Question Number 167512 by MikeH last updated on 18/Mar/22
$$\int_{\mathrm{0}} ^{{r}} \sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:{dx} \\ $$
Commented by mr W last updated on 18/Mar/22
$$={area}\:{of}\:{a}\:{quater}\:{circle}=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{4}} \\ $$
Commented by MikeH last updated on 18/Mar/22
$$\mathrm{thanks}\:\mathrm{sir}.\:\mathrm{please}\:\mathrm{can}\:\mathrm{you}\: \\ $$$$\mathrm{look}\:\mathrm{at}\:\mathrm{Q167520} \\ $$
Answered by LEKOUMA last updated on 18/Mar/22
![Posons x=rsin t ⇒ dt=rcos tdt si x=0 ⇒ t=0 x=r ⇒ t=(( π)/2) =∫_0 ^(π/2) (√(r^2 −r^2 sin^2 t))×rcos tdt =∫_0 ^(π/2) (√(r^2 (1−sin^2 t)))×rcos tdt =∫_0 ^(π/2) r^2 cos^2 tdt=r^2 ∫_0 ^(π/2) cos^2 tdt =r^2 ∫_0 ^(π/2) (((1+cos 2t)/2))dt=(1/2)r^2 (∫_0 ^(π/2) dt+∫_0 ^(π/2) cos 2tdt) =(1/2)r^2 ([t]_0 ^(π/2) +[(1/2)sin 2t]_0 ^(π/2) ) =(1/2)r^2 ((π/2))=((r^2 π)/4)](Q167514.png)
$${Posons}\:\:{x}={r}\mathrm{sin}\:{t}\:\Rightarrow\:{dt}={r}\mathrm{cos}\:{tdt} \\ $$$${si}\:{x}=\mathrm{0}\:\Rightarrow\:{t}=\mathrm{0} \\ $$$${x}={r}\:\Rightarrow\:{t}=\frac{\:\pi}{\mathrm{2}} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{r}^{\mathrm{2}} −{r}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} {t}}×{r}\mathrm{cos}\:{tdt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{r}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {t}\right)}×{r}\mathrm{cos}\:{tdt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {r}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} {tdt}={r}^{\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\:^{\mathrm{2}} {tdt} \\ $$$$={r}^{\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{t}}{\mathrm{2}}\right){dt}=\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dt}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\:\mathrm{2}{tdt}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \left(\left[{t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\right)=\frac{{r}^{\mathrm{2}} \pi}{\mathrm{4}} \\ $$
Commented by MikeH last updated on 18/Mar/22
$$\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot} \\ $$