A hemispherical bowl with a lead ball inside

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Question from Jean, a student:

"(a) Water is being poured into a hemispherical bowl of radius 3 inch
at the rate of 1 inch^3/s. How fast is the water level rising when the
water is 1 inch deep ?

(b) In (a), suppose that the bowl contains a lead ball 2 inch in
diameter, and find how fast the water level is rising when the ball is
half submerged."

Solution :
(a) V = (pi/3)* h^2*(3R-h) (volume of a segment of height h of a sphere)

=> dV/dt = pi*h*(2R-h)*dh/dt
=> dh/dt = 1/(pi*h*(2R-h))*dV/dt,

Now R = 3 inch ; h = 1 inch ; dV/dt = 1 inch^3/s,
so dh/dt = 1/(5*pi) inch/s

(b) ????

Hi Jean,

When the lead ball is in the bowl it makes a hole in the water in the shape of a segment of a sphere, this time of radius 1 inch but still at the height of h inches. You know the volume of this hole so subtract it from the volume of the segment in part (a) to get the volume of the water around the ball. Now do the calculus to find the rate at which the water level is rising.

Penny

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.