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 28427 by abdo imad last updated on 25/Jan/18

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

Commented by abdo imad last updated on 27/Jan/18

$$letputI=\int {\int }_D\sqrt{2-x^2-y^2}letusethech.x=rcos\theta and$$$$y=rsin\theta x^2+y^2⩽\sqrt{2}\Rightarrow 0<r^2⩽\sqrt{2}\Rightarrow 0<r⩽{}^4\sqrt{2}$$$$I=\int {\int }_{0<r⩽{}^4\sqrt{2}and0⩽\theta ⩽2\pi }\sqrt{2-r^2}rdrd\theta$$$$I=\left({\int }_0^{4_{\sqrt{2}}}r\sqrt{2-r^2}dr\right)\left({\int }_0^{2\pi }\theta \right)=2\pi {\int }_0^{4_{\sqrt{2}}}r\sqrt{2-r^2}dr$$$$=-\frac{2\pi }{3}{\int }_0^{4_{\sqrt{2}}}3(-r){(2-r^2)}^{\frac{1}{2}}dr$$$$=-\frac{2\pi }{3}{\left[{(2-r^2)}^{\frac{3}{2}}\right]}_0^{4_{\sqrt{2}}}=\frac{-2\pi }{3}({(2-{\left({}^4\sqrt{2)}\right)}^2)}^{\frac{3}{2}}-2^{\frac{3}{2}})$$$$=\frac{-2\pi }{3}((2-\sqrt{2)}{}^{\frac{3}{2}}-2\sqrt{2})$$$$=\frac{-2\pi }{3}\left((2-\sqrt{2)}\sqrt{2-\sqrt{2}}-2\sqrt{2})\right).$$

Answered by ajfour last updated on 27/Jan/18

$$I={\int }_0^{2\pi }\left[\frac{1}{2}{\int }_0^{\sqrt{2}}\sqrt{2-\left(r^2\right)}d\left(r^2\right)\right]d\theta$$$$=\left[\frac{2\pi (2-r^2)\sqrt{2-r^2}}{3}\right]{\mid }_{r^2=\sqrt{2}}^0$$$$I=\frac{2\pi }{3}[2\sqrt{2}-(2-\sqrt{2})\sqrt{2-\sqrt{2}}].$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com