Questions: Consider the following. The double integral of xy dA, where D is enclosed by the quarter-circle y = sqrt(25-x^2), x >= 0, and the axes Rewrite the above as an iterated integral. The integral from 0 to sqrt(25-x^2) of xy dy dx Evaluate the double integral.

Solution Steps

To evaluate the given double integral, we need to set up the integral in terms of the given region and then compute it. The region \(D\) is enclosed by the quarter-circle \(y = \sqrt{25 - x^2}\) and the axes \(x \geq 0\). We will integrate with respect to \(y\) first and then \(x\).

1. Set up the iterated integral with the given bounds.
2. Integrate the inner integral with respect to \(y\).
3. Integrate the resulting expression with respect to \(x\).

We start with the inner integral, which is given by:

\[ \int_{0}^{\sqrt{25 - x^2}} x y \, dy \]

Calculating this integral, we find:

\[ \int_{0}^{\sqrt{25 - x^2}} x y \, dy = x \cdot \frac{y^2}{2} \bigg|_{0}^{\sqrt{25 - x^2}} = x \cdot \frac{(25 - x^2)}{2} \]

Thus, the inner integral simplifies to:

\[ \frac{x(25 - x^2)}{2} \]

Next, we set up the outer integral:

\[ \int_{0}^{5} \frac{x(25 - x^2)}{2} \, dx \]

Now we evaluate the outer integral:

\[ \int_{0}^{5} \frac{x(25 - x^2)}{2} \, dx = \frac{1}{2} \int_{0}^{5} (25x - x^3) \, dx \]

Calculating this integral, we have:

\[ \frac{1}{2} \left( \frac{25x^2}{2} - \frac{x^4}{4} \right) \bigg|_{0}^{5} = \frac{1}{2} \left( \frac{25 \cdot 25}{2} - \frac{625}{4} \right) \]

This simplifies to:

\[ \frac{1}{2} \left( \frac{625}{2} - \frac{625}{4} \right) = \frac{1}{2} \left( \frac{1250}{4} - \frac{625}{4} \right) = \frac{1}{2} \cdot \frac{625}{4} = \frac{625}{8} \]

Final Answer