is your integral of the volume, do you have any questions here so far or should I continue?

Answer 16

Aww , you deleted it!

Answer 17

Because it was wrong :P

Answer 18

Ohhhh you have lied to us!

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There are two possible ways to have a volume bounded by a sphere and cone since the cone cuts the sphere into two parts!

Answer 19

There's this part and this part