Answer:
The volume of the solid is 243[tex]\sqrt{2} \ \pi[/tex]
Step-by-step explanation:
From the information given:
BY applying sphere coordinates:
0 ≤ x² + y² + z² ≤ 81
0 ≤ ρ² ≤ 81
0 ≤ ρ ≤ 9
The intersection that takes place in the sphere and the cone is:
[tex]x^2 +y^2 ( \sqrt{x^2 +y^2 })^2 = 81[/tex]
[tex]2(x^2 + y^2) =81[/tex]
[tex]x^2 +y^2 = \dfrac{81}{2}[/tex]
Thus; the region bounded is: 0 ≤ θ ≤ 2π
This implies that:
[tex]z = \sqrt{x^2+y^2}[/tex]
ρcosФ = ρsinФ
tanФ = 1
Ф = π/4
Similarly; in the X-Y plane;
z = 0
ρcosФ = 0
cosФ = 0
Ф = π/2
So here; [tex]\dfrac{\pi}{4} \leq \phi \le \dfrac{\pi}{2}[/tex]
Thus, volume: [tex]V = \iiint_E \ d V = \int \limits^{\pi/2}_{\pi/4} \int \limits ^{2\pi}_{0} \int \limits^9_0 \rho ^2 \ sin \phi \ d\rho \ d \theta \ d \phi[/tex]
[tex]V = \int \limits^{\pi/2}_{\pi/4} \ sin \phi \ d \phi \int \limits ^{2\pi}_{0} d \theta \int \limits^9_0 \rho ^2 d\rho[/tex]
[tex]V = \bigg [-cos \phi \bigg]^{\pi/2}_{\pi/4} \bigg [\theta \bigg]^{2 \pi}_{0} \bigg [\dfrac{\rho^3}{3} \bigg ]^{9}_{0}[/tex]
[tex]V = [ -0+ \dfrac{1}{\sqrt{2}}][2 \pi -0] [\dfrac{9^3}{3}- 0 ][/tex]
V = 243[tex]\sqrt{2} \ \pi[/tex]
