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Question Number 167512 by MikeH last updated on 18/Mar/22

$${\int }_0^r\sqrt{r^2-x^2}dx$$

Commented by mr W last updated on 18/Mar/22

$$=areaofaquatercircle=\frac{\pi r^2}{4}$$

Commented by MikeH last updated on 18/Mar/22

$$\mathrm{thanks}\mathrm{sir}.\mathrm{please}\mathrm{can}\mathrm{you}$$$$\mathrm{look}\mathrm{at}\mathrm{Q}167520$$

Answered by LEKOUMA last updated on 18/Mar/22

$$Posonsx=r\sin t\Rightarrow dt=r\cos tdt$$$$six=0\Rightarrow t=0$$$$x=r\Rightarrow t=\frac{\pi }{2}$$$$={\int }_0^{\frac{\pi }{2}}\sqrt{r^2-r^2\sin {}^{2}t}\times r\cos tdt$$$$={\int }_0^{\frac{\pi }{2}}\sqrt{r^2(1-\sin {}^{2}t)}\times r\cos tdt$$$$={\int }_0^{\frac{\pi }{2}}{r}^2\cos {}^{2}tdt=r^2{\int }_0^{\frac{\pi }{2}}\cos {}^{2}tdt$$$$=r^2{\int }_0^{\frac{\pi }{2}}\left(\frac{1+\cos 2t}{2}\right)dt=\frac{1}{2}{r}^2({\int }_0^{\frac{\pi }{2}}dt+{\int }_0^{\frac{\pi }{2}}\cos 2tdt)$$$$=\frac{1}{2}{r}^2({\left[t\right]}_0^{\frac{\pi }{2}}+{\left[\frac{1}{2}\sin 2t\right]}_0^{\frac{\pi }{2}})$$$$=\frac{1}{2}{r}^2\left(\frac{\pi }{2}\right)=\frac{r^2\pi }{4}$$

Commented by MikeH last updated on 18/Mar/22

$$\mathrm{thanks}\mathrm{a}\mathrm{lot}$$

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