Question Number 214742 by mr W last updated on 18/Dec/24
Commented by mr W last updated on 18/Dec/24
$${find}\:{the}\:{volume}\:{of}\:{water}\:{in}\:{the} \\ $$$${cylinderical}\:{cup}. \\ $$
Commented by ajfour last updated on 18/Dec/24
https://youtu.be/pAkC2541W4Q?si=Pn7NfbYt0vSZwa-5 A video lecture of a projectile motion question by me.
Answered by ajfour last updated on 18/Dec/24
Commented by ajfour last updated on 18/Dec/24
$$\frac{{x}}{{L}}=\frac{{y}}{{R}} \\ $$$${y}=\left(\frac{{R}}{{L}}\right){x}\:\:\:\: \\ $$$$\mathrm{cos}\:\theta=\frac{{R}−{y}}{{R}}=\mathrm{1}−\frac{{x}}{{L}} \\ $$$${x}=\mathrm{0}\:\:\Leftrightarrow\:\theta=\mathrm{0}\:\:\:,\:{x}={L}\:\Leftrightarrow\:\theta=\frac{\pi}{\mathrm{2}} \\ $$$${dx}={L}\mathrm{sin}\:\theta \\ $$$$\int{dV}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \left({R}^{\mathrm{2}} \theta−{R}^{\mathrm{2}} \mathrm{sin}\:\theta\mathrm{cos}\:\theta\right)\left({L}\mathrm{sin}\:\theta\right) \\ $$$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \theta\mathrm{sin}\:\theta{d}\theta=\left(−\theta\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\frac{{V}}{{R}^{\mathrm{2}} {L}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\frac{{V}}{\pi{R}^{\mathrm{2}} {L}}=\frac{\mathrm{2}}{\mathrm{3}\pi}\:\approx\:\mathrm{0}.\mathrm{2122} \\ $$$$ \\ $$
Commented by mr W last updated on 19/Dec/24
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Answered by aleks041103 last updated on 20/Dec/24
![Let the base be on the Oxy plane and the base′s center be at x=y=z=0. Obviously then, the axis of the cyllinder is the z axis. Now, the water level plane is z=((Ly)/r) Then the water occupies : { ((x^2 +y^2 ≤r^2 )),((y≥0)),((0≤z≤(L/r)y)) :} Therefore: V=∫_(y=0) ^(y=r) ∫_(x=−(√(r^2 −y^2 ))) ^(x=(√(r^2 −y^2 ))) ∫_(z=0) ^(z=Ly/r) dzdxdy= =∫_(y=0) ^(y=r) (∫_(x=−(√(r^2 −y^2 ))) ^(x=(√(r^2 −y^2 ))) dx)(∫_(z=0) ^(z=Ly/r) dz)dy= =∫_0 ^r (L/r)2y(√(r^2 −y^2 ))dy= =(L/r)∫_0 ^r^2 (√(r^2 −t))dt= =(L/r)[−(2/3)(r^2 −x)^(3/2) ]_0 ^r^2 = =((2L)/(3r))r^3 ⇒V=(2/3)Lr^2](Q214804.png)
$${Let}\:{the}\:{base}\:{be}\:{on}\:{the}\:{Oxy}\:{plane}\:{and}\:{the} \\ $$$${base}'{s}\:{center}\:{be}\:{at}\:{x}={y}={z}=\mathrm{0}. \\ $$$${Obviously}\:{then},\:{the}\:{axis}\:{of}\:{the}\:{cyllinder} \\ $$$${is}\:{the}\:{z}\:{axis}. \\ $$$${Now},\:{the}\:{water}\:{level}\:{plane}\:{is} \\ $$$${z}=\frac{{Ly}}{{r}} \\ $$$${Then}\:{the}\:{water}\:{occupies}\:: \\ $$$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant{r}^{\mathrm{2}} }\\{{y}\geqslant\mathrm{0}}\\{\mathrm{0}\leqslant{z}\leqslant\frac{{L}}{{r}}{y}}\end{cases} \\ $$$${Therefore}: \\ $$$${V}=\underset{{y}=\mathrm{0}} {\overset{{y}={r}} {\int}}\:\:\underset{{x}=−\sqrt{{r}^{\mathrm{2}} −{y}^{\mathrm{2}} }} {\overset{{x}=\sqrt{{r}^{\mathrm{2}} −{y}^{\mathrm{2}} }} {\int}}\:\underset{{z}=\mathrm{0}} {\overset{{z}={Ly}/{r}} {\int}}{dzdxdy}= \\ $$$$=\underset{{y}=\mathrm{0}} {\overset{{y}={r}} {\int}}\:\:\left(\underset{{x}=−\sqrt{{r}^{\mathrm{2}} −{y}^{\mathrm{2}} }} {\overset{{x}=\sqrt{{r}^{\mathrm{2}} −{y}^{\mathrm{2}} }} {\int}}\:{dx}\right)\left(\underset{{z}=\mathrm{0}} {\overset{{z}={Ly}/{r}} {\int}}{dz}\right){dy}= \\ $$$$=\underset{\mathrm{0}} {\overset{{r}} {\int}}\frac{{L}}{{r}}\mathrm{2}{y}\sqrt{{r}^{\mathrm{2}} −{y}^{\mathrm{2}} }{dy}= \\ $$$$=\frac{{L}}{{r}}\underset{\mathrm{0}} {\overset{{r}^{\mathrm{2}} } {\int}}\sqrt{{r}^{\mathrm{2}} −{t}}{dt}= \\ $$$$=\frac{{L}}{{r}}\left[−\frac{\mathrm{2}}{\mathrm{3}}\left({r}^{\mathrm{2}} −{x}\right)^{\mathrm{3}/\mathrm{2}} \right]_{\mathrm{0}} ^{{r}^{\mathrm{2}} } = \\ $$$$=\frac{\mathrm{2}{L}}{\mathrm{3}{r}}{r}^{\mathrm{3}} \\ $$$$ \\ $$$$\Rightarrow{V}=\frac{\mathrm{2}}{\mathrm{3}}{Lr}^{\mathrm{2}} \\ $$