The images you've uploaded involve solving a double integral by converting Cartesian coordinates to polar coordinates and simplifying the resulting integrals. Let's go through the key steps of solving this integral.

Problem Statement:

You're given the following boundaries: $0 \leq x \leq b - y$ $0 \leq y \leq a$ The goal is to evaluate the integral: $\int \int \sqrt{a^2 - x^2} \, dx \, dy$

Step-by-Step Solution:

1. Identify the Region: The limits suggest that the region of integration lies in the first quadrant, bounded by the lines $x = 0$, $y = 0$, $x = b - y$, and $y = a$. This region is suitable for conversion into polar coordinates since the limits and the function have symmetry.

2. Convert to Polar Coordinates: In polar coordinates, the transformation from Cartesian coordinates is: $x = r \cos \theta, \quad y = r \sin \theta$ The Jacobian determinant for this transformation introduces an extra factor of $r$, so the area element $dx \, dy$ becomes $r \, dr \, d\theta$.

   The given integral then converts into: $\int_0^\frac{\pi}{2} \int_0^a r \sqrt{a^2 - r^2} \, r \, dr \, d\theta$

3. Substitute the Integral: To simplify the inner integral, let $a^2 - r^2 = t$. Differentiating both sides with respect to $r$: $-2r \, dr = dt \quad \Rightarrow \quad r \, dr = -\frac{1}{2} dt$ The limits of integration for $r$ are from $0$ to $a$, and accordingly, $t$ ranges from $a^2$ to $0$.

4. Solve the Integral: Substituting into the integral, we get: $\int_0^\frac{\pi}{2} \int_{a^2}^0 \sqrt{t} \left( -\frac{1}{2} \right) dt \, d\theta$ Solving the inner integral: $\int \sqrt{t} \, dt = \frac{2}{3} t^{3/2}$ Evaluating this from $a^2$ to 0: $\frac{2}{3} (0^{3/2} - (a^2)^{3/2}) = -\frac{2}{3} a^3$ So the integral simplifies to: $\frac{a^3}{3} \int_0^\frac{\pi}{2} d\theta = \frac{a^3}{3} \cdot \frac{\pi}{2}$

5. Final Answer: The result of the double integral is: $\frac{a^3 \pi}{6}$

This approach involves using polar coordinates to simplify the given integral, substituting appropriate limits, and evaluating the resulting integrals step by step.

Do you have any questions about this? Here are five related questions:

1. How can we verify the limits of integration when converting to polar coordinates?
2. What conditions make polar coordinates more suitable than Cartesian coordinates for integration?
3. How does the Jacobian affect the transformation of coordinates in multiple integrals?
4. Can this technique of substitution be applied to any integral with square roots?
5. What are the general steps to follow when converting double integrals from Cartesian to polar coordinates?

Tip: When solving integrals involving square roots, look for symmetry in the limits or the integrand to simplify using polar or other coordinate systems.