Question

How do you Evaluate the integral by changing to cylindrical coordinates?

`\int_(-2)^2 \int_(-\sqrt(4-y^(2)))^(\sqrt(4-y^(2)))int_(sqrt(x^2+y^2))^2 (xz) dzdxdy`

Answer 1

`\int_(-2)^2 \int_(-\sqrt(4-y^(2)))^(\sqrt(4-y^(2)))int_(sqrt(x^2+y^2))^2 (xz) \ dz \ dx\ dy`

The limit on the innermost integral defines the volume inside the cone, vertex at origin, concentric with z-axis, radius 2 at `z = 2`.

The other limits lie on or outside this cone and so they can be simplified as follows in cylindrical:

`= \int_(0)^(2pi) \int_0^2 int_(r)^2 underbrace(r cos theta \ z)_(= xz) * underbrace(r \ dz\ dr \ d theta)_(equiv dA)`

• Note that for every positive `xz` isnside the cone, there is a corresponding negative `xz` value, so one would expect the integration to result in zero.

`= \int_(0)^(2pi) \int_0^2 [ r^2 cos theta z^2/2 ]_(r)^2 \ dr \ d theta`

`= \int_(0)^(2pi) \int_0^2 2 r^2 cos theta - r^4/2 cos theta \ dr \ d theta`

`= \int_(0)^(2pi) [ 2/3 r^3 cos theta - r^5/10 cos theta ]_0^2 \ d theta`

`=32/15 \int_(0)^(2pi) cos theta \ d theta = 0`