We are tasked with solving the given double integral by converting it into polar coordinates. Here's how to proceed:

Step 1: Analyze the given integral

The integral is:

$\int_{0}^{1} \int_{0}^{\sqrt{a^2 - y^2}} \left( x^2 + y^2 \right) \, dy \, dx$

The region of integration is bounded by:

1. $0 \leq x \leq 1$,
2. $0 \leq y \leq \sqrt{a^2 - y^2}$.

This region corresponds to a quarter of a circle in the first quadrant with radius $a$, bounded by the line $x = 1$.

Step 2: Convert to polar coordinates

In polar coordinates, we make the substitutions: $x = r \cos\theta, \quad y = r \sin\theta, \quad x^2 + y^2 = r^2, \quad dx \, dy = r \, dr \, d\theta.$

Now rewrite the bounds:

• In polar coordinates, the circular region in the first quadrant is defined by:
  • $0 \leq r \leq a$,
  • $0 \leq \theta \leq \frac{\pi}{2}$.

The integral becomes: $\int_{0}^{\frac{\pi}{2}} \int_{0}^{a} r^2 \cdot r \, dr \, d\theta = \int_{0}^{\frac{\pi}{2}} \int_{0}^{a} r^3 \, dr \, d\theta.$

Step 3: Solve the integral

First, evaluate the inner integral with respect to $r$: $\int_{0}^{a} r^3 \, dr = \left[ \frac{r^4}{4} \right]_{0}^{a} = \frac{a^4}{4}.$

Now substitute this into the outer integral: $\int_{0}^{\frac{\pi}{2}} \frac{a^4}{4} \, d\theta = \frac{a^4}{4} \int_{0}^{\frac{\pi}{2}} 1 \, d\theta = \frac{a^4}{4} \cdot \frac{\pi}{2}.$

Final Answer:

$\int_{0}^{1} \int_{0}^{\sqrt{a^2 - y^2}} \left( x^2 + y^2 \right) \, dy \, dx = \frac{\pi a^4}{8}.$

Let me know if you have any questions or would like a more detailed explanation of any step!

Related Questions:

1. What is the geometric interpretation of this integral?
2. How do the Jacobian and area element $r \, dr \, d\theta$ arise in polar coordinates?
3. How would the solution differ if the region of integration was a full circle instead of a quarter circle?
4. Can this integral be computed directly in Cartesian coordinates? If so, how?
5. What would the bounds look like in polar coordinates for other quadrants?

Tip:

When converting to polar coordinates, always rewrite the bounds carefully and ensure to include the Jacobian ($r$) in the integrand to avoid errors.