Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 28160 by abdo imad last updated on 21/Jan/18

$$find\int {\int }_D\sqrt{xy}dxdywithD=\{(x,y)\in R^{}/{(x^2+y^2)}^2⩽xy\}$$

Commented by abdo imad last updated on 26/Jan/18

$$letputI=\int {\int }_D\sqrt{xy}dxdyletusehech.x=rcos\theta and$$$$y=rsin\theta \Rightarrow 0<r^4<r^{2}cos\theta sin\theta \Rightarrow 0<r^2<\frac{1}{2}sin\left(2\theta \right)$$$$sowemusthavesin\left(2\theta \right)>0\Rightarrow 0<\theta <\frac{\pi }{2}and$$$$0<r<\frac{1}{\sqrt{2}}\sqrt{sin\left(2\theta \right)}$$$$I={\int }_0^{\frac{\pi }{2}}\left({\int }_0^{\frac{1}{\sqrt{2}}\sqrt{sin\left(2\theta \right)}}\sqrt{r^{2}sin\theta cos\theta }rdr\right)d\theta$$$$=\frac{1}{\sqrt{2}}{\int }_0^{\frac{\pi }{2}}\left({\int }_0^{\frac{1}{\sqrt{2}}\sqrt{sin\left(2\theta \right)}}{r}^{2}dr\right)\sqrt{sin\left(2\theta \right)}d\theta$$$$=\frac{1}{3\sqrt{2}}{\int }_0^{\frac{\pi }{2}}{\left[r^3\right]}_0^{\frac{1}{\sqrt{2}}\sqrt{sin\left(2\theta \right)}}\sqrt{sin\left(2\theta \right)}d\theta$$$$=\frac{1}{3\sqrt{2}}{\int }_0^{\frac{\pi }{2}}\left(\frac{1}{2}sin\left(2\theta \right)\frac{1}{\sqrt{2}}\sqrt{sin\left(2\theta \right)}\right)\sqrt{sin\left(2\theta \right)})d\theta$$$$=\frac{1}{12}{\int }_0^{\frac{\pi }{2}}{sin}^2\left(2\theta \right)d\theta =\frac{1}{12}{\int }_0^{\frac{\pi }{2}}\frac{1-cos\left(4\theta \right)}{2}d\theta$$

Commented by abdo imad last updated on 26/Jan/18

$$I=\frac{\pi }{48}-\frac{1}{24}{\int }_0^{\frac{\pi }{2}}cos\left(4\theta \right)d\theta =\frac{\pi }{48}-\frac{1}{4\times 24}{\left[sin\left(4\theta \right)\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{48}-0$$$$\Rightarrow I=\frac{\pi }{48}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com