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A hemispherical bowl of radius 8 in. is being filled with water at a constant rate. If the water level is rising at the rate of 1/3 in./s at the instant when the water is 6 in. deep, find how fast the water is flowing in by using the fact that if V is the volume of the water at time t, then dV/dt = (pi)(r)^2(dh/dt)

I just need someone to check my work. From my picture, I have:

dV/dt = (pi)(64-(8-h)^2) dh/dt.

just plug in 6 for h and 1/3 for dh/dt and solve right? Thanks.

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# Related rates prob.

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