Question Number 228414 by fantastic2 last updated on 11/Apr/26
Commented by fantastic2 last updated on 11/Apr/26
$${suppose}\:{the}\:{shape}\:{of}\:{the}\:{red}\:{jelly}\:{is} \\ $$$$\mathrm{3}{d}\:{version}\:{of}\:{eq}\:{y}={b}^{\mathrm{2}} −{x}^{\mathrm{2}} \:{and}\:{it}\:{is}\:{sliced} \\ $$$${like}\:{the}\:{blue}\:{line}\:{in}\:{the}\:{picture} \\ $$$${find}\:{the}\:{volume}\:{of}\:{the}\:{small}\:{part} \\ $$
Commented by Frix last updated on 12/Apr/26
$$\mathrm{It}\:\mathrm{could}\:\mathrm{be}\:\frac{{b}^{\mathrm{4}} \pi}{\mathrm{32}}\:\mathrm{but}\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}… \\ $$
Commented by fantastic2 last updated on 12/Apr/26
$${great}! \\ $$
Answered by mr W last updated on 13/Apr/26
Commented by mr W last updated on 13/Apr/26
![(y/H)=((r/R))^2 ⇒r=R(√(y/H)) (y/H)=(b/R) ⇒b=((Ry)/H) cos θ=(b/r)=(√(y/H)) a=(√(r^2 −b^2 ))=(R/H)(√(y(H−y))) dA=R^2 (y/H) cos^(−1) (√(y/H))−((Ry)/H)×(R/H)(√(y(H−y))) dV=[R^2 (y/H) cos^(−1) (√(y/H))−((Ry)/H)×(R/H)(√(y(H−y)))]dy V=∫_0 ^H [R^2 (y/H) cos^(−1) (√(y/H))−((Ry)/H)×(R/H)(√(y(H−y)))]dy =R^2 H∫_0 ^H {(y/H)[cos^(−1) (√(y/H))−(√((y/H)(1−(y/H))))]}d((y/H)) =R^2 H∫_0 ^1 {ξ[cos^(−1) (√ξ)−(√(ξ(1−ξ)))]}dξ =((πR^2 H)/(32))](https://www.tinkutara.com/question/Q228434.png)
$$\frac{{y}}{{H}}=\left(\frac{{r}}{{R}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{r}={R}\sqrt{\frac{{y}}{{H}}} \\ $$$$\frac{{y}}{{H}}=\frac{{b}}{{R}} \\ $$$$\Rightarrow{b}=\frac{{Ry}}{{H}} \\ $$$$\mathrm{cos}\:\theta=\frac{{b}}{{r}}=\sqrt{\frac{{y}}{{H}}} \\ $$$${a}=\sqrt{{r}^{\mathrm{2}} −{b}^{\mathrm{2}} }=\frac{{R}}{{H}}\sqrt{{y}\left({H}−{y}\right)} \\ $$$${dA}={R}^{\mathrm{2}} \frac{{y}}{{H}}\:\mathrm{cos}^{−\mathrm{1}} \sqrt{\frac{{y}}{{H}}}−\frac{{Ry}}{{H}}×\frac{{R}}{{H}}\sqrt{{y}\left({H}−{y}\right)} \\ $$$${dV}=\left[{R}^{\mathrm{2}} \frac{{y}}{{H}}\:\mathrm{cos}^{−\mathrm{1}} \sqrt{\frac{{y}}{{H}}}−\frac{{Ry}}{{H}}×\frac{{R}}{{H}}\sqrt{{y}\left({H}−{y}\right)}\right]{dy} \\ $$$${V}=\int_{\mathrm{0}} ^{{H}} \left[{R}^{\mathrm{2}} \frac{{y}}{{H}}\:\mathrm{cos}^{−\mathrm{1}} \sqrt{\frac{{y}}{{H}}}−\frac{{Ry}}{{H}}×\frac{{R}}{{H}}\sqrt{{y}\left({H}−{y}\right)}\right]{dy} \\ $$$$\:\:={R}^{\mathrm{2}} {H}\int_{\mathrm{0}} ^{{H}} \left\{\frac{{y}}{{H}}\left[\mathrm{cos}^{−\mathrm{1}} \sqrt{\frac{{y}}{{H}}}−\sqrt{\frac{{y}}{{H}}\left(\mathrm{1}−\frac{{y}}{{H}}\right)}\right]\right\}{d}\left(\frac{{y}}{{H}}\right) \\ $$$$\:\:={R}^{\mathrm{2}} {H}\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\xi\left[\mathrm{cos}^{−\mathrm{1}} \sqrt{\xi}−\sqrt{\xi\left(\mathrm{1}−\xi\right)}\right]\right\}{d}\xi \\ $$$$\:\:=\frac{\pi{R}^{\mathrm{2}} {H}}{\mathrm{32}} \\ $$
Commented by fantastic2 last updated on 13/Apr/26
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Answered by fantastic2 last updated on 12/Apr/26
Commented by fantastic2 last updated on 12/Apr/26
 h=b^2 −bx x=((b^2 −h)/b) OP=((b^2 −h)/b) α=cos^(−1) ((√(b^2 −h))/b) area of green region ((2α)/(2π))πr^2 −(1/2)r^2 sin (2α) =(b^2 −h)(cos^(−1) ((√(b^2 −h))/b)−((sin (2cos^(−1) ((√(b^2 −h))/b)))/2)) volume=∫_0 ^b^2 (b^2 −h)(cos^(−1) ((√(b^2 −h))/b)−((sin (2cos^(−1) ((√(b^2 −h))/b)))/2))dh =((b^4 π)/(32)) u^3](https://www.tinkutara.com/question/Q228427.png)
$${see}\:{pic}\:{at}\:{right} \\ $$$${at}\:{y}={h}\: \\ $$$${x}=\sqrt{{b}^{\mathrm{2}} −{h}} \\ $$$$\therefore{r}=\sqrt{{b}^{\mathrm{2}} −{h}} \\ $$$${now}\:{circle}\:{formed}\:{at}\:{y}={h}\left({right}\:{pic}\right)\:{can}\:{be}\:{imagined} \\ $$$${as}\:{y}={h}\left({left}\:{pic}\right)\:{and}\:{the}\:{plane}\:{cutting}\:{the}\:{circle}\:\left({right}\:{pic}\left(\right.\right. \\ $$$${can}\:{be}\:{imagined}\:{as}\:{y}={b}^{\mathrm{2}} −{bx}\left[\frac{{x}}{{b}}+\frac{{y}}{{b}^{\mathrm{2}} }=\mathrm{1}\right]\left({left}\:{pic}\right) \\ $$$${h}={b}^{\mathrm{2}} −{bx} \\ $$$${x}=\frac{{b}^{\mathrm{2}} −{h}}{{b}} \\ $$$${OP}=\frac{{b}^{\mathrm{2}} −{h}}{{b}} \\ $$$$\alpha=\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{{b}^{\mathrm{2}} −{h}}}{{b}} \\ $$$${area}\:{of}\:{green}\:{region} \\ $$$$\frac{\mathrm{2}\alpha}{\mathrm{2}\pi}\pi{r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \mathrm{sin}\:\left(\mathrm{2}\alpha\right) \\ $$$$=\left({b}^{\mathrm{2}} −{h}\right)\left(\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{{b}^{\mathrm{2}} −{h}}}{{b}}−\frac{\mathrm{sin}\:\left(\mathrm{2cos}^{−\mathrm{1}} \frac{\sqrt{{b}^{\mathrm{2}} −{h}}}{{b}}\right)}{\mathrm{2}}\right) \\ $$$${volume}=\int_{\mathrm{0}} ^{{b}^{\mathrm{2}} } \left({b}^{\mathrm{2}} −{h}\right)\left(\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{{b}^{\mathrm{2}} −{h}}}{{b}}−\frac{\mathrm{sin}\:\left(\mathrm{2cos}^{−\mathrm{1}} \frac{\sqrt{{b}^{\mathrm{2}} −{h}}}{{b}}\right)}{\mathrm{2}}\right){dh} \\ $$$$=\frac{{b}^{\mathrm{4}} \pi}{\mathrm{32}}\:{u}^{\mathrm{3}} \\ $$
Commented by fantastic2 last updated on 13/Apr/26
