We are asked to solve the double integral over the region $D$ defined by the following inequalities:

1. $x^2 + y^2 \geq 2y$
2. $x + y^2 \leq 2x$
3. $y \geq 0$

Step 1: Understand the Region $D$

The first task is to interpret the constraints and find the region $D$ over which the double integral will be evaluated. Let's analyze each inequality one by one.

First inequality: $x^2 + y^2 \geq 2y$

We can rewrite this as: $x^2 + y^2 - 2y \geq 0$ Complete the square on the $y$-terms: $x^2 + (y - 1)^2 \geq 1$ This represents the region outside (or on the boundary of) a circle centered at $(0, 1)$ with radius 1.

Second inequality: $x + y^2 \leq 2x$

Rearranging this inequality: $y^2 \leq x$ This describes the region to the right of the parabola $y^2 = x$, i.e., the region where $x \geq y^2$.

Third inequality: $y \geq 0$

This simply restricts the region to the upper half-plane, where $y \geq 0$.

Step 2: Sketch the Region $D$

The region $D$ is the area where all the above inequalities hold simultaneously:

• The points are outside or on the boundary of the circle $x^2 + (y - 1)^2 \geq 1$.
• The points lie to the right of the parabola $x = y^2$.
• The points are in the upper half-plane, i.e., $y \geq 0$.

Step 3: Set Up the Double Integral

We need to set up the limits for the double integral based on the described region. To do so, let’s examine the geometry of the region more closely. We need to express $x$ and $y$ in such a way that they follow the constraints of the region.

Let’s consider the possible range for $y$. Since $y \geq 0$ and the first inequality involves a circle centered at $(0,1)$, the values of $y$ should start from 0 and extend upwards until the circle's boundary. The exact range for $x$ would depend on the values of $y$, constrained by the second inequality $x \geq y^2$.

The limits for $x$:

For each fixed value of $y$, the value of $x$ ranges from $y^2$ (from the parabola) to the right boundary of the circle, which can be determined by solving the equation $x^2 + (y - 1)^2 = 1$ for $x$: $x = \sqrt{1 - (y - 1)^2}$

Thus, the double integral becomes: $\int_0^1 \int_{y^2}^{\sqrt{1 - (y - 1)^2}} dx \, dy$

Step 4: Evaluate the Integral

1. Integrate with respect to $x$ first: $\int_{y^2}^{\sqrt{1 - (y - 1)^2}} dx = \sqrt{1 - (y - 1)^2} - y^2$

2. Integrate with respect to $y$: Now, we need to integrate this expression with respect to $y$ from 0 to 1. This will give us the value of the double integral over the region $D$.

Final Answer:

The double integral simplifies to: $\int_0^1 \left( \sqrt{1 - (y - 1)^2} - y^2 \right) dy$

Now, this integral can be computed numerically or by using appropriate substitution methods.

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Would you like to go over the steps for computing this integral in more detail or need help with the numerical evaluation? Let me know if you'd like further explanations.

Related Questions:

1. How do we evaluate integrals involving square roots like $\sqrt{1 - (y - 1)^2}$?
2. Can the given double integral be simplified using polar coordinates?
3. How do the inequalities $x^2 + y^2 \geq 2y$ and $x + y^2 \leq 2x$ influence the shape of the region?
4. What is the geometric interpretation of the inequality $x^2 + (y - 1)^2 \geq 1$?
5. How do you handle the numerical evaluation of an integral with non-trivial bounds?

Tip:

When facing integrals over complex regions, always start by clearly understanding the boundaries first. A good sketch of the region helps immensely when setting up the limits for the integral.