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Question Number 43252 by ajfour last updated on 08/Sep/18

Commented by MJS last updated on 08/Sep/18

$$\mathrm{circle}{x}^2+y^2=1$$$$y=\sqrt{1-x^2}$$$$\mathrm{parabola}x={ay}^2-1$$$$y=\sqrt{a(x+1)};a>0$$$$\mathrm{circle}\cap \mathrm{parabola}$$$$1-x^2=a(x+1)$$$$x^2+ax+(a-1)=0$$$$x_1=-1\left[\mathrm{trivial}\right]$$$$x_2=1-a$$$$y_2=\sqrt{a(2-a)}$$$$$$$$\frac{\mathrm{area}\mathrm{above}\mathrm{parabola}}{\mathrm{area}\mathrm{below}\mathrm{parabola}}=2$$$$\frac{\underset{-1}{\stackrel{1-a}{\int }}\sqrt{1-x^2}dx-\underset{-1}{\stackrel{1-a}{\int }}\sqrt{a(x+1)}dx}{\underset{-1}{\stackrel{1-a}{\int }}\sqrt{a(x+1)}dx+\underset{1-a}{\stackrel{1}{\int }}\sqrt{1-x^2}dx}=2$$$$\frac{3}{2}(\arcsin (1-a)+(1-a)\sqrt{a(2-a)})-2\sqrt{a{(2-a)}^3}-\frac{\pi }{4}=0$$$$a\approx .0802770$$$$x_2\approx .919723$$$$y_2\approx .392568$$$$\alpha =2\arctan \frac{y_2}{x_2}\approx 46.2288°$$

Commented by ajfour last updated on 09/Sep/18

$$thankyouSir.$$

Answered by MrW3 last updated on 08/Sep/18

$$R=radius$$$$2\times \frac{2}{3}\times R\sin \frac{\alpha }{2}\times (R+R\cos \frac{\alpha }{2})+\frac{R^2}{2}(\alpha -\sin \alpha )=\frac{\pi R^2}{3}$$$$\Rightarrow 8\sin \frac{\alpha }{2}+\sin \alpha +3\alpha =2\pi$$$$\Rightarrow \alpha \approx 0.8068rad=46.23°$$

Commented by MrW3 last updated on 09/Sep/18

Commented by MrW3 last updated on 09/Sep/18

$$A_1=\frac{2}{3}\times ab=\frac{2}{3}\times R\sin \frac{\alpha }{2}\times (R+R\cos \frac{\alpha }{2})$$$$A_2=\frac{R^2}{2}(\alpha -\sin \alpha )$$$$A_{shaded}=2A_1+A_2=\frac{\pi R^2}{3}$$

Commented by ajfour last updated on 09/Sep/18

$$straightandwiseSir,thanks.$$

Commented by MrW3 last updated on 09/Sep/18

$$thankyousir.$$$$whatabouttheresultifthequestion$$$$isextendedtospacialvolume?$$

Commented by MrW3 last updated on 09/Sep/18

$$yesasphereisdividedintotwoequal$$$$parts.$$

Commented by ajfour last updated on 09/Sep/18

$$thentherearetworegions,not$$$$three!$$

Answered by ajfour last updated on 09/Sep/18

$$3Dextension:$$$$sphere:r^2+{(z-R)}^2=R^2$$$$paraboloid:z={Ar}^2$$$$letA(a,{Aa}^2)$$$$\Rightarrow a^2+{({Aa}^2-R)}^2=R^2$$$$\Rightarrow {Aa}^2=R+\sqrt{R^2-a^2}.....\left(i\right)$$$$or\frac{{Aa}^2}{R}=1+\sqrt{1-{\left(\frac{a}{R}\right)}^2}...{\left(i\right)}^{\ast }$$$${\int }_0^a(z_{sphere}-z_{para})\left(2\pi rdr\right)=\frac{2\pi R^3}{3}$$$$\Rightarrow {\int }_0^a[R+\sqrt{R^2-r^2}-{Ar}^2]rdr=\frac{R^3}{3}$$$$....\left(ii\right)$$$$\frac{a^{2}R}{2}-\frac{{(R^2-a^2)}^{3/2}}{3}+\frac{R^3}{3}-\frac{{Aa}^4}{4}=\frac{R^3}{3}$$$$\frac{a}{R}=x=\sin \frac{\alpha }{2}$$$$\Rightarrow 6x^2-4{(1-x^2)}^{3/2}=3x^2(1+\sqrt{1-x^2})$$$$\Rightarrow 6\sin {}^2\frac{\alpha }{2}-4\cos {}^3\frac{\alpha }{2}=3\left(\sin {}^2\frac{\alpha }{2}\right)(1+\cos \frac{\alpha }{2})$$$$let\cos \frac{\alpha }{2}=p$$$$\Rightarrow 6(1-p^2)-4p^3=3(1-p^2)(1+p)$$$$\Rightarrow p^3+3p^2+3p+1=4$$$$\Rightarrow {(p+1)}^3=4$$$$\mathit{p}=\sqrt[3]{4}-1$$$$\mathit{\alpha }={2\cos }^{-1}\mathit{p}={2\cos }^{-1}(\sqrt[3]{4}-1)$$$$\approx 108.0545.$$

Commented by MrW3 last updated on 09/Sep/18

$$wow,suchacleansolution!thankyou$$$$somuch!$$

Answered by MrW3 last updated on 09/Sep/18

$$3Dvolume:$$$$volumeofsphere=\frac{4\pi R^3}{3}$$$$volumeofparaboloid=\frac{\pi a^{2}b}{2}=\frac{\pi }{2}{\left(R\sin \frac{\alpha }{2}\right)}^2(R+R\cos \frac{\alpha }{2})$$$$=\frac{\pi R^3}{2}{\sin }^2\frac{\alpha }{2}(1+\cos \frac{\alpha }{2})$$$$$$$$volumeofcap=\pi h^2(R-\frac{h}{3})=\pi {(R-R\cos \frac{\alpha }{2})}^2(R-\frac{R-R\cos \frac{\alpha }{2}}{3})$$$$=\frac{\pi }{3}{R}^3{(1-\cos \frac{\alpha }{2})}^2(2+\cos \frac{\alpha }{2})$$$$$$$$\frac{\pi R^3}{2}{\sin }^2\frac{\alpha }{2}(1+\cos \frac{\alpha }{2})+\frac{\pi }{3}{R}^3{(1-\cos \frac{\alpha }{2})}^2(2+\cos \frac{\alpha }{2})=\frac{1}{2}\times \frac{4\pi R^3}{3}$$$${3\sin }^2\frac{\alpha }{2}(1+\cos \frac{\alpha }{2})+2{(1-\cos \frac{\alpha }{2})}^2(2+\cos \frac{\alpha }{2})=4$$$$3(1-{\cos }^2\frac{\alpha }{2})(1+\cos \frac{\alpha }{2})+2{(1-\cos \frac{\alpha }{2})}^2(2+\cos \frac{\alpha }{2})=4$$$$(1-\cos \frac{\alpha }{2})({\cos }^2\frac{\alpha }{2}+4\cos \frac{\alpha }{2}+7)=4$$$$\Rightarrow \cos \frac{\alpha }{2}\approx 0.5874$$$$\Rightarrow \alpha \approx 108.05°$$

Commented by ajfour last updated on 09/Sep/18

$$yessir.Thanks.$$

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