To solve this problem, we need to convert the given integral into polar coordinates and evaluate it. The integral is as follows:

$\int_0^1 \int_y^{\sqrt{2y - y^2}} 4 \, dx \, dy$

Step 1: Interpret the Region in Cartesian Coordinates

The integral is defined over the region bounded by $y$ and $\sqrt{2y - y^2}$ in the $x$-direction, and from $y = 0$ to $y = 1$ in the $y$-direction. To convert this to polar coordinates, let's analyze the boundary $x = \sqrt{2y - y^2}$.

Rewrite $x = \sqrt{2y - y^2}$ as follows: $x^2 = 2y - y^2$ $x^2 + y^2 = 2y$ $x^2 + y^2 - 2y = 0$ Complete the square: $x^2 + (y - 1)^2 = 1$

This equation represents a circle of radius 1 centered at $(0, 1)$ in Cartesian coordinates.

Step 2: Set up the Integral in Polar Coordinates

In polar coordinates:

• $x = r \cos \theta$
• $y = r \sin \theta$
• The area element $dx \, dy = r \, dr \, d\theta$

The circle centered at $(0, 1)$ with radius 1 implies that $r$ ranges from 0 to 2\cosθ