Question

Find the volume of the region bounded by the sphere x^2+y^2+z^2=9 and the cone z=sqrt(x^2+y^2).

I have already turned into polar coordinates with 0<=rho<=3, 0<=phi<=pi/4, and 0<=theta<=pi/2 integrating (rho^4)(sin(phi)) drho dphi dtheta. Can someone help me get the answer out of this? I keep coming up with the incorrect answer, so calculations are off somewhere (it is possible that i have my boundaries wrong...).

Answer 1

just off the bat, you that in spherical coordinates\[dV=rdrd\theta\phi\]you pick up an extra r

Answer 2

in spherical it picks up the p^2sinphi instead of the r right?, the r is only added in cylinderical coordinates.

Answer 3

p being rho

Answer 4

oh my bad, ok let me actually look at the question...

Answer 5

ok, why is theta only going to from 0 to pi/2 and not 2pi ?

Answer 6

it doesn't say "in the first octant"

Answer 7

no, I typed entire question and it is kicking my butt! I am so good at doing them individually, but putting them the sphere with the cone just throws me off.

Answer 8

I think it should be\[\int_{0}^{2\pi}\int_{0}^{\frac\pi2}\int_{0}^{3}\rho^2\sin\phi d\rho d\phi d\theta\]

Answer 9

I don't know why you cut everything in half is what I'm saying....

Answer 10

I began with 2pi but changed it when I found a similar problem on internet. I have found those to not be as helpful as I had hoped.

Answer 11

so 2pi would be the only incorrect thing you see in the integrand limits?

Answer 12

I also have going to pi/4 in for phi but that gives zero, which is the integral but not the actual vulume

Answer 13

phi to pi/2 rather

Answer 14

I kept getting final volume of 0 when I tried with pi/2. maybe my calculations are wrong

Answer 15

I have the choices between:
A. (27/4)pi(2-sqrt2)
B.9pi(2-sqrt3)
C. 27/4)pi(2-sqrt3)
D.9pi(2-sqrt2)

Answer 16

yes, because of the sin part
I think that is normal because the object is symmetric about z, but that is clearly not the volume
hey looky what I found pretty similar to this, but without a function being integrated over it and radius rho=1
http://tutorial.math.lamar.edu/Classes/CalcIII/TISphericalCoords.aspx

Answer 17

That is the page I found earlier that made me set the limits to pi/2 for theta and pi/4 for phi. Do you think I should stick with the integration limits that they have?

Answer 18

ok we know your rho bounds are right so let's narrow it down\[9\int_{0}^{?}\int_{0}^{?}\sin\phi d\phi d\theta\]that logic seems to hold in this situation to me, but since we want actual volume...

Answer 19

where did you get the bound pi/4 ?

Answer 20

i took rho(cosphi)=sqrt(9/2) (substituding in cosphi for z when you get to the part where you see z^2+z^2=rho

Answer 21

it ended up equaling pi/4

Answer 22

ok, so then my only suggestion is to change the bounds on theta to 2pi
let's see if that gives an option...

Answer 23

I will try real quick and let you know. Thank you!

Answer 24

I get D that way

Answer 25

I have attached my work so far, but changing the theta integrand limits doesn't seem to get me to the right place. Could you take a look and see if I am doing something wrong along the way?

Answer 26

I can't zoom in and see it :(

Answer 27

I tried re-doing the file, see if this helps...

Answer 28

not much...
I'm typing it out hold on

Answer 29

\[\int\limits_{0}^{2\pi}-243/5(\sqrt2)+243/5\]
is where I ended up at.

Answer 30

\[\int_{0}^{3}\int_{0}^{\frac\pi4}\int_{0}^{2\pi}\rho^2\sin\phi d\rho d\theta d\phi=9\int_{0}^{\frac\pi4}\int_{0}^{2\pi}\sin\phi d\theta d\phi\]\[=18\pi\int_{0}^{\frac\pi4}\sin\phi d\phi =-18\pi\cos\phi|_{0}^{\frac\pi4}\]\[=-18\pi(\frac{\sqrt2}2-1)=9\pi(2-\sqrt2)\]tadah!
(I have to go soon)

Answer 31

those bounds on the first integrals are switched a bit...

Answer 32

\[\int_{0}^{\frac\pi4}\int_{0}^{2\pi}\int_{0}^{3}\rho^2\sin\phi d\rho d\theta d\phi=9\int_{0}^{\frac\pi4}\int_{0}^{2\pi}\sin\phi d\theta d\phi\]\[=18\pi\int_{0}^{\frac\pi4}\sin\phi d\phi =-18\pi\cos\phi|_{0}^{\frac\pi4}\]\[=-18\pi(\frac{\sqrt2}2-1)=9\pi(2-\sqrt2)\]there we go

Answer 33

Genius, thank you!!!

Answer 34

I think you just made some algebra mistakes, your setup was good
but thanks, and you're welcome :)