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Question Number 207170 by hardmath last updated on 08/May/24

$$\mathrm{Find}:{\int }_0^4\sqrt{16-{\mathrm{x}}^2}\mathrm{dx}=?$$

Commented by MM42 last updated on 08/May/24

$$x=4sin\theta \Rightarrow dx=4cos\theta d\theta$$$${\int }_0^{\frac{\pi }{2}}16{cos}^2\theta d\theta ={\int }_0^{\frac{\pi }{2}}8(1+cos2\theta )d\theta$$$${=8(\theta +\frac{1}{2}sin2\theta )]}_0^{\frac{\pi }{2}}$$$$=4\pi ✓$$$$$$

Commented by hardmath last updated on 08/May/24

$$\mathrm{Answer}:4\pi$$

Commented by MM42 last updated on 08/May/24

$$★$$$${\int }_0^a\sqrt{a^2-x^2}\stackrel{x=asin\theta }{=}{\int }_0^{\frac{\pi }{2}}{a}^2{cos}^2\theta d\theta$$$$=\left(\frac{\pi a^2}{4}\right)✓$$$$$$

Commented by hardmath last updated on 10/May/24

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}$$

Answered by Skabetix last updated on 09/May/24

$$={\int }_0^4\sqrt{4^2-x^2}$$$$Or{\int }_0^a\sqrt{a^2-x^2}dx=\frac{a^2}{4}\pi$$$$Sohere{\int }_0^4\sqrt{4^2-x^2}dx=\frac{4^2}{4}\pi =\frac{16}{4}\pi =4\pi$$

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