Change limits of integration into spherical polar coordinates

Baron

Junior Member
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Oct 3, 2010
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Please refer to the attachment for the question

My work (I tried using latex but had some difficulty)

ϕ is the angle from z axis to y, θ is angle from x axis to y
x = ρ sinϕ cos θ
y =
ρ sinϕ sin θ
z =
ρ cos ϕ
ρ^2 = x^2 + y^2 + z^2

So my integral in spherical coordinates so far is:

\(\displaystyle \int_{}\;\ \int_{}\;\ \int_{} \) ​ρ^5 * p^2 * sin ϕ dρ dθ dϕ or
\(\displaystyle \int_{}\;\ \int_{}\;\ \int_{} \) ​ρ^7 * sin ϕ dρ dθ dϕ

The limits of integration of theta are the easiest as the xy plane is a circle centered at (0,0) with radius of 1 so theta is between 0 and 2pi because y = sqrt(1-x^2) and y = -sqrt(1+x^2)

To find the limits of integration of rho, I used algebra
z =
ρ cos ϕ = 1 + sqrt(1-x^2-y^2)
ρ cos ϕ - 1 = sqrt(1 - (ρ sinϕ cos θ)^2 - (ρ sinϕ sin θ)^2)
(ρ cos ϕ - 1)^2 = 1 - ρ2 sin2 ϕ cos2 θ - ρ2 sin2 ϕ sin2 θ
ρ2 cos2 ϕ - 2 ρ cos ϕ + 1 = 1 - ρ2 sin2 ϕ (cos2 θ + sin2 θ)
ρ2 cos2 ϕ - 2 ρ cos ϕ = -ρ2 (1-cos2 ϕ)(1)
ρ2 cos2 ϕ - 2 ρ cos ϕ = - ρ2 + ρ2 cos2 ϕ
2 ρ cos ϕ = ρ2
2 cos ϕ = ρ

so
ρ varies between 0 and 2 cos ϕ

Is my work correct so far? How do I find the limits of integration for phi? I assume the lower limit of integration for phi is 0 because the figure starts from the z axis. I also know phi is always between 0 to pi but I have no idea what the upper limit of integration is

I'm having difficulty visualizing what is happening in the yz plane as I have no idea what the figure looks like with z = 1 + sqrt(1-x^2-y^2) and
z = 1 - sqrt (1
-x^2-y^2)

 

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  • triple integral in spherical polar.JPG
    triple integral in spherical polar.JPG
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