A cylindrical tank with radius 2 m and height 5 m is filled with water. Find the work required to pump all the water to the top of the tank.

Step 1: Define the problem and variables

We need to calculate the work done to pump water out of the cylindrical tank. The work done ($W$) is the integral of the force required to lift a small slice of water multiplied by the distance it is lifted. Here, $\rho$ is the density of water, $g$ is the acceleration due to gravity, $A(y)$ is the cross-sectional area of the tank at height $y$, and $(H-y)$ is the distance each slice of water at height $y$ needs to be lifted to the top of the tank.

Step 2: Identify given values and constants

The radius of the cylindrical tank is $r = 2$ m, and its height is $H = 5$ m. The density of water is $\rho = 1000 \text{ kg/m}^3$, and the acceleration due to gravity is $g = 9.8 \text{ m/s}^2$. The water is filled to the top, so we will integrate from $y=0$ to $y=H$.

Step 3: Calculate the cross-sectional area

For a cylindrical tank, the cross-sectional area $A(y)$ is constant at any height $y$. It is given by the formula for the area of a circle, $\pi r^2$. Substituting the given radius $r=2$ m, we get $A(y) = 4\pi \text{ m}^2$.

Step 4: Set up the integral for work

Now we substitute all the known values into the work integral. The limits of integration are from $y=0$ (bottom of the tank) to $y=5$ (top of the tank). The distance to lift each slice is $(H-y) = (5-y)$.

Step 5: Evaluate the integral

First, we pull the constants out of the integral: $1000 \times 9.8 \times 4\pi = 39200\pi$. Then, we integrate $(5-y)$ with respect to $y$, which gives $5y - \frac{y^2}{2}$.

Step 6: Calculate the final work

Now we evaluate the definite integral by plugging in the upper and lower limits. This gives us $39200\pi \left( 25 - \frac{25}{2} \right) = 39200\pi \left( \frac{25}{2} \right)$. Multiplying these values, we get the total work done.

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