https://file.aiquickdraw.com/mathbot/file/1726391708324rxs74j85.jpg

The image shows a double integral of the form:

\[
\int_0^a \int_0^{\sqrt{a^2 - y^2}} dx \, dy
\]

This is an expression for calculating the area of a quarter circle in the first quadrant, with a radius \(a\). Here's how we approach solving this:

### Step 1: Solve the inner integral (with respect to \(x\))

\[
\int_0^{\sqrt{a^2 - y^2}} dx
\]

This is a simple integration:

\[
\int_0^{\sqrt{a^2 - y^2}} dx = \sqrt{a^2 - y^2}
\]

### Step 2: Substitute this result into the outer integral

Now, the problem reduces to:

\[
\int_0^a \sqrt{a^2 - y^2} \, dy
\]

This is a standard integral, which represents the area of a quarter circle. The solution is:

\[
\frac{1}{4} \pi a^2
\]

### Conclusion:

The result of the double integral is \( \frac{1}{4} \pi a^2 \), which corresponds to the area of a quarter circle with radius \(a\).

Would you like more details on this solution or need further clarifications?

### Follow-up questions:
1. How can we apply polar coordinates to simplify double integrals over circular regions?
2. How would the limits of integration change if this integral described a semicircle instead of a quarter circle?
3. What is the general method for finding the area of a circle using double integrals?
4. Can this type of integral be used to compute volumes as well?
5. How would the solution change if the radius of the circle were different in the \(x\)- and \(y\)-directions?

### Tip:
For integrals involving circular regions, switching to polar coordinates often simplifies the computation.

Evaluate the double integral \( \int_0^a \int_0^{\sqrt{a^2 - y^2}} dx \, dy \).

Learn how to solve a double integral representing the area of a quarter circle with radius a. This problem covers integral calculus and geometric concepts, including the evaluation of \( \int_0^a \int_0^{\sqrt{a^2 - y^2}} dx \, dy \).

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