Evaluate ∫∫(x2 ​​​​​​​- y2)xdx dy over the area in the first quadrant bounded by the circle x2 + y2 =a2.

Answer: \(\dfrac{a^{5}}{15}\)

The region is the quarter circle \(x^{2}+y^{2}\le a^{2}\), \(x\ge0\), \(y\ge0\). Change to polar coordinates: \(x = r\cos\theta\), \(y = r\sin\theta\), \(dx\,dy = r\,dr\,d\theta\), with \(0\le r\le a\), \(0\le\theta\le\pi/2\).

The integrand becomes

\[(x^{2}-y^{2})x = r^{2}\cos2\theta\cdot r\cos\theta = r^{3}\cos2\theta\cos\theta\]

So

\[I = \int_{0}^{\pi/2}\!\!\int_{0}^{a} r^{4}\cos2\theta\cos\theta\,dr\,d\theta\]

\[= \frac{a^{5}}{5}\int_{0}^{\pi/2}\cos2\theta\cos\theta\,d\theta\]

Using \(2\cos A\cos B = \cos(A+B)+\cos(A-B)\):

\[\int_{0}^{\pi/2}\cos2\theta\cos\theta\,d\theta = \frac{1}{2}\int_{0}^{\pi/2}(\cos3\theta+\cos\theta)\,d\theta\]

\[= \frac{1}{2}\left[\frac{\sin3\theta}{3}+\sin\theta\right]_{0}^{\pi/2} = \frac{1}{2}\left(-\frac{1}{3}+1\right) = \frac{1}{3}\]

\[I = \frac{a^{5}}{5}\times\frac{1}{3} = \frac{a^{5}}{15}\]

Check: \(\iint x^{3}\,dA = \frac{a^{5}}{5}\cdot\frac{2}{3}\) and \(\iint xy^{2}\,dA = \frac{a^{5}}{5}\cdot\frac{1}{3}\); their difference is \(\frac{a^{5}}{15}\).