Question Number 124666 by benjo_mathlover last updated on 05/Dec/20
$$\:{Find}\:{the}\:{volume}\:{that}\:{remains} \\ $$$${after}\:{the}\:{hole}\:{of}\:{radius}\:\mathrm{1}\:{bored}\: \\ $$$${through}\:{the}\:{center}\:{of}\:{a}\:{solid} \\ $$$${sphere}\:{of}\:{radius}\:\mathrm{3}. \\ $$$$\left({a}\right)\:\mathrm{18}\pi\:\:\:\:\left({b}\right)\:\frac{\mathrm{28}}{\mathrm{3}}\pi\:\:\:\:\left({c}\right)\:\mathrm{36}\pi\:\:\:\:\left({d}\right)\:\frac{\mathrm{56}\pi}{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$
Commented by ajfour last updated on 05/Dec/20
Commented by ajfour last updated on 05/Dec/20
![cap volume V. V=∫_h ^( R) πr^2 dy r=Rsin θ , y=Rcos θ = ∫_0 ^( α) π(R^2 sin^2 θ)Rsin θdθ sin α=(1/3) , cos α=((2(√2))/3) = πR^3 ∫_((2(√2))/3) ^( 1) (1−t^2 )dt ; t=cos θ = πR^3 (t−(t^3 /3))∣_((2(√2))/3) ^1 =27π[(2/3)−(((2(√2))/3)−((16(√2))/(81)))] =27π[(2/3)−((38(√2))/(81))] remaining volume = (4/3)π(27)−2π(1)^2 (2(√2))−2V =36π−4(√2)π−36π+((76(√2))/3)π V_(remaining) =((64(√2)π)/3) .](https://www.tinkutara.com/question/Q124676.png)
$${cap}\:{volume}\:{V}. \\ $$$${V}=\int_{{h}} ^{\:{R}} \pi{r}^{\mathrm{2}} {dy}\:\:\: \\ $$$${r}={R}\mathrm{sin}\:\theta\:\:\:,\:{y}={R}\mathrm{cos}\:\theta\: \\ $$$$\:\:\:=\:\int_{\mathrm{0}} ^{\:\alpha} \pi\left({R}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta\right){R}\mathrm{sin}\:\theta{d}\theta \\ $$$$\mathrm{sin}\:\alpha=\frac{\mathrm{1}}{\mathrm{3}}\:,\:\:\mathrm{cos}\:\alpha=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$$\:\:=\:\pi{R}^{\mathrm{3}} \int_{\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}} ^{\:\:\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right){dt}\:\:\:\:\:;\:\:{t}=\mathrm{cos}\:\theta \\ $$$$\:\:=\:\pi{R}^{\mathrm{3}} \left({t}−\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\right)\mid_{\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}} ^{\mathrm{1}} \\ $$$$\:\:\:=\mathrm{27}\pi\left[\frac{\mathrm{2}}{\mathrm{3}}−\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}−\frac{\mathrm{16}\sqrt{\mathrm{2}}}{\mathrm{81}}\right)\right] \\ $$$$\:\:\:=\mathrm{27}\pi\left[\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{38}\sqrt{\mathrm{2}}}{\mathrm{81}}\right]\: \\ $$$${remaining}\:{volume} \\ $$$$\:\:=\:\frac{\mathrm{4}}{\mathrm{3}}\pi\left(\mathrm{27}\right)−\mathrm{2}\pi\left(\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2}\sqrt{\mathrm{2}}\right)−\mathrm{2}{V} \\ $$$$\:=\mathrm{36}\pi−\mathrm{4}\sqrt{\mathrm{2}}\pi−\mathrm{36}\pi+\frac{\mathrm{76}\sqrt{\mathrm{2}}}{\mathrm{3}}\pi \\ $$$$\:{V}_{{remaining}} =\frac{\mathrm{64}\sqrt{\mathrm{2}}\pi}{\mathrm{3}}\:. \\ $$
Commented by benjo_mathlover last updated on 05/Dec/20
$${nothing}\:{available}\:{choice}\:{answer} \\ $$$${is}\:{correct}? \\ $$
Commented by mr W last updated on 05/Dec/20
$${yes}.\:{all}\:{answers}\:{given}\:{are}\:{wrong}. \\ $$$${i}\:{got}\:{a}\:{formula}\:{for}\:{the}\:{remaining} \\ $$$${volume}\:{which}\:{is} \\ $$$${V}_{{remain}} =\frac{\mathrm{4}\pi\left({R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}} \\ $$
Answered by mr W last updated on 05/Dec/20
$${Method}\:{I} \\ $$$${R}={radius}\:{of}\:{sphere} \\ $$$${r}={radius}\:{of}\:{hole} \\ $$$${h}=\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }=\sqrt{\mathrm{3}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${volume}\:{of}\:{sphere} \\ $$$${V}=\frac{\mathrm{4}\pi{R}^{\mathrm{3}} }{\mathrm{3}} \\ $$$${volume}\:{of}\:{cylinder}\:{with}\:{height}\:\mathrm{2}{h} \\ $$$${V}_{\mathrm{1}} =\mathrm{2}{h}\pi{r}^{\mathrm{2}} \\ $$$${volume}\:{of}\:{each}\:{cap} \\ $$$${V}_{\mathrm{2}} =\pi\left(\Delta{h}\right)^{\mathrm{2}} \left({R}−\frac{\Delta{h}}{\mathrm{3}}\right)\:{with}\:\Delta{h}={R}−{h} \\ $$$$ \\ $$$${volume}\:{of}\:{remaining}\:{sphere}\:{part} \\ $$$${V}_{{Remain}} ={V}−{V}_{\mathrm{1}} −\mathrm{2}{V}_{\mathrm{2}} \\ $$$$=\frac{\mathrm{4}\pi{R}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{2}\pi{hr}^{\mathrm{2}} −\mathrm{2}\pi\left({R}−{h}\right)^{\mathrm{2}} \frac{\left(\mathrm{2}{R}+{h}\right)}{\mathrm{3}} \\ $$$$=\frac{\mathrm{4}\pi\mathrm{3}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{2}\pi\mathrm{2}\sqrt{\mathrm{2}}×\mathrm{1}^{\mathrm{2}} −\mathrm{2}\pi\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \frac{\left(\mathrm{2}×\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{3}} \\ $$$$=\frac{\mathrm{108}−\mathrm{12}\sqrt{\mathrm{2}}−\mathrm{4}×\mathrm{27}+\mathrm{4}×\mathrm{19}\sqrt{\mathrm{2}}}{\mathrm{3}}\pi \\ $$$$=\frac{\mathrm{64}\sqrt{\mathrm{2}}\pi}{\mathrm{3}} \\ $$
Commented by mr W last updated on 05/Dec/20
Commented by mr W last updated on 05/Dec/20
![Method II r≤x≤R h(x)=(√(R^2 −x^2 )) dV=2h(x)2πx=4πx(√(R^2 −x^2 )) V_(Remain) =∫_r ^R 4πx(√(R^2 −x^2 ))dx =−2π∫_r ^R (√(R^2 −x^2 ))d(R^2 −x^2 ) =((4π)/3)[(R^2 −x^2 )^(3/2) ]_R ^r =((4π(R^2 −r^2 )^(3/2) )/3) =((4π(3^2 −1^2 )^(3/2) )/3) =((64(√2)π)/3)](https://www.tinkutara.com/question/Q124683.png)
$${Method}\:{II} \\ $$$${r}\leqslant{x}\leqslant{R} \\ $$$${h}\left({x}\right)=\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$${dV}=\mathrm{2}{h}\left({x}\right)\mathrm{2}\pi{x}=\mathrm{4}\pi{x}\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$${V}_{{Remain}} =\int_{{r}} ^{{R}} \mathrm{4}\pi{x}\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx} \\ $$$$=−\mathrm{2}\pi\int_{{r}} ^{{R}} \sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} }{d}\left({R}^{\mathrm{2}} −{x}^{\mathrm{2}} \right) \\ $$$$=\frac{\mathrm{4}\pi}{\mathrm{3}}\left[\left({R}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{{R}} ^{{r}} \\ $$$$=\frac{\mathrm{4}\pi\left({R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}} \\ $$$$=\frac{\mathrm{4}\pi\left(\mathrm{3}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}} \\ $$$$=\frac{\mathrm{64}\sqrt{\mathrm{2}}\pi}{\mathrm{3}} \\ $$
Commented by benjo_mathlover last updated on 05/Dec/20
$${typo}\:{sir}.\:\frac{\mathrm{4}\pi×\mathrm{8}\sqrt{\mathrm{8}}}{\mathrm{3}}\:=\:\frac{\mathrm{64}\pi\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$