If a hemispherical bowl of radius 6cm contains water to a depth of h cm, the volume of the water is 1/3πh^2(18-h). Water is poured into the bowl at a rate 4 cm^3/s . Find the rate at ehich the water level is rising when the depth is 2 cm.

V = (1/3)πh^2 (18-h)

= 6πh^2 - (1/3)πh^3

dV/dt = 12πh dh/dt - πh^2 dh/dt

4 = dh/dt(12πh - πh^2)
when h = 2

4 = dh/dt(24π - 4π)
dh/dt = 4/(20π) = 1/(5π) cm/s

To find the rate at which the water level is rising, we need to find the derivative of the volume with respect to time.

Given that the volume of the water in the bowl is given by the equation V = (1/3)πh^2(18 - h), where V is the volume and h is the depth of the water.

Taking the derivative of V with respect to time:

dV/dt = (1/3)π * 2h * dh/dt + (1/3)πh^2 * (-dh/dt)

Where dh/dt represents the rate at which the depth is changing (or the rate at which the water level is rising).

We are given that dh/dt = 4 cm^3/s, and we want to find dh/dt when h = 2 cm.

Substituting the given values into the derivative equation:

dV/dt = (1/3)π * 2(2) * 4 + (1/3)π * (2^2) * (-4)
= (8/3)π + (4/3)π
= (12/3)π
= 4π cm^3/s

Therefore, when the depth is 2 cm, the rate at which the water level is rising is 4π cm^3/s.

To find the rate at which the water level is rising when the depth is 2 cm, we need to find the derivative of the volume with respect to time, and then substitute the given values.

Given:
Radius of the hemispherical bowl = 6 cm
Volume of the water = 1/3πh^2(18-h)
Rate at which water is poured into the bowl = 4 cm^3/s

1. Differentiate the volume formula with respect to time:
dV/dt = d/dt(1/3πh^2(18-h))

2. Simplify the derivative:
dV/dt = 1/3π * (2h * dh/dt) * (18 - h) + 1/3π * h^2 * (-1 * dh/dt)

3. Substitute the given values:
dh/dt = 4 cm^3/s
h = 2 cm

4. Calculate dV/dt when h = 2 cm and dh/dt = 4 cm^3/s:
dV/dt = 1/3π * (2(2) * 4) * (18 - 2) + 1/3π * (2^2) * (-1 * 4)

5. Simplify the expression:
dV/dt = 1/3π * (16) * (16) + 1/3π * (4) * (-4)

6. Calculate the final answer:
dV/dt = (256/3)π + (-16/3)π
dV/dt = (240/3)π
dV/dt = 80π cm^3/s

Therefore, the rate at which the water level is rising when the depth is 2 cm is 80π cm^3/s.