double integrals

Question

Find the volume of the solid that lies between z = 10−2x^ 2−2y^2 and z = ( x^2 + y^2)^(1/2) .

Answer 1

Question: Find the volume of the solid that lies between the surfaces z = 10 - 2x² - 2y² and z = √(x² + y²).

First, we need to set up the problem. What do the surfaces look like?

The first equation can be rewritten z = 10 - 2(x² + y²). In both equations, z is a function of (x² + y²), which is the square of the distance, r, from (0,0) to (x,y). So we should transform from Cartesian coordinates (x,y) to polar coordinates (r,φ), where φ is the angle, in radians, from the positive x-axis to the line from (0,0) to (x,y). The range of φ is 0 to 2π.

Restated in polar coordinates, the question becomes "Find the volume of the solid that lies between the surfaces z = 10 - 2r² and z = r."

The equation z = r is a cone whose axis is the positive z-axis, with its apex at the origin (0,0,0).

The equation z = 10 - 2r² is the parabola z = 10 - x² rotated around the z-axis. The parabolic surface points down and has its maximum at (x,y,z) = (0,0,10), or equivalently at (r,z) = (0,10).

Now that we have a mental picture of these two surfaces, where do they intersect? Where the z-coordinates of the two surfaces are equal: z = 10 - 2r² = r, or just 10 - 2r² = r. This is a quadratic equation, rewrite it as 2r² + r - 10 = 0 = (2r + 5)(r - 2). This has two solutions: r = 2, and r = -5/2. But r is non-negative, so the only valid solution is r = 2.

So the surfaces intersect in a circle with radius of 2 and centered on the z-axis. It may not be obvious, but we only care about the volume between these two surfaces inside this circle. Otherwise, we have to consider all values of r from 0 to infinity, and that will give us an infinite answer.

Within this circle, which surface is above the other? The rotated parabola is above the cone. Remember, the cone's apex is at (0,0,0), and the maximum z-value of the parabolic surface is 10.

Now to set up our integral. The volume element in 3 Cartesian dimensions is dxdydz. But in the x-y plane, instead of using Cartesian coordinates with area element dxdy, we're using polar coordinates (r,φ), whose area element is (dr)(rdφ). In 3 dimensions, the volume element is (dz)(dr)(rdφ). Our first shot at the expression for the volume integral is V = ∫dr ∫rdφ ∫dz. I've put the integral over z to the far right because z is a function of r. The integral over φ is in the middle because (rdφ) also depends on r, but not on z.

But r does not depend on φ, so we can factor that out of the middle integral, giving us V = ∫rdr ∫dφ ∫dz. It's hard to show limits of integration in the integrals here, so I won't try, I'll just state them. The integral ∫dz is from z = r (lower limit) to z = 10 - 2r² (upper limit). Doing this integral gives us (10 - 2r² - r).

So we have

V = ∫rdr ∫dφ (10 - 2r² - r).

The r-dependence can come out of the integral over φ:

V = ∫(10 - 2r² - r)(rdr) ∫dφ

= ∫(10r - 2r³ - r²)dr ∫dφ.

The integral ∫dφ is from φ = 0 (lower limit) to φ = 2π (upper limit), so ∫dφ = 2π, and

V = 2π∫(10r - 2r³ - r²)dr.

The limits of integration are from 0 to 2 (remember our circle of radius 2?). Doing the integral gives us

V = 2π[10r²/2 - 2r⁴/4 - r³/3](r=2) - 2π[10r²/2 - 2r⁴/4 - r³/3](r=0)

= 2π[10(2)²/2 - 2(2)⁴/4 - (2)³/3] - 2π[10(0)²/2 - 2(0)⁴/4 - (0)³/3]

= 2π[10×4/2 - 2×16/4 - 8/3]

= 2π[20 - 8 - 8/3]

= 2π[12 - 8/3]

= 2π[36/3 - 8/3]

= 2π[36 - 8]/3

= 2π(28)/3

= 56π/3