Problem 28 - Calculus Early Transcendentals

Question

A cylindrical water tank has height 8 m and radius \(2 \mathrm{m}\) (see figure). a. If the tank is full of water, how much work is required to pump the water to the level of the top of the tank and out of the tank? b. Is it true that it takes half as much work to pump the water out of the tank when it is half full as when it is full? Explain.

Step-by-Step Solution

Answer

Question: Compare the work required to pump water out of a cylindrical tank that is full and a tank that is half full. Answer: To pump the water out of the tank when it is half full requires only one-fourth of the work compared to when the tank is full.

Step 1: Find the Volume of a Cylindrical Slice

Consider a small cylindrical slice of height \(\mathrm{dy}\) at a height of \(y\) from the bottom of the tank. The volume of this slice will be \(\mathrm{dV}=A\,\mathrm{dy}\), where \(A\) is the area of the base of the cylinder. Since the radius of the tank is constant, the area of the base is given by \(A=\pi\, r^2\). So, \(\mathrm{dV}=\pi\, r^2\,\mathrm{dy}\).

Step 2: Calculate the Work Done on a Single Slice

The weight of the water in the slice is given by \(\mathrm{dW}=\rho \,g\, \mathrm{dV}\), where \(\rho\) is the density of water and \(g\) is the acceleration due to gravity. To pump the water in the slice up to the top of the tank, it needs to be raised a distance \((8\, \mathrm{m} - y)\). Thus, the work done on the slice is given by \(\mathrm{dw}=\mathrm{dW} \times (8 - y) = \rho\, g\, \mathrm{dV} \times (8-y)\).

Step 3: Integrate the Work Done Over the Height of the Tank

To find the total work done for the full tank, integrate the expression for \(\mathrm{dw}\) from \(0\) to \(8\): $$ W_\text{full} = \int_0^8 \rho\, g\, (\pi\, r^2\, \mathrm{dy})\cdot (8-y) $$ For the case when the tank is half full, we integrate from \(0\) to \(4\): $$ W_\text{half} = \int_0^4 \rho\, g\, (\pi\, r^2\, \mathrm{dy})\cdot (8-y) $$

Step 4: Evaluate the Integrals

The density of the water, radius, and gravity are constant. So we can evaluate both integrals: $$ W_\text{full} = \rho\, g\, \pi\, r^2 \int_0^8 (8-y)\, \mathrm{dy} $$ $$ W_\text{half} =\rho\, g\, \pi\, r^2 \int_0^4 (8-y)\, \mathrm{dy} $$ Calculating the definite integrals, we get: $$ W_\text{full} = 256\,\pi\,\rho\,g\,r^2 $$ $$ W_\text{half} = 64\,\pi\,\rho\,g\,r^2 $$

Step 5: Compare the Work for Full and Half Full Tanks

We compute the ratio between the work required for the half-full tank and the full tank: $$ \frac{W_\text{half}}{W_\text{full}} = \frac{64\,\pi\,\rho\,g\,r^2}{256\,\pi\,\rho\,g\,r^2} = \frac{1}{4} $$

Step 6: Conclusions

a. To pump all the water out of the tank when it is full requires \(256\,\pi\,\rho\,g\,r^2\) units of work. b. To pump the water out of the tank when it is half full is not half as much work as when it is full. Instead, it is only one-fourth of the work.

Key Concepts

Cylindrical Tank

A cylindrical tank is shaped like a cylinder, which is essentially a 3D shape with two circular bases connected by a curved surface. In our problem, the tank has a height of 8 meters and a radius of 2 meters. Understanding the geometry of a cylindrical tank is crucial when calculating how much work is needed to pump water.

The volume of the tank is determined by the formula for the volume of a cylinder, which is given by:

• Volume = \( \pi r^2 h \)

where \(r\) is the radius and \(h\) is the height.

Each slice of water within the tank can be viewed as a thin cylinder itself. By dividing the tank into these infinitesimally small horizontal slices, we can calculate the volume of each slice and hence the work needed to move that slice to the top of the tank.

Integration in Calculus

In calculus, integration is a fundamental tool used to find accumulations, such as area under a curve or, in our context, the total work required to pump water. Here, we use integration to sum up the work done on each tiny slice of water within the tank.

For our water tank problem, work done on a slice is calculated using the integral:

• Work (W) = \[\int_{0}^{h} \rho g (\pi r^2) (h-y) \, dy \]

where \(\rho\) is the density of water, \(g\) is acceleration due to gravity, \(r\) is the radius, \(h\) is the height of the tank, and \(y\) is the vertical position of the slice within the tank.

This integral essentially accumulates the work done for each horizontal slice from the bottom to the top of the tank. Integration allows us to account for how much work is required to lift each slice a different distance to the top.

Hydrostatics

Hydrostatics focuses on fluids at rest and the forces and pressures associated with them. In our exercise, when considering each slice of water within the cylindrical tank, we take into account the weight of the water and gravity.

The force needed for moving a slice is determined by the weight of the slice, which is the product of:

• Density of water (\(\rho\))
• Gravity (\(g\))
• Volume of the slice (small cylindrical section)

which results in the expression \(\rho g \pi r^2 \, dy\).

It's fascinating how hydrostatics reveals that not only does the depth of the water matter but also its distribution affects the work necessary to move it upwards. Hydrostatics principles guide us in acknowledging how water pressure varies with depth and consequently influences the work required to pump water from any given depth.