Multivariable calculus, Integral using spherical coordinates

[wildleaf]

Homework Statement

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.

Homework 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)

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.

 

[tiny-tim]

hi wildleaf!

(have a a pi: π and try using the X² icon just above the Reply box )

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.

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)?

 

[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 don't want to help to much.

 

[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!

 

[wildleaf]

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

 

[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 .

 

[wildleaf]

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

 

[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.

 

[wildleaf]

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

 

[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