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The first thing we need to do is define a frame of reference. We let x represent the vertical distance below the top of the tank. That is, we orient the x -axis vertically, with the origin at the top of the tank and the downward direction being positive (see the following figure).

This figure is a right circular cylinder that is vertical. It represents a tank of water. The radius of the cylinder is 4 m, the height of the cylinder is 10 m. The height of the water inside the cylinder is 8 m. There is also a horizontal line on top of the tank representing the x=0. A line is drawn vertical beside the cylinder with a downward arrow labeled x.
How much work is needed to empty a tank partially filled with water?

Using this coordinate system, the water extends from x = 2 to x = 10 . Therefore, we partition the interval [ 2 , 1 0 ] and look at the work required to lift each individual “layer” of water. So, for i = 0 , 1 , 2 ,… , n , let P = { x i } be a regular partition of the interval [ 2 , 1 0 ] , and for i = 1 , 2 ,… , n , choose an arbitrary point x i * [ x i 1 , x i ] . [link] shows a representative layer.

This figure is a right circular cylinder representing a tank of water. Inside of the cylinder is a layer of water with thickness delta x. The thickness begins at xsub(i-1) and ends at xsubi.
A representative layer of water.

In pumping problems, the force required to lift the water to the top of the tank is the force required to overcome gravity, so it is equal to the weight of the water. Given that the weight-density of water is 9800 N/m 3 , or 62.4 lb/ft 3 , calculating the volume of each layer gives us the weight. In this case, we have

V = π ( 4 ) 2 Δ x = 16 π Δ x .

Then, the force needed to lift each layer is

F = 9800 · 16 π Δ x = 156,800 π Δ x .

Note that this step becomes a little more difficult if we have a noncylindrical tank. We look at a noncylindrical tank in the next example.

We also need to know the distance the water must be lifted. Based on our choice of coordinate systems, we can use x i * as an approximation of the distance the layer must be lifted. Then the work to lift the i th layer of water W i is approximately

W i 156,800 π x i * Δ x .

Adding the work for each layer, we see the approximate work to empty the tank is given by

W = i = 1 n W i i = 1 n 156,800 π x i * Δ x .

This is a Riemann sum, so taking the limit as n , we get

W = lim n i = 1 n 156,800 π x i * Δ x = 156,800 π 2 10 x d x = 156,800 π [ x 2 2 ] | 2 10 = 7,526,400 π 23,644,883.

The work required to empty the tank is approximately 23,650,000 J.

For pumping problems, the calculations vary depending on the shape of the tank or container. The following problem-solving strategy lays out a step-by-step process for solving pumping problems.

Problem-solving strategy: solving pumping problems

  1. Sketch a picture of the tank and select an appropriate frame of reference.
  2. Calculate the volume of a representative layer of water.
  3. Multiply the volume by the weight-density of water to get the force.
  4. Calculate the distance the layer of water must be lifted.
  5. Multiply the force and distance to get an estimate of the work needed to lift the layer of water.
  6. Sum the work required to lift all the layers. This expression is an estimate of the work required to pump out the desired amount of water, and it is in the form of a Riemann sum.
  7. Take the limit as n and evaluate the resulting integral to get the exact work required to pump out the desired amount of water.

We now apply this problem-solving strategy in an example with a noncylindrical tank.

A pumping problem with a noncylindrical tank

Assume a tank in the shape of an inverted cone, with height 12 ft and base radius 4 ft. The tank is full to start with, and water is pumped over the upper edge of the tank until the height of the water remaining in the tank is 4 ft. How much work is required to pump out that amount of water?

The tank is depicted in [link] . As we did in the example with the cylindrical tank, we orient the x -axis vertically, with the origin at the top of the tank and the downward direction being positive (step 1).

This figure is an upside-down cone. The cone has an axis through the center. The top of the cone on the axis is labeled x=0.
A water tank in the shape of an inverted cone.

The tank starts out full and ends with 4 ft of water left, so, based on our chosen frame of reference, we need to partition the interval [ 0 , 8 ] . Then, for i = 0 , 1 , 2 ,… , n , let P = { x i } be a regular partition of the interval [ 0 , 8 ] , and for i = 1 , 2 ,… , n , choose an arbitrary point x i * [ x i 1 , x i ] . We can approximate the volume of a layer by using a disk, then use similar triangles to find the radius of the disk (see the following figure).

This figure has two images. The first has the x-axis. Below the axis, on a slant is a line segment extending up to the x-axis. Beside the line segment is a horizontal right circular cylinder. The second image has a triangle. The right triangle mirrors the first image with the hypotenuse the line segment in the first image. The top of the triangle is 4 units. the length of the vertical side is 12 units. The vertical side is also divided into two parts; the first is xsubi, the second is 12-xsubi. It is divided at the level where the first image has the cylinder.
Using similar triangles to express the radius of a disk of water.

From properties of similar triangles, we have

r i 12 x i * = 4 12 = 1 3 3 r i = 12 x i * r i = 12 x i * 3 = 4 x i * 3 .

Then the volume of the disk is

V i = π ( 4 x i * 3 ) 2 Δ x (step 2).

The weight-density of water is 62.4 lb/ft 3 , so the force needed to lift each layer is approximately

F i 62.4 π ( 4 x i * 3 ) 2 Δ x (step 3).

Based on the diagram, the distance the water must be lifted is approximately x i * feet (step 4), so the approximate work needed to lift the layer is

W i 62.4 π x i * ( 4 x i * 3 ) 2 Δ x (step 5).

Summing the work required to lift all the layers, we get an approximate value of the total work:

W = i = 1 n W i i = 1 n 62.4 π x i * ( 4 x i * 3 ) 2 Δ x (step 6).

Taking the limit as n , we obtain

W = lim n i = 1 n 62.4 π x i * ( 4 x i * 3 ) 2 Δ x = 0 8 62.4 π x ( 4 x 3 ) 2 d x = 62.4 π 0 8 x ( 16 8 x 3 + x 2 9 ) d x = 62.4 π 0 8 ( 16 x 8 x 2 3 + x 3 9 ) d x = 62.4 π [ 8 x 2 8 x 3 9 + x 4 36 ] | 0 8 = 10,649.6 π 33,456.7.

It takes approximately 33,450 ft-lb of work to empty the tank to the desired level.

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x-xcosx/sinsq.3x
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x-xcosx/sin^23x
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sin x - sin x cos^2 x sin x (1-cos^2 x) note the identity:sin^2 x + cos^2 x = 1 thus, sin^2 x = 1 - cos^2 x now substitute this into the above: sin x (sin^2 x), now multiply, yielding: sin^3 x Q.E.D.
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Practice Key Terms 4

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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