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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

Convert the integral \( \int_0^1 \int_y^{\sqrt{2y - y^2}} 4 \, dx \, dy \) to polar coordinates and evaluate the integral.

Learn how to convert the integral \( \int_0^1 \int_y^{\sqrt{2y - y^2}} 4 \, dx \, dy \) to polar coordinates and evaluate it. This problem involves key calculus concepts like double integrals and Cartesian to polar transformations, making it suitable for college calculus students.

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