Question Number 17473 by Arnab Maiti last updated on 06/Jul/17
$$\mathrm{Prove}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\:\mathrm{2a}} \sqrt{\mathrm{2ax}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=\frac{\Pi\mathrm{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$
Answered by sma3l2996 last updated on 06/Jul/17
$${A}=\int_{\mathrm{0}} ^{\mathrm{2}{a}} \sqrt{\mathrm{2}{ax}−{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{2}{a}} \sqrt{{a}^{\mathrm{2}} −{a}^{\mathrm{2}} +\mathrm{2}{ax}−{x}^{\mathrm{2}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}{a}} \sqrt{{a}^{\mathrm{2}} −\left({x}−{a}\right)^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{2}{a}} {a}\sqrt{\mathrm{1}−\left(\frac{{x}−{a}}{{a}}\right)^{\mathrm{2}} }{dx} \\ $$$${t}=\frac{{x}−{a}}{{a}}\Rightarrow{adt}={dx} \\ $$$${A}={a}^{\mathrm{2}} \int_{−\mathrm{1}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt} \\ $$$${t}={sin}\left({u}\right)\Rightarrow{dt}={cos}\left({u}\right){du}\Leftrightarrow\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt}=\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left({u}\right)}{cos}\left({u}\right){du} \\ $$$$\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt}={cos}^{\mathrm{2}} \left({u}\right){du} \\ $$$${A}={a}^{\mathrm{2}} \int_{\frac{−\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}} \left({u}\right){du}={a}^{\mathrm{2}} \int_{\frac{−\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left(\mathrm{2}{u}\right)+\mathrm{1}}{\mathrm{2}}{du} \\ $$$${A}={a}^{\mathrm{2}} \left[\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{u}\right)+\frac{\mathrm{1}}{\mathrm{2}}{u}\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} =\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right) \\ $$$${A}=\frac{{a}^{\mathrm{2}} \pi}{\mathrm{2}} \\ $$