Bounded by the paraboloid z = 4 + 2x2 + 2y2 and the plane z = 10 in the first octant; use polar coordinates to find the volume?

Convert to polar:

z = 4 + 2x² + 2y²

z = 4 + 2(x² + y²)

z = 4 + 2r²

At z=10:

10 = 4 + 2r²

6 = 2r²

r = √3

So the limits are:

4+2r² ≤ z ≤ 10

0 ≤ r ≤ √3

0 ≤ θ ≤ π/2

Integrating:

V = ∫∫∫ dV

V = ∫∫∫ r dz dr dθ

V = ∫∫ r (z)|ᵣ¹⁰ dr dθ

V = ∫∫ r (10 - (4 + 2r²)) dr dθ

V = ∫∫ r (6 - 2r²) dr dθ

V = ∫∫ (6r - 2r³) dr dθ

V = ∫ (3r² - ½ r⁴)|₀ʳ dθ

V = ∫ (9 - ⁹/₂) dθ

V = ∫ ⁹/₂ dθ

V = ⁹/₂ (π/2 - 0)

V = 9π / 4

When they keep giving these problems with second-degree equations, I find that I can usually derive the answer using geometric relations alone, no calculus.

In this case the two curve boundaries define a paraboloid of revolution. The boundaries meet at a circular base of radius 3π. The vertex is (0, 0, 6), which is 6 units below the plane of the base. The volume of a paraboloid is one-half the base area times the height.

1/2(3π)(6) = 9π

The required region is only that part of the paraboloid lying in the first octant. By symmetry, that is one-fourth the volume of the paraboloid.

9π/4

I offer this only as confirmation of the answer given by Some Body.

Sketch it. Element of volume in polar coords is r dz dr dθ.

z from 4 + 2r² to 10.

r from 0 to √3 [why?]

θ from 0 to [you should know].

Integrate wrt z, then r then θ. [The θ integration can be done any time here though].