Converting to Spherical coordinates and Solving the Triple Intergral

Question

 

Convert the integral

∫(-1 < x < 1)  ∫(0 < y< sqrt(1-x²)  ∫(( -sqrt(1-x²-y²) < z < 0)      z²(x²+y²+z²)^(1/2)    dzdydx

 

into an equivalent integral in spherical coordinates and
evaluate the integral.

 

what I learned in class was to convert to the spherical coordinates is that

p = sqrt(x^2+y^2+z^2)

 

x = psinΦcosθ

 

y=psinΦsinθ

 

z = pcosΦ

 

I already know how to solve a triple integral but what I need to know is how do you convert the integral into spherical coordinates and then solve it??

Answer 1

There are 3 steps:

 

1. Convert the integrand

2. Convert the differentials

3. Convert the limits

 

Here's what I get when I follow these steps:

 

 

1. Apply the conversion formulas that you stated, directly to the integrand, and grind through the algebra to come up with a new integrand in terms of ρ, Φ, θ. 

 

z²(x²+y²+z²)^1/2 = (ρcos Φ)² ρ = ρ³ cos² Φ

 

2. dz dy dx can be converted directly to ρ² sin Φ dρ dΦ dθ

So now the full integrand is ρ³ cos² Φ ρ² sin Φ dρ dΦ dθ = ρ⁵ cos² Φ sin Φ dρ dΦ dθ

 

3.  You have to visualize this, but the original limits describe a quarter of a sphere of radius 1, in the positive half in the y direction and the negative half in the z direction.  That means that ρ goes from 0 to 1, Φ goes from 0 to π, and θ goes from π/2 to 3π/2.

 

 

So the final integral is:

 

∫_π/2 ^3π/2 ∫₀ ^π ∫₀ ¹  ρ⁵ cos² Φ sin Φ dρ dΦ dθ

 

Because the new limits are all constants, you can then rearrange terms in order to solve it:

 

∫_π/2 ^3π/2 ∫₀ ^π ∫₀ ¹  ρ⁵ cos² Φ sin Φ dρ dΦ dθ

= ∫_π/2 ^3π/2 ∫₀ ^π cos² Φ sin Φ (∫₀ ¹  ρ⁵)  dρ dΦ dθ

= ∫_π/2 ^3π/2 ∫₀ ^π cos² Φ sin Φ [p⁶ / 6]₀¹ dΦ dθ

= ∫_π/2 ^3π/2 ∫₀ ^π cos² Φ sin Φ (1/6 - 0) dΦ dθ

= 1/6 ∫_π/2 ^3π/2 ∫₀ ^π cos² Φ sin Φ dΦ dθ

 

...etc...

 

I'll leave the rest of the integration to you.

 

 

I'm not 100% sure of every step, so please review it carefully and see if you agree.