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Question Number 105476 by ajfour last updated on 29/Jul/20

Commented by ajfour last updated on 29/Jul/20

$$Findthemassofthesquareplate$$$$OABC,ifmassdensityperunit$$$$areaisproportionaltothe$$$$distancefromcornerO,given$$$$by\mathit{\rho }=\mathit{k}\mathit{r}/\mathit{a}.$$

Answered by mr W last updated on 29/Jul/20

$$M=2{\int }_0^{\frac{\pi }{4}}{\int }_0^{\frac{a}{\cos \theta }}\rho rdrd\theta$$$$=\frac{2k}{a}{\int }_0^{\frac{\pi }{4}}{\int }_0^{\frac{a}{\cos \theta }}{r}^{2}drd\theta$$$$=\frac{2a^{2}k}{3}{\int }_0^{\frac{\pi }{4}}\frac{d\theta }{{\cos }^3\theta }$$$$=\frac{2a^{2}k}{3}{\int }_0^{\frac{\pi }{4}}\frac{d\left(\sin \theta \right)}{{(1-{\sin }^2\theta )}^2}$$$$=\frac{a^{2}k}{6}{[\ln \frac{1+\sin \theta }{1-\sin \theta }-\frac{1}{1+\sin \theta }+\frac{1}{1-\sin \theta }]}_0^{\frac{\pi }{4}}$$$$=\frac{a^{2}k}{6}[\ln \frac{1+\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}}-\frac{1}{1+\frac{1}{\sqrt{2}}}+\frac{1}{1-\frac{1}{\sqrt{2}}}]$$$$=\frac{a^{2}k}{3}[\ln (\sqrt{2}+1)+\sqrt{2}]$$

Commented by ajfour last updated on 29/Jul/20

$$Fantastic,Sir.Thanksalot.$$

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