How to Calculate the Volume of a Solid Bounded by a Sphere and a Cone?

[TheSpaceGuy]

Homework Statement

Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 =4, above the xy-plane, and below the cone z=sqrt(x^2 + y^2).

The Attempt at a Solution

Use Cylindrical Coordinates.

Note that r ≤ z ≤ √(4 - r^2).

These sphere and cone intersect when x^2 + y^2 + (x^2 + y^2) = 4
==> x^2 + y^2 = 2, a circle with radius √2.
So, the projection onto the xy-plane is a disk centered at the origin with radius √2.

Thus, the volume equals
∫∫∫ 1 dV
= ∫(t = 0 to 2π) ∫(r = 0 to √2) ∫(z = r to √(4 - r^2)) 1 (r dz dr dt)
= 2π ∫(r = 0 to √2) r (√(4 - r^2) - r) dr
= π ∫(r = 0 to √2) (2r √(4 - r^2) - 2r^2) dr
= π [(-2/3)(4 - r^2)^(3/2) - 2r^3/3] {for r = 0 to √2}
= (-2π/3) [r^3 + (4 - r^2)^(3/2)] {for r = 0 to √2}
= (-2π/3) [(2√2 + 2√2) - (0 + 8)]
= (π/3) (16 - 8√2).

Is this correct?

 

[icystrike]

why the boundary of dz is from r to 4-r^2?

 

[TheSpaceGuy]

Isn't it? Thats what I thought. What is it then?

 

[LCKurtz]

Homework Statement

Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 =4, above the xy-plane, and below the cone z=sqrt(x^2 + y^2).

The Attempt at a Solution

Use Cylindrical Coordinates.

Note that r ≤ z ≤ √(4 - r^2).

These sphere and cone intersect when x^2 + y^2 + (x^2 + y^2) = 4
==> x^2 + y^2 = 2, a circle with radius √2.
So, the projection onto the xy-plane is a disk centered at the origin with radius √2.

Thus, the volume equals
∫∫∫ 1 dV
= ∫(t = 0 to 2π) ∫(r = 0 to √2) ∫(z = r to √(4 - r^2)) 1 (r dz dr dt)
= 2π ∫(r = 0 to √2) r (√(4 - r^2) - r) dr
= π ∫(r = 0 to √2) (2r √(4 - r^2) - 2r^2) dr
= π [(-2/3)(4 - r^2)^(3/2) - 2r^3/3] {for r = 0 to √2}
= (-2π/3) [r^3 + (4 - r^2)^(3/2)] {for r = 0 to √2}
= (-2π/3) [(2√2 + 2√2) - (0 + 8)]
= (π/3) (16 - 8√2).

Is this correct?

Unfortunately, no. I didn't check your calculations but your setup gives the volume above the cone, not below it. Of course you could subtract that from the volume of the hemisphere but that ducks the issue of setting it up directly and correctly.

Try spherical coordinates; it is a more natural choice.