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# Triple Integral - Spherical?

Integrate the function f(x,y,z)=5x+7y over the solid given by the "slice" of an ice-cream cone in the first octant bounded by the planes x=0 and y=sqrt(19/125)x and contained in a sphere centered at the origin with radius 13 and a cone opening upwards from the origin with top radius 12.

### 1 Answer

- kbLv 76 years agoFavorite Answer
Using spherical coordinates:

y = x√(19/125)

==> ρ sin θ sin φ = ρ cos θ sin φ √(19/125)

==> θ = arctan√(19/125)

Note that x = 0 yields θ = π/2 as a bound

Sphere: x^2 + y^2 + z^2 = 13^2 ==> ρ = 13.

Cone: z^2 = a(x^2 + y^2) for some a.

Since this intersects the sphere:

x^2 + y^2 + a(x^2 + y^2) = 13^2

==> x^2 + y^2 = 169/(a+1).

We need 169/(a+1) = 12^2 (since the top radius is 12)

==> a = 25/144.

So, the cone has equation z^2 = (25/144)(x^2 + y^2)

==> (ρ cos φ)^2 = (25/144)(ρ sin φ)^2

==> tan φ = 12/5.

Thus, ∫∫∫ (5x + 7y) dV

= ∫(θ = arctan√(19/125) to π/2) ∫(φ = 0 to arctan(12/5)) ∫(ρ = 0 to 13) (5ρ cos θ sin φ + 7ρ sin θ sin φ) * (ρ^2 sin φ dρ dφ dθ)

= ∫(θ = arctan√(19/125) to π/2) (5 cos θ + 7 sin θ) dθ

* ∫(φ = 0 to arctan(12/5)) sin^2(φ) dφ * ∫(ρ = 0 to 13) ρ^3 dρ

= (5 sin θ - 7 cos θ) {for θ = arctan√(19/125) to π/2}

* ∫(φ = 0 to arctan(12/5)) (1/2)(1 - cos(2φ)) dφ * (1/4)13^4

= (1/4)13^4 * [5 - (5/12)√19 + (7/12)√125)] *

(1/2)(φ - sin(2φ)/2) {for φ = 0 to arctan(12/5)}

= (1/4)13^4 * (1/12) (60 - 5√19 + 35√5) *

(1/2)(φ - sin φ cos φ) {for φ = 0 to arctan(12/5)}

= (1/96) 13^4 * (60 - 5√19 + 35√5) * (arctan(12/5) - 12/13 * 5/13)

= (169/96) (60 - 5√19 + 35√5) (169 arctan(12/5) - 60).

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I hope this helps!