I need help with this, the topic is tripple integrals

Question

Find the volume of the solid that lies within the sphere x^2+y^2+z^2=49 above the xy plane and outside the cone z= 3sqrt(x^2+y^2)

Answer 1

1) find the intersection of the cone and the sphere to be a circle with radius of 0.7sqrt(10) and z=sqrt(2.1)

2) find the volume of a hemisphere using spherical coordinates ... dv=ρ²sin(θ)dρdΘdΦ and the limits on ρ are 0 to 7, θ are 0-2pi and Φ are 0 to pi/2. Answer is 2/3Pi(7)^3.

3) find the volume of the cone using cylindrical coordinates ... dv=ρdρdΦdzΦ, limits on ρ are 0 to .7sqrt(10), Φ 0 to 2pi, z from 0 to sqrt(2.1).

4) Subtract the volume of the cone from the volume of the sphere

Answer 2

In essence, you are looking for difference between the volumes of hemisphere with radius R= 7 and the volume of a cone (vertice at (0,0) and r=3z [r is the radius of the cone's intersection with the plane z above the (x,y) plane] and the hight such as to touch the hemisphere from inside.

If you combine the equation for the sphere and that of the cone, you'd find the hight of the cone. Then you could even avoid any integration by using the formulas V_hsph=(2/3)πR³/3 for the volume of a hemisphere and for the cone V_cone=1/3 πr²z and get their difference

If you would be required to get those volumes through integration, let me know and I could help you with that, too

Dr.Ariel B