Rate of increase (Calculus) surface of a bowl

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The interior surface of a bowl is made by rotating the curve x^2 = 4y about the y-axis, the units on both axis being in cm. Water is poured into the bowl at a constant rate of 36cm^3/sec. When the depth of the water is h cm, the area of the surface of the water is S cm^2. Find:
i) The rate of increase in h when S = 12
ii) The rate of increase in S when h = 9.

thanks in advance
 
volume of revolution using disks about the y-axis ...

\(\displaystyle \L x^2 = 4y\)

let h = depth of water

as a function of h, the volume may be defined by the integral ...

\(\displaystyle \L V = \pi \int_0^h 4y dy\)

take the derivative w/r to time ...

\(\displaystyle \L \frac{d}{dt}[V = \pi \int_0^h 4y dy]\)

\(\displaystyle \L \frac{dV}{dt} = \pi 4h \frac{dh}{dt}\)

you were given \(\displaystyle \L \frac{dV}{dt} = 36 \frac{cm^3}{s}\) ...

\(\displaystyle \L 36 = \pi 4h \frac{dh}{dt}\)

\(\displaystyle \L \frac{dh}{dt} = \frac{9}{\pi h}\)

now, the surface area as a function of h is just \(\displaystyle \L S = 4\pi h\) ...

can you finish?
 
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