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

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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...Show more

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....Show more

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]....Show more