Question

use spherical coordinates to evaluate the integral S[-1,1]S[sqrt(1-x^2), 0] S[sqrt(1-x^2-y^2),0] e^-(x^2+y^2+z^2)^(3/2)

Answer 1

do you know the formula to conversion from cartesian coordinates to spherical ones?

Answer 2

i know rho squared = x^2 + y^2 + z^2 but i dont know how to convert the bounds

Answer 3

identify the shape you are integrating over, what is it?

Answer 4

would theta be 0 to pi and rho be from 0 to 1?

Answer 5

its a cone?

Answer 6

not is is a sphere
not sure what you mean by theta, some people use theta and phi intechangeably
byt your bounds for rho are correct

Answer 7

no, it is...

Answer 8

*

Answer 9

\[0\le z\le\sqrt{1-x^2-y^2}\implies z^2+y^2+x^2=1\]which is a sphere of radius 1

Answer 10

yeah I'm stuck as to how i would change the bounds since dv = rho^2 sin(phi) d(rho) d(phi) d(theta)

Answer 11

dV is not really waht we look at in changing the bounds
sorry, brb
in the meantime think about what the bounds would be to represent a sphere

Answer 12

x=1 y=1 z = 1 all of them are plus and minus

Answer 13

I mean in terms of rho, theta, and phi

Answer 14

theta would be 0 to pi and phi would be 0 to pi/4?

Answer 15

wait would theta be 0 to 2pi?

Answer 16

theta is zero to 2pi and rho is from 0 to pi I think
let me double-check when I get off the phone

Answer 17

alright thanks

Answer 18

whoops the bounds are messed up the corrected ? is use spherical coordinates to evaluate the integral S[-1,1]S[0, sqrt(1-x^2)] S[0,sqrt(1-x^2-y^2)] e^-(x^2+y^2+z^2)^(3/2)

Answer 19

oh I figured that much

Answer 20

so rho is 0 to 1 and theta is from 0 to 2pi right? and phi's what we're missing

Answer 21

if we are doing the whole sphere phi must be either -pi/2 to pi/2
or 0 to pi
(I actually don't think it makes a difference, but I may have forgotten that detail)

Answer 22

i dont think it makes a difference since the range ends up being the same

Answer 23

I agree

Answer 24

so changing the values would be S[0,2pi]S[0,pi]S[0,1] rho^2 sin(phi) d(rho) d(phi) d(theta)?

Answer 25

well S[0,2pi]S[0,pi]S[0,1] (e^-(rho^3))* rho^2 sin(phi) d(rho) d(phi) d(theta)

Answer 26

right, which is pretty easy to integrate, eh?

Answer 27

lol right...

Answer 28

I think we got it then.
Good luck!

Answer 29

thanks

Answer 30

welcome

Answer 31

hey on my ? isn't it a quarter circle due to the sqrts so shouldn't the limits be 0 ≤ Θ ≤ π, 0 ≤ φ ≤ π/2, and 0 ≤ ρ ≤ 1?
thanks

Answer 32

if we take only the positive sqrt it would be just the upper half of the sphere\[0\le\theta\le2\pi\]and\[0\le\phi\le\frac\pi2\]rereading I think you are actually right, and that we are only doing the upper half

Answer 33

I think example2 here is similar:
http://tutorial.math.lamar.edu/Classes/CalcIII/TISphericalCoords.aspx