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=∫_0 ^(2a) (√(2ax−x^2 ))dx=∫_0 ^(2a) (√(a^2 −a^2 +2ax−x^2 ))dx =∫_0 ^(2a) (√(a^2 −(x−a)^2 ))dx=∫_0 ^(2a) a(√(1−(((x−a)/a))^2 ))dx t=((x−a)/a)⇒adt=dx A=a^2 ∫_(−1) ^1 (√(1−t^2 ))dt t=sin(u)⇒dt=cos(u)du⇔(√(1−t^2 ))dt=(√(1−sin^2 (u)))cos(u)du (√(1−t^2 ))dt=cos^2 (u)du A=a^2 ∫_((−π)/2) ^(π/2) cos^2 (u)du=a^2 ∫_((−π)/2) ^(π/2) ((cos(2u)+1)/2)du A=a^2 [(1/4)sin(2u)+(1/2)u]_(−(π/2)) ^(π/2) =(a^2 /2)((π/2)+(π/2)) A=((a^2 π)/2)](https://www.tinkutara.com/question/Q17500.png)
$${A}=\int_{\mathrm{0}} ^{\mathrm{2}{a}} \sqrt{\mathrm{2}{ax}−{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{2}{a}} \sqrt{{a}^{\mathrm{2}} −{a}^{\mathrm{2}} +\mathrm{2}{ax}−{x}^{\mathrm{2}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}{a}} \sqrt{{a}^{\mathrm{2}} −\left({x}−{a}\right)^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{2}{a}} {a}\sqrt{\mathrm{1}−\left(\frac{{x}−{a}}{{a}}\right)^{\mathrm{2}} }{dx} \\ $$$${t}=\frac{{x}−{a}}{{a}}\Rightarrow{adt}={dx} \\ $$$${A}={a}^{\mathrm{2}} \int_{−\mathrm{1}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt} \\ $$$${t}={sin}\left({u}\right)\Rightarrow{dt}={cos}\left({u}\right){du}\Leftrightarrow\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt}=\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left({u}\right)}{cos}\left({u}\right){du} \\ $$$$\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt}={cos}^{\mathrm{2}} \left({u}\right){du} \\ $$$${A}={a}^{\mathrm{2}} \int_{\frac{−\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}} \left({u}\right){du}={a}^{\mathrm{2}} \int_{\frac{−\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left(\mathrm{2}{u}\right)+\mathrm{1}}{\mathrm{2}}{du} \\ $$$${A}={a}^{\mathrm{2}} \left[\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{u}\right)+\frac{\mathrm{1}}{\mathrm{2}}{u}\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} =\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right) \\ $$$${A}=\frac{{a}^{\mathrm{2}} \pi}{\mathrm{2}} \\ $$