Uppgift 6

Med polära koordinater $x=r\cos t$, $y=r\sin t$ erhålles
$$\begin{aligned} \iint_Dx^2\,dx\,dy &=\int_0^{\sqrt2}\left(\int_{-\pi/4}^{3\pi/4}r^2\cos^2t\cdot r\,dt\right)dr\\ &=\int_0^{\sqrt2}r^3\,dr\cdot\int_{-\pi/4}^{3\pi/4}\frac{1+\cos2t}2\,dt\\ &=\left[\frac{r^4}4\right]_0^{\sqrt2}\cdot\left[\frac t2+\frac{\sin2t}4\right]_{-\pi/4}^{3\pi/4}\\ &=\frac44\left(\left(\frac{3\pi}8-\frac14\right)-\left(-\frac\pi8-\frac14\right)\right)=\frac\pi2. \end{aligned}$$