How fast the water level is rising when the ball is half submerged. Question:


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*Water is being poured into a hemispherical bowl of radius 3 in at the rate of 1 in3/s. How fast is the water level rising when the water is 1 in deep.


*In Problem 1, suppose that the bowl contains a lead ball 2 inches in diameter, and find how fast the water level is rising when the ball is half submerged.

For the 1st question:
Volume of the cap of a sphere of radius $R$ at height $\displaystyle h = {πh^2(3R-h^2)\over 3}$
$\displaystyle V = πh^2R - {πh^3\over 3}$
$\displaystyle {d\over dt} V = (2πhR - πh^2){dh\over dt}$
Given that $ \displaystyle  {d\over dt} V = 1\text{ in}^3/\text s$ and that $ \displaystyle R = 3\text{ in } \implies {dh\over dt} = {1\over 6πh -πh^2}$
So at $h=1, \displaystyle {dh\over dt} = {1\over 5π}$
For 2nd question I have drawn the figure below:

I can't write down the volume in terms of variables. So I am stuck there.
Can you please help me out of this? That is give that formula for volume in terms of variables.
 A: 
$R_{bowl} = 3 in$
$R_{ball} = 1 in$
At height H from bottom the volume of the bowl = $V_1$ =  ${1\over 3}πH^2(3R_{bowl} - H)$
Volume occupied by ball at height H = $V_2$ = ${1\over 3}πH^2(3R_{ball}-H)$
Effective volume $V_{eff}$
= $V_1$ - $V_2$
= ${1\over 3}πH^2(3R_{bowl} - 3R_{ball})$
= $πH^2(R_{bowl} - R_{ball})
= πH^2(3 - 1) = πH^2$
V = $2πH^2$ ,
${d\over dt}V$ = $4πH{d\over dt}H$
Given that ${d\over dt}V$
= ${1 in^3\over s}$
$1{ in^3\over s}$ = $4πH{d\over dt}H$
⇒${d\over dt}H$ =   ${1{in^3\over s}\over 4πH}$
When ball is half submerged so H = $R_{ball}$ = 1 in
So rate of increase of height at H = 1 in
${d\over dt}H$
$$ = {1{ in\over s}\over 4π(1)}$$
$$ = {1\over 4π} {in\over s} $$
A: While it is convenient to apply the "spherical cap" formula, it is not needed (and is in fact obtained by integrating the relation for the volume of an "infinitesimal layer" we produce in this problem).

If the hemispherical bowl of radius $ \ R \ $ is filled to a depth $ \ h \ \ , $ then the distance from the "center" of the hemisphere to the water's surface is $ \ R - h \ \ . $  The "infinitesimal layer" of water at the surface of the filled portion of the bowl has "thickness" $ \ dh \ $ and a radius [marked by the red horizontal line segment] given by  $ \ r \ = \ \sqrt{R^2 - (R - h)^2} \ $ $ = \ \sqrt{R^2 - (R^2 - 2Rh + h^2)} \ = \ \sqrt{ 2Rh - h^2 } \ \ . $  This makes the "infinitesimal volume" of the surface layer of water $ \ dV \   = \   \pi   ·    r^2   ·   dh \   = \   \pi   ·   (2Rh - h^2)   ·   dh  \ \ $
and the "related-rate" equation is then
$$ \frac{dV}{dt}  \ \ = \ \ \pi \ · \ (2Rh - h^2) \ · \ \frac{dh}{dt} \ \ \Rightarrow \ \ \frac{dh}{dt} \ \ = \ \  \frac{\frac{dV}{dt}}{\pi   · h ·  (2R  - h )}  \ \ , $$
as you found.
[You will learn later -- or may even be asked in an exercise -- to find the volume of a spherical cap from $ \ V \ = \ \int_0^{h} \  \pi   ·   (2Ry - y^2)  \ \   dy \ \ . \ ] $
For the second part, in which a submerged sphere of radius $ \ \rho \ $ is placed in the center of the bottom of the bowl, we can develop a similar expression to find the radius of the circular area that the sphere "blocks off" in the surface layer of water [marked by the pale red horizontal segment].  For $ \ 0 \ \le \ h \ \le \ \rho \ \ , $
we have $ \ \mathbf{r} \ = \ \sqrt{\rho^2 - (\rho - h)^2} \  = \  \sqrt{ 2\rho h - h^2 } \ \ . $  A disk of this radius is then "subtracted" from the infinitesimal surface layer, yielding   $ \ dV' \ = \  [ \pi ·  r^2  - \pi · \mathbf{r}^2 ]  ·   dh \   = \   \pi   ·   (2Rh - h^2 - 2 \rho h + h^2)  ·   dh \ $ $ = \ 2\pi · h · ( R  -   \rho )·dh  \ \ . $  The modified related-rate equation is now
$$   \frac{dh}{dt} \ \ = \ \  \frac{\frac{dV'}{dt}}{2\pi   · h ·  ( R  - \rho )}  \ \ . $$
(The numerical results found by using the quantities stated in the problem agree with the values you give.)
Were we to continue filling the bowl to its brim, we would observe a "piecewise" result for the rate at which the water level rises. As we immerse the upper half of the ball, the "obstructed disk" shrinks, with its radius found from
$ \ \mathbf{r} \ = \ \sqrt{\rho^2 - (h - \rho)^2} \  = \  \sqrt{ 2\rho h - h^2 } \ \ , $ actually the same expression we found above.  Once the ball is completely under water, filling the rest of the bowl proceeds using the argument for the first part of the problem.  For a constant "in-flow rate" of water $ \ \frac{dV}{dt} \  $ into the bowl, we have
$$   \frac{dh}{dt} \ \ = \ \  \left\{ \begin{array}{rcl}
\large{\frac{\frac{dV }{dt}}{2\pi \  · \ h \ · \ ( R  - \rho )}} \ \ , & \mbox{for} \ \ 0 \ \le \ h \ \le \ 2·\rho
  \\ \large{\frac{\frac{dV}{dt}}{\pi \  · \ h \ · \ (2R  - h )}} \ \ , & \mbox{for} \ \ 2·\rho \ \le \ h \ \le \ R
\end{array}\right.  \ \ . $$
[Note that the two rates agree at $ \ h \ = \ 2·\rho \ \ , $ so there is a smooth transition as the ball becomes submerged.]
A: The change in the water level times the surface area of the water level equals the rate of flow into the bowl.
In problem 1. $SA = 5\pi\text { in}^2$
$\frac {1 \frac {\text{in}^3}{s}}{ 5\pi \text { in}^2}$
problem 2.  There is a disc of radius 1 taking up space in your bowl,  $SA = 5\pi - \pi = 4\pi\text { in}^2$
$\frac {1}{4\pi} \frac {\text {in}}{s}$
