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Question Number 30543 by mondodotto@gmail.com last updated on 23/Feb/18

Commented by abdo imad last updated on 23/Feb/18

$$thech.x-\frac{1}{2}=sintgive$$$$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}4\pi (\frac{1}{2}+sint)\sqrt{1-{sin}^{2}t}costdt$$$$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}4\pi (\frac{1}{2}+sint){cos}^{2}tdt=2\pi {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{cos}^{2}tdt$$$$+4\pi {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}sint{cos}^{2}tdtbutthefunctiont\to {sintcos}^{2}tisodd$$$$\Rightarrow {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}sint{cos}^{2}tdt=0\Rightarrow I=4\pi {\int }_0^{\frac{\pi }{2}}\left(\frac{1+cos\left(2t\right)}{2}\right)dt$$$$I=2\pi {\int }_0^{\frac{\pi }{2}}dt+2\pi {\int }_0^{\frac{\pi }{2}}cos\left(2t\right)dt={\pi }^2+2\pi {\left[\frac{1}{2}sin\left(2t\right)\right]}_0^{\frac{\pi }{2}}$$$$={\pi }^2+0\Rightarrow I={\pi }^2.$$

Answered by ajfour last updated on 23/Feb/18

$$I=4\pi {\int }_{-1/2}^{3/2}(1-x)\sqrt{1-{(\frac{1}{2}-x)}^2}dx$$$$\Rightarrow 2I=4\pi {\int }_{-1/2}^{3/2}\sqrt{1-{(x-\frac{1}{2})}^2}dx$$$$I=2\pi [\frac{1}{2}(x-\frac{1}{2})\sqrt{1-{(x-\frac{1}{2})}^2}+$$$$\frac{1}{2}{\sin }^{-1}(x-\frac{1}{2})]{\mid }_{-1/2}^{3/2}$$$$I=2\pi \times \frac{1}{2}[{\sin }^{-1}1-{\sin }^{-1}(-1)]$$$$\Rightarrow I={\mathit{\pi }}^2.$$

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