Evaluation of a Double Integral in Cartesian or Polar Coordinates

Hello. Today we are going to evaluate a double integral. The integral is given in Cartesian coordinates, but looking at the integrand and the limits, we will see that switching to polar coordinates makes it much easier.

Let's first write down the integral as it is described. It is the integral from y equals zero to one, and the integral from x equals the square root of three y to the square root of four minus y squared, of the square root of x squared plus y squared, d x d y.

We need to identify the region of integration. The limits for x are between the line x equals square root of three times y and the circular arc x equals the square root of four minus y squared.

Let's visualize this region. The circle has a radius of two. The line makes an angle with the x-axis. Since y equals one over square root three times x, the tangent of the angle theta is one over square root three.

In polar coordinates, x squared plus y squared equals r squared, so the integrand becomes r. The differential area element d x d y becomes r d r d theta.

Now for the limits. The region is bounded by the x-axis where theta is zero, and the line y equals one over square root three x, which corresponds to theta equals pi over six.

The radial distance r goes from the origin out to the circle of radius two.