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

my working:"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

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)