Question

Set up a double integral in polar coordinates that represents the volume of the solid inside the sphere x² + y² + ² = 4 and above the cone z = √(3x² + 3y²)​

Answer 1

The sphere and cone meet in a cylinder,

`x^2 + y^2 + \left(√(3x^2+3y^2)\right)^2 = 4x^2+4y^2 = 4 \implies x^2+y^2=1`

Then the volume of the region is given in Cartesian coordinates by

`\displaystyle \int_(-1)^1 \int_(-√(1-x^2))^(√(1-x^2)) \int_(√(3x^2+3y^2))^(√(4-x^2-y^2)) dz \, dy \, dx`

which reduces to the double integral

`\displaystyle \int_(-1)^1 \int_(-√(1-x^2))^(√(1-x^2)) \left(√(4-x^2-y^2) - √(3x^2+3y^2)\right) \, dy \, dx`

Now convert to polar coordinates. We're integrating over the unit circle, so the integral in polar is

`\displaystyle \int_0^(2\pi) \int_0^1 \left(√(4-r^2) - √(3r^2)\right) r \, dr \, d\theta = \boxed{\int_0^(2\pi) \int_0^1 \left(r √(4-r^2) - \sqrt3\,r^2\right) \, dr \, d\theta}`