Homework Help: Multivariable calculus, Integral using spherical coordinates

1. Aug 7, 2011

wildleaf

1. The problem statement, all variables and given/known data
Using spherical coordinates, set up but DO NOT EVALUATE the triple integral of f(x,y,z) = x(x^2+y^2+z^2)^(-3/2) over the ball x^2 + y^2 + z^2 ≤ 16 where 2 ≤ z.

2. Relevant equations
x = ρ sin ϕ cos θ
y = ρ sin ϕ sin θ
z = ρ cos ϕ
ρ^2 = x^2 + y^2 + z^2

∫∫∫w f(x,y,z) dxdydz
= ∫from θ1 to θ2 ∫from ϕ1 to ϕ2 ∫from ρ1 to ρ2 f(ρ*sinϕcosθ,ρ*sinϕsinθ,ρ*cosϕ) (ρ^2*sinϕ dρ dϕ dθ)

3. The attempt at a solution
First, I graphed xy, xz, yz plane, and from there I tried to find θ, ϕ, and ρ. For ρ, I got 0≤ρ≤4, I am not too sure about the bottom bound (0). For ϕ, I got 0 ≤ ϕ≤ pi/4, I was told that the top bound (pi/4) is wrong. For θ, I got 0 ≤ θ ≤ pi, I think both the bounds are right for this one.

Then I plugged them into the triple intergal and changed x(x^2+y^2 + z^2)^(-3/2) into spherical coordinate to get "ρ^-2*sinϕcosθ"

∫from 0 to pi ∫from 0 to pi/4 ∫from 0 to 4 ρ^-2*sinϕcosθ (ρ^2*sinϕ dρ dϕ dθ)
= ∫from 0 to pi ∫from 0 to pi/4 ∫from 0 to 4 (sinϕ)^2 cosθ dρ dϕ dθ

Can you please help me find the correct bounds for θ, ϕ, and ρ and I believe the function in spherical coodinates is right. Thanks in advance.

2. Aug 8, 2011

tiny-tim

hi wildleaf!

(have a a pi: π and try using the X2 icon just above the Reply box )
yes, in this as in many integrals over a sphere, you can have either 0 ≤ θ ≤ π or 0 ≤ θ ≤ 2π … which one you use for θ will affect the limits you use for ϕ

before we go any further, how would you describe the region (in words)?

3. Aug 8, 2011

cragar

ρ doesn't go from 0 because you have that plane at z=2 and figure out where that intersects the sphere it will help with the angle. I dont want to help to much.

4. Aug 9, 2011

wildleaf

In the xy plane, the region is a circle with radius of 4. The yz and xz plane looks similar, it also a circle with radius 4 but there is a line z = 2, and we want the top of the circle. HELP!!! ME!!!

5. Aug 9, 2011

wildleaf

In the xyz space, it would want the top half of the circle.

6. Aug 9, 2011

cragar

on the bounds of rho , we have z=2 and $z=\rho cos(\phi)$
so $\rho$ will go from $2sec(\phi)$ to 4 and that will make sure we are in the top half of the sphere .

7. Aug 9, 2011

wildleaf

Ohhh...
This is if I choose 0 ≤ θ ≤ π ?

8. Aug 9, 2011

cragar

θ will go from 0 to 2π . ϕ will go from 0 to where the plane intersects the sphere. which would be. and rho will go to what i said above.

9. Aug 9, 2011

wildleaf

0 ≤ θ ≤ 2n, 2sec(ϕ) ≤ ρ ≤ 4, 0 ≤ ϕ ≤ pi/4?

10. Aug 10, 2011

cragar

everything looks good except that top bound for ϕ , cos(ϕ)=1/2
what angle has a cosine of 1/2 , i got the 1/2 from 2/4