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Question Number 214742 by mr W last updated on 18/Dec/24

Commented by mr W last updated on 18/Dec/24

$f i n d t h e v o l u m e o f w a t e r i n t h e$ $c y l i n d e r i c a l c u p .$
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

	$\frac{x}{L} = \frac{y}{R}$ $y = (\frac{R}{L}) x$ $\cos θ = \frac{R - y}{R} = 1 - \frac{x}{L}$ $x = 0 \Leftrightarrow θ = 0, x = L \Leftrightarrow θ = \frac{π}{2}$ $d x = L \sin θ$ $\int d V = \int_{0}^{\frac{π}{2}} (R^{2} θ - R^{2} \sin θ \cos θ) (L \sin θ)$ $\int_{0}^{\frac{π}{2}} θ \sin θ d θ = (- θ \cos θ + \sin θ) ∣_{0}^{\frac{π}{2}} = 1$ $\frac{V}{R^{2} L} = 1 - \frac{1}{3} = \frac{2}{3}$ $\frac{V}{π R^{2} L} = \frac{2}{3 π} \approx 0.2122$
	Commented by mr W last updated on 19/Dec/24
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	Answered by aleks041103 last updated on 20/Dec/24
	$Let the base be on the Oxy plane and the base′s center be at x=y=z=0. Obviously then, the axis of the cyllinder is the z axis. Now, the water level plane is z=((Ly)/r) Then the water occupies : { ((x^2 +y^2 ≤r^2 )),((y≥0)),((0≤z≤(L/r)y)) :} Therefore: V=∫_(y=0) ^(y=r) ∫_(x=−(√(r^2 −y^2 ))) ^(x=(√(r^2 −y^2 ))) ∫_(z=0) ^(z=Ly/r) dzdxdy= =∫_(y=0) ^(y=r) (∫_(x=−(√(r^2 −y^2 ))) ^(x=(√(r^2 −y^2 ))) dx)(∫_(z=0) ^(z=Ly/r) dz)dy= =∫_0 ^r (L/r)2y(√(r^2 −y^2 ))dy= =(L/r)∫_0 ^r^2 (√(r^2 −t))dt= =(L/r)[−(2/3)(r^2 −x)^(3/2) ]_0 ^r^2 = =((2L)/(3r))r^3 ⇒V=(2/3)Lr^2$
	$L e t t h e b a s e b e o n t h e O x y p l a n e a n d t h e$ ${b a s e}^{'} s c e n t e r b e a t x = y = z = 0 .$ $O b v i o u s l y t h e n, t h e a x i s o f t h e c y l l i n d e r$ $i s t h e z a x i s .$ $N o w, t h e w a t e r l e v e l p l a n e i s$ $z = \frac{L y}{r}$ $T h e n t h e w a t e r o c c u p i e s :$ ${\begin{cases} x^{2} + y^{2} ⩽ r^{2} \\ y ⩾ 0 \\ 0 ⩽ z ⩽ \frac{L}{r} y \end{cases}$ $T h e r e f o r e :$ $V = \underset{y = 0}{\int^{y = r}} \underset{x = - \sqrt{r^{2} - y^{2}}}{\int^{x = \sqrt{r^{2} - y^{2}}}} \underset{z = 0}{\int^{z = L y / r}} d z d x d y =$ $= \underset{y = 0}{\int^{y = r}} (\underset{x = - \sqrt{r^{2} - y^{2}}}{\int^{x = \sqrt{r^{2} - y^{2}}}} d x) (\underset{z = 0}{\int^{z = L y / r}} d z) d y =$ $= \underset{0}{\int^{r}} \frac{L}{r} 2 y \sqrt{r^{2} - y^{2}} d y =$ $= \frac{L}{r} \underset{0}{\int^{r^{2}}} \sqrt{r^{2} - t} d t =$ $= \frac{L}{r} {[- \frac{2}{3} {(r^{2} - x)}^{3 / 2}]}_{0}^{r^{2}} =$ $= \frac{2 L}{3 r} r^{3}$ $\Rightarrow V = \frac{2}{3} {L r}^{2}$
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