Question

evaluate the double integrals of xy dx dy over the positive quadrant of the circle x^2+y^2=a^2

Answer 1

by using above method u can solve it
hope it helps..

Answer 2

Given:

A double integral.

To find:

Evaluate the double integral over the positive quadrant of the circle.

Solution:

$\begin{array}{l}\iint x y d x d y=\int_{x=0}^{a}\left[\int_{y=0}^{\sqrt{a^{2}-x^{2}}} x y d y\right] d x \\\\\left\\because x^{2}+y^{2}=a^{2} \\\\\Rightarrow y^{2}=a^{2}-x^{2} \\\\y=\sqrt{a^{2}-x^{2}} \\\\=\int_{0}^{a} x \cdot\left[\frac{y^{2}}{2}\right]_{0}^{\sqrt{a^{2}-x^{2}}} d x \\\\=\int_{0}^{a} x\left(\frac{a^{2}-x^{2}}{2}\right) d x \\\\=\frac{1}{2} \int_{0}^{a}\left(-x^{3}\right+ a^{2} x) d x \\\\=\frac{1}{2}\left[a^{2} \frac{x^{2}}{2}-\frac{x^{4}}{4}\right]_{0}^{a} = \frac{a^{4}}{8} \end$

Therefore, $\frac{a^{4}}{8}$ is the required solution.