double integrals (pls anwer)

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

The easiest way to solve this to use a polar transformation: dx dy = r dr dθ

Then, we have z ranging from r to 10 - 2r²

10 - r - 2r² = 0

(5 + 2r)(2 - r) = 0

r = 2

r ranges from 0 to 2

As we have no restrictions on θ, we just multiply by 2π

We integrate

2π(10 - r - 2r²)r dr from 0 to 2, or 2π(10r - r² - 2r³)dr from 0 to 2, or

2π(5r² - r³/3 - r⁴/2) evaluated at 0 and 2; as the evaluation at 0 is 0, we have

2π(5*4 - 8/3 - 16/2) = 2π(20 - 8/3 - 8) = 2π(12 - 8/3) = 56/3 π