FInd the volume of the solid that lies within the sphere x^2+y^2+z^2=36 , above the xy plane, and outside the cone z=5sqrt(x^2+y^2)

Answer 1

It will be convenient to do think about this problem in spherical coordinates : (r , θ , φ), where :

r = (x² + y² + z²)^1/2 , θ = cos^-1 (z / r) is the polar angle (measured from the z-axis) and φ = tan^-1(y / x) is the azimuthal angle (counter-clockwise from x-axis looking from the +ve z-direction). The reverse transformations are : z = r cos θ , x = r sin θ cos φ and y = r sin θ sin φ. The reason this is most convenient is that the limits of the r , θ and φ integrals will be independent of the coordinates themselves (so that the volume integral we would have to evaluate will reduce to a product of 3 single integrals)

In spherical representation, volume is given by the integral

V = ∫∫∫ r² dr sin θ dθ dφ = ∫∫∫ r² dr d(cos θ) dφ , where the integral is taken over the volume we are interested in.

Now, in this problem, we wish to find the volume enclosed above the x-y plane inside the sphere r = 6, but outside the cone z = 5 (x² + y²)^1/2, which in spherical coordinates, translates to cot θ = 5 ⇒

cos θ = 5/√26.

The 'inside' of the cone is ranges from θ = 0 = cos^-1(1) to θ = cos^-1(5/√26). The volume between the sphere and the cone is then reduces to :

V_in = (∫₀⁶ r² dr)·(_∫5/√26¹ d(cos θ))·(∫₀^2π dφ), which can be easily evaulated.

The remainder is V_out = (1/2) V_sphere - V_in is the volume we are interested in. This may also be written as the integral V_out = (∫₀⁶ r² dr)·(_∫0^5/√26 d(cos θ))·(∫₀^2π dφ) as θ = π / 2 = cos^-1(0) on the x-y plane

Hope this helps.