Triple Integration from Rectangular to Spherical Coordinates

[enwarnock]

Homework Statement

Convert the integral from rectangular coordinates to spherical coordinates

2 √(4-x^2) 4
∫ ∫ ∫ x dz dy dx
-2 -√(4-x^2) x^2+y^2

Homework Equations

x=ρ sin∅ cosθ
y=ρ sin∅ cosθ
z=ρ cos∅

In case the above integrals cannot be understood:
-2 ≤ x ≤ 2
-√(4-x^2) ≤ y ≤ √(4-x^2)
x^2+y^2 ≤ z ≤ 4

The Attempt at a Solution

I figured that 0≤θ≤2∏ but and that the x converts to ρ sin∅ cosθ
and I know you have to multiply the original converted function (ρ sin∅ cosθ) to ρ^2 sin∅
but that's all i figured out

 

[LCKurtz]

Homework Statement

Convert the integral from rectangular coordinates to spherical coordinates

2 √(4-x^2) 4
∫ ∫ ∫ x dz dy dx
-2 -√(4-x^2) x^2+y^2

Homework Equations

x=ρ sin∅ cosθ
y=ρ sin∅ cosθ
z=ρ cos∅

The Attempt at a Solution

I figured that 0≤θ≤2∏ but and that the x converts to ρ sin∅ cosθ
and I know you have to multiply the original converted function (ρ sin∅ cosθ) to ρ^2 sin∅
but that's all i figured out

I assume that integtration means$$
\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{x^2+y^2}^4x\, dzdydx$$The first thing you should do is draw a picture of the 3D region described by the limits. You will need it to do the spherical limits. Are you sure you aren't asked to put it into cylindrical coordinates? That would be the natural choice.

 

[enwarnock]

I assume that integtration means$$
\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{x^2+y^2}^4x\, dzdydx$$The first thing you should do is draw a picture of the 3D region described by the limits. You will need it to do the spherical limits. Are you sure you aren't asked to put it into cylindrical coordinates? That would be the natural choice.

Actually the homework problem asked to convert to both cylindrical and spherical coordinates and I already finished the cylindrical.
And yes thank you for cleaning up my equation!

 

[HallsofIvy]

\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{x^2+y^2}^4x\, dzdydx
First, mark vertical lines, on an xy- graph, at x= -2 and x= 2. Of course, y= -\sqrt{4- x^2} and y= \sqrt{4- x^2} are halves of the circle x^2+ y^2= 4 that lies between those vertical lines.

Then z= x^2+ y^2 to z= 4 can be written, in spherical coordinates, as \rho cos(\phi)= \rho^2 sin^2(\phi) or \rho= cot(\phi)csc(\phi) and \rho cos(\phi)= 4 or \rho= sec(\phi)

 

[LCKurtz]

@enwarnock: So how are you coming on the spherical coordinate limits? Do you see that depending on what values $\phi$ takes that your $\rho$ is a two piece function and you are going to need two triple integrals to express the volume?