# Math Help - Hemispherical bowl question.

1. ## Hemispherical bowl question.

I am really keen to know where I went wrong in this question, so any help is appreciated.

"A hemispherical bowl of depth and radius R has volume (2*pi*R^3)/3. If it contains water to a depth h, the volume of water is:
pi(R-h)(R^2-(1/3)*((R-h)^2)).If water is poured into the bowl at a rate of V'(t), then what is h'(t)?" (THE CARTOON GUIDE TO CALCULUS, p.124, ISBN=9780061689093
my working:

Since h'(t) represented the change in water height level at an instantaneous point of time, and V'(t) represented the change of volume of water at at an instantaneous point of time, I guessed that I had to use implicit differentiation to make a differential equation such that i can solve for h'(t).
Thus, (using newton's notation)

V = pi[(R-h)(R^2-(1/3)*((R-h)^2))]

V' = pi[(R-h)(R^2-(1/3)*((R-h)^2))]' {[Cf(x)]' = Cf'(x)}

V' = pi[(R-h)'(R^2-(1/3)*((R-h)^2)) + (R-h)(R^2-(1/3)*((R-h)^2))'] {product rule}

V' = pi[-h'(R^2-(1/3)*((R-h)^2)) + (R-h)((-1/3)*(2(R-h))*-h')] {R is a constant, h and V is differentiated with respect to time}

V' = pi[-h'(R^2-(1/3)*((R-h)^2)) + (R-h)((1/3)(2(R-h))*h')]

V' = pi[-h'(R^2-(1/3)*((R-h)^2)) + h'(2/3)((R-h)^2)] { Simplification }

-V' = pi[h'(R^2-(1/3)*((R-h)^2)) - h'(2/3)((R-h)^2)] { *-1 on both sides }

-V' = pi*h'[(R^2-(1/3)*((R-h)^2)) - (2/3)((R-h)^2)]

-V' = pi*h'[(R^2-((R-h)^2)]

-V' = pi*h'[(R^2-(R^2 - 2Rh + h^2)] { (a - b)^2 = a^2 - 2ab + b^2 }

-V' = pi*h'(R^2 - R^2 + 2Rh - h^2)

-V' = pi*h'(2Rh - h^2)

h' = -V'/(pi(2Rh - h^2)), where h' is the rate of height increase, V' is rate of volume increase, R is radius of hemispherical bowl, and h is height at the instantaneous point of time.

However, i plugged my test variables into the equation, i got my h' to be -ve, when my V', R and h are +ve. Please help!!!
(sorry for long post)

2. ## Re: Hemispherical bowl question.

I get the same result for h' as you, but I think that the original expression for the volume is wrong.
I think that the volume of water should be
$V=\pi\left[Rh^{2}-\frac{h^{3}}{3}\right].$

3. ## Re: Hemispherical bowl question.

rotate the circle $x^2+y^2=R^2$ about the y-axis to form a sphere of radius $R$ ...

$V = \pi \int_{-R}^h (R^2-y^2) \, dy$

$\frac{dV}{dt} = \pi(R^2 - h^2) \cdot \frac{dh}{dt}$

4. ## Re: Hemispherical bowl question.

skeeter, are u saying how to derive the expression for knowing the volume of water at height h?thanks

btw, how do you type in math expressions nicely without use of parenthesis and symbols? sry, new here.

5. ## Re: Hemispherical bowl question.

I get the same result for h' as you, but I think that the original expression for the volume is wrong.
I think that the volume of water should be
V=\pi\left[Rh^{2}-\frac{h^{3}}{3}\right].
could u tell me where u got that info,Bob, thanks soo much.

6. ## Re: Hemispherical bowl question.

skeeter, are u saying how to derive the expression for knowing the volume of water at height h?thanks
Opps, sry skeeter, i didn't realise that u were solving for h'! I don't understand what u did though, 'cause the problem expected me to use implicit differentiation to solve, not integral. Anyway, i took BobP's equation and manipulated it, giving me two different solutions for h'

BobP:

V=pi(Rh^2-(1/3)h^3)
V'=pi(Rh^2-(1/3)h^3)'
V'=pi(2hh'-hh')
h'=V'/pi(h)

Skeeter:

h'=V'/pi(R^2-h^2)

?? Thanks for the help!

7. ## Re: Hemispherical bowl question.

Originally Posted by jonwen97
could u tell me where u got that info,Bob, thanks soo much.
I got the expression for the volume by an integration, (like skeeter, though I disagree with his limits, I think the top one should be $-R+h)$).

I rotated around the x-axis rather than the y-axis.

$V=\pi \int_{R-h}^{R}R^{2}-x^{2} dx,$

and I checked this against the result given in Schaum's Mathematical Handbook. (Schaum's formula is the same as the one I derived.)

The other point is that your differentiation in your last post is wrong.
Look again at that line number 3.

8. ## Re: Hemispherical bowl question.

BobP:
V=pi(Rh^2-(1/3)h^3)
V'=pi(Rh^2-(1/3)h^3)'
V'=pi(2hh'-hh')
h'=V'/pi(h)
Opps, thanks for pointing out my careless mistake. AND IT TOTALLY WORKS NOW!! It was just like my last equation, but -ve. Now the -ve values turned +ve, and it makes sense!!

V=pi(Rh^2-(1/3)h^3)
V'=pi(Rh^2-(1/3)h^3)'
V'=pi(R(2h)(h')-(1/3)(3h^2)(h'))
V'=pi(2Rhh'-(h^2)h')
V'=h'pi(2Rh-h^2)
h'=V'/pi(2Rh-h^2)

!!THANK YOU SOOOOO MUCH!!

9. ## Re: Hemispherical bowl question.

Originally Posted by BobP
I got the expression for the volume by an integration, (like skeeter, though I disagree with his limits, I think the top one should be $-R+h)$).
my mistake ... I was thinking of h as a y-value rather than a height above the bottom of the sphere.

$V = \int_{-R}^{h-R} (R^2 - y^2) \, dy$

$\frac{dV}{dt} = [R^2 - (h-R)^2] \cdot \frac{dh}{dt}$

$\frac{dV}{dt} = [R^2 - (h^2-2hR+R^2)] \cdot \frac{dh}{dt}$

$\frac{dV}{dt} = [2hR-h^2] \cdot \frac{dh}{dt}$

better?