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Question Number 27595 by abdo imad last updated on 10/Jan/18

$$find\int {\int }_{D}xy\sqrt{x^2+y^2}dxdywith$$$$D=\{(x,y)\in R^2/x^2+2y^2⩽1,x⩾0,y⩾0\}$$

Commented by abdo imad last updated on 15/Jan/18

$$weusethepolarcoordinateletusethech.x=rcos\theta$$$$andy=\frac{1}{\sqrt{2}}sin\theta duetothediffeomorphismewemusthave$$$$0⩽\theta ⩽\frac{\pi }{2}and0⩽r⩽1$$$$I=\int {\int }_{D}xy\sqrt{x^2+2y^2}dxdy=\int {\int }_w\mathrm{\Phi }of/j_{\mathrm{\Phi }}/drd\theta$$$$(r,\theta )-(f_1(r,\theta ),f_2(r,\theta ))=\left(xy\right)=(rcos\theta ,\frac{r}{\sqrt{2}}sin\theta )$$$$M_j=\left({}_{\frac{1}{\sqrt{2}}sin\theta }^{cos\theta }{}_{\frac{r}{\sqrt{2}}cos\theta }^{-rsin\theta }\right)$$$$I=\int {\int }_{0⩽r⩽1and0⩽\theta ⩽\frac{\pi }{2}}rcos\theta .\frac{r}{\sqrt{2}}sin\theta r.\frac{r}{\sqrt{2}}drd\theta$$$$I=\frac{1}{2}{\int }_{0^{}}^1{r}^4{\int }_0^{\frac{\pi }{2}}cos\theta sin\theta d\theta =\frac{1}{4}{\left[\frac{1}{5}{r}^5\right]}_0^1{\int }_0^{\frac{\pi }{2}}sin\left(2\theta \right)d\theta$$$$=\frac{1}{40}{[-cos\left(2\theta \right)]}_0^{\frac{\pi }{2}}=\frac{1}{20}.$$

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