Is my solution correct? (Related rates)

[SarahT]

A leaky hemispherical water tank with a radius of 8 meters is filled to a depth of h meters. The volume of the water tank is given by V=pi*h^2*[r-(h/3)]. When h=5 meters the water level is dropping at the rate of 0.8 per hour. At what rate is water leaking out of the tank at that instant?

V=pi*h^2*(8-(h/3))
V=pi/3*h^2*(24-h)

dV/dt= ((2pih/3)*(24-h)-(pi(h^2)/3))*dh/dt
dV/dt=(pi*h*(16-h))*dh/dt

when h=5, dV/dt=pi*5*(16-5)*(-0.8)
=-44pi m^3/hr

Is my solution correct?
Thanks for your help.

 

[srmichael]

A leaky hemispherical water tank with a radius of 8 meters is filled to a depth of h meters. The volume of the water tank is given by V=pi*h^2*[r-(h/3)]. When h=5 meters the water level is dropping at the rate of 0.8 per hour. At what rate is water leaking out of the tank at that instant?

V=pi*h^2*(8-(h/3))
V=pi/3*h^2*(24-h)

dV/dt= ((2pih/3)*(24-h)-(pi(h^2)/3))*dh/dt
dV/dt=(pi*h*(16-h))*dh/dt

when h=5, dV/dt=pi*5*(16-5)*(-0.8)
=-44pi m^3/hr

Is my solution correct?
Thanks for your help.

No, that is not correct. You substituted 8 in for the radius r when in fact r is a variable as well, dependent on the height, h. You will need to find a formula for r in terms of h that you can then substitue in for r in the volume formula, then you can take the derivative of V with respect to t.