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17Calculus Integrals - Work - Moving Fluid

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These problems are almost always word problems. If you haven't already, we suggest you read the article You Can Successfully Solve Word Problems before going on.

The idea of these problems is to move fluid out of a tank by raising it against gravity. The hard part is how to set up the differential \(dW\).

Here is a quick video clip giving an overview.

PatrickJMT - Moving Fluid [1min-11secs]

video by PatrickJMT

A few comments are in order.
1. Since we are told to calculate the work involved raising the water against gravity, we do not have to worry about the distance moving the water side-to-side. That means you need to only think about the vertical component of the work to raise the water to the level that the problem asks for.

2. When solving these problems, one thing you need to do in all problems is to set up a coordinate system. We believe that, in most cases, the best place to set up your coordinate system is at the bottom of the tank. This should yield the easiest equations and you will usually get an integral that you can solve. However, some instructors recommend setting the coordinate system at the top of the tank. Check with your instructor to see what they require. (We recommend working these problems several times, setting up the coordinate system in different places. This will give you different integrals to solve but the answer is the same.)

Practice

Unless otherwise instructed, solve these problems giving your answers in exact terms.
Note - Although we recommend in the article on successfully solving word problems to organize your problems so that you work several similar problems one right after another, we have not arranged them like that here. We let you do it, if you choose to.

A tank of water is in the shape of a cylinder of height 25 meters and radius of 7 meters. If the tank is completely filled with water how much work is required to pump all of the water to the top of the tank. Assume that the density of water is 1000 kg/m3.

Problem Statement

A tank of water is in the shape of a cylinder of height 25 meters and radius of 7 meters. If the tank is completely filled with water how much work is required to pump all of the water to the top of the tank. Assume that the density of water is 1000 kg/m3.

Final Answer

\(150,062,500\pi\) N-m

Problem Statement

A tank of water is in the shape of a cylinder of height 25 meters and radius of 7 meters. If the tank is completely filled with water how much work is required to pump all of the water to the top of the tank. Assume that the density of water is 1000 kg/m3.

Solution

The general work integral is \(\displaystyle{W = \int_c^d{F(y) ~ dy } }\)
We set up our coordinate system at the bottom of the tank with the y-axis up through the middle of the tank, which is why we used y as our variable of integration.
The volume of a thin slice of water is \(\pi r^2 dy = 49 \pi ~dy\) m3. We move this thin slice of water a distance y metres to the top of the tank.
We are given the density of water which we need to multiply by gravity (9.8m/sec2) to get the weight. So the integrand is \((1000 kg/m^3)(9.8m/sec^2)(y~m)(49\pi ~dy ~m^3)\). Checking our units, we have kg-m2/sec2 = kg-m/sec2 - m = N-m.

\(\displaystyle{ W = \int_0^{25}{ (1000)(9.8)(49\pi y ~dy) } }\)

\(\displaystyle{ W = 480200\pi[ y^2/2]_0^{25} }\)

\(\displaystyle{ W = 480200\pi[ (25)^2/2] - 0 = 150062500\pi }\) N-m

Looking at this number, it seems really large. However, if we think about the size of the tank and the amount of water, it makes sense.

Final Answer

\(150,062,500\pi\) N-m

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A tank of water is in the shape of an inverted pyramid that is 18 feet tall and whose top is a square with sides 4 feet long. If there is initially 12 feet of water in the tank determine the amount of work needed to pump all of the water to the top of the tank. Assume that the density of water is 62 lb/ft3.

Problem Statement

A tank of water is in the shape of an inverted pyramid that is 18 feet tall and whose top is a square with sides 4 feet long. If there is initially 12 feet of water in the tank determine the amount of work needed to pump all of the water to the top of the tank. Assume that the density of water is 62 lb/ft3.

Final Answer

W = 15872 lb-ft

Problem Statement

A tank of water is in the shape of an inverted pyramid that is 18 feet tall and whose top is a square with sides 4 feet long. If there is initially 12 feet of water in the tank determine the amount of work needed to pump all of the water to the top of the tank. Assume that the density of water is 62 lb/ft3.

Solution

The general work integral is \(\displaystyle{W = \int_c^d{F(y) ~ dy } }\)
We set up our coordinate system at the bottom of the tank with the y-axis up through the middle of the tank, which is why we used y as our variable of integration.

Figure 1

A volume of a thin slice of water is the area of a square times dy. However, the area of the square depends on where in the tank we are talking about. At the very top it is 4x4=16 ft2. At the bottom of the tank it is zero. So we need to determine an equation relating the y-value with the length of a side. This is most easily done using similar triangles. We have plotted the triangle in Figure 1.
When y=18, l=4, so what is l for any y? Using similar triangles, we can come up with the equation \(\displaystyle{\frac{18}{4} = \frac{y}{l} }\). Solving for l we get \(l=2y/9\). We can check this with the triangle in Figure 1.
So the area is \(A=l^2 = (2y/9)^2\) giving the volume of a thin slice of water at any point as \((2y/9)^2~dy\).
Now the problem states that the tank is filled up to 12 feet, so the range on y is [0,12]. In some cases we would say that our force equation involved just y. This would be true if the tank was initially completely full. However, in this case, we need to raise the water not just to the top of the initial water level but another 6 feet to the top of the tank. So how do we write this idea in equation form?
Let's think about several levels in the tank. First, the very top slice. In this case y=12 and we need to move it a distance of 6 feet.
Next, consider the very bottom slice when y=0. We need to move this slice 18 feet. This is easy to see that we need to move it a distance 18-y. Does this also hold for the top slice? Yes, when y=12, the distance is 18-y=18-12=6 feet. We are now ready to set up our integral and evaluate it.

\(\displaystyle{W = \int_0^{12}{(62)(18-y)(4y^2/81 ~dy) } }\)

\(\displaystyle{W = \frac{248}{81} \int_0^{12}{(18-y)(y^2 ~dy) } }\)

\(\displaystyle{W = \frac{248}{81} \int_0^{12}{18y^2 - y^3 ~dy } }\)

\(\displaystyle{W = \frac{248}{81} \left[\frac{18y^3}{3} - \frac{y^4}{4} \right]_0^{12} }\)

\(\displaystyle{W = \frac{248}{81} \left[\frac{18(12)^3}{3} - \frac{(12)^4}{4} \right] - 0 }\)

\(\displaystyle{W = \frac{248}{81} \left[10368 - 5184 \right] }\)

\(W = 15872 \) lb-ft

Final Answer

W = 15872 lb-ft

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We have a 10ft long trough with cross section as a triangle with base 3ft and height 3ft (point of triangle at the bottom of the tank). Calculate the work required to drain the entire tank by pumping the water out of a pipe 2ft above the top of the tank. (density of water is 62.5lb/ft3)

Problem Statement

We have a 10ft long trough with cross section as a triangle with base 3ft and height 3ft (point of triangle at the bottom of the tank). Calculate the work required to drain the entire tank by pumping the water out of a pipe 2ft above the top of the tank. (density of water is 62.5lb/ft3)

Solution

1207 video

video by PatrickJMT

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We have a cylindrical tank, half full of water, which is 10m high and radius 5m. Calculate the work required to pump the water out of the tank to a level 4m above the top of the tank. Use 9800 N/m3 for the weight of water.

Problem Statement

We have a cylindrical tank, half full of water, which is 10m high and radius 5m. Calculate the work required to pump the water out of the tank to a level 4m above the top of the tank. Use 9800 N/m3 for the weight of water.

Solution

1210 video

video by Dragonometry

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Calculate the work required to empty a conical tank, 10m tall with a radius of 4m at the base. The tank is initially filled with water up to 2m from the top, with the point down, and is to be emptied out the top.

Problem Statement

Calculate the work required to empty a conical tank, 10m tall with a radius of 4m at the base. The tank is initially filled with water up to 2m from the top, with the point down, and is to be emptied out the top.

Hint

Problem Statement

Calculate the work required to empty a conical tank, 10m tall with a radius of 4m at the base. The tank is initially filled with water up to 2m from the top, with the point down, and is to be emptied out the top.

Hint

Solution

2504 video

video by blackpenredpen

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An aquarium 2m long, 1m wide and 1m deep is full of water. Find the work needed to pump half of the water out of the aquarium. Use the density of water 1000 kg/m3.

Problem Statement

An aquarium 2m long, 1m wide and 1m deep is full of water. Find the work needed to pump half of the water out of the aquarium. Use the density of water 1000 kg/m3.

Solution

2505 video

video by blackpenredpen

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A circular swimming pool has a diameter of 24ft, the sides are 5ft high and the depth of the water is 4ft. How much work is required to pump all the water out over the side? Water weighs 62.5 lb/ft3.

Problem Statement

A circular swimming pool has a diameter of 24ft, the sides are 5ft high and the depth of the water is 4ft. How much work is required to pump all the water out over the side? Water weighs 62.5 lb/ft3.

Solution

2506 video

video by blackpenredpen

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We have a 8m long trough with cross section as a triangle with base 3m and height 3m (point of triangle at the bottom of the tank). Calculate the work required to drain the entire tank by pumping the water out of a pipe 2m above the top of the tank.

Problem Statement

We have a 8m long trough with cross section as a triangle with base 3m and height 3m (point of triangle at the bottom of the tank). Calculate the work required to drain the entire tank by pumping the water out of a pipe 2m above the top of the tank.

Solution

2507 video

video by blackpenredpen

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For a spherical tank of radius 3m full of water, calculate the work required to pump the water out to a point 1m above the tank.

Problem Statement

For a spherical tank of radius 3m full of water, calculate the work required to pump the water out to a point 1m above the tank.

Solution

2508 video

video by blackpenredpen

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A tank of water is the shape of the lower half of a sphere of radius 6 meters. If the initial depth of the water is 4 meters how much work is required to pump all the water to the top of the tank. Assume that the density of water is 1000 kg/m3.

Problem Statement

A tank of water is the shape of the lower half of a sphere of radius 6 meters. If the initial depth of the water is 4 meters how much work is required to pump all the water to the top of the tank. Assume that the density of water is 1000 kg/m3.

Hint

It will help you work this problem if you first go through the previous practice problem. In that video, he does a great job of explaining how to set up the integral.

Problem Statement

A tank of water is the shape of the lower half of a sphere of radius 6 meters. If the initial depth of the water is 4 meters how much work is required to pump all the water to the top of the tank. Assume that the density of water is 1000 kg/m3.

Final Answer

\(2508800\pi\) N-m

Problem Statement

A tank of water is the shape of the lower half of a sphere of radius 6 meters. If the initial depth of the water is 4 meters how much work is required to pump all the water to the top of the tank. Assume that the density of water is 1000 kg/m3.

Hint

It will help you work this problem if you first go through the previous practice problem. In that video, he does a great job of explaining how to set up the integral.

Solution

Like he does in the previous practice problem, we will set up the coordinate system at the center of the sphere (if we had a full sphere). Our problem has only the bottom half but the same ideas hold.

\(\displaystyle{W = \int_{-6}^{-2}{(-y)(9.8)(1000)\pi(36-y^2) ~dy } }\)

\(\displaystyle{W = 9800\pi \int_{-6}^{-2}{y^3-36y ~dy } }\)

\(\displaystyle{W = 9800\pi \left[\frac{y^4}{4} - \frac{36y^2}{2} \right]_{-6}^{-2} }\)

\(\displaystyle{W = 9800\pi \left[\frac{(-2)^4}{4} - \frac{36(-2)^2}{2} \right] - \left[\frac{(-6)^4}{4} - \frac{36(-6)^2}{2} \right] }\)

\(\displaystyle{W = 9800\pi \left[4 - 72 \right] - \left[324 - 648 \right] }\)

\(\displaystyle{W = 9800\pi \left[256 \right] = 2508800\pi }\) N-m

Final Answer

\(2508800\pi\) N-m

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For a truncated cone 8ft tall with bottom radius 3ft and top radius 6ft, calculate the work required to pump an initially full tank of water out of the top.

Problem Statement

For a truncated cone 8ft tall with bottom radius 3ft and top radius 6ft, calculate the work required to pump an initially full tank of water out of the top.

Solution

2509 video

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We have a cylindrical tank that is laying on its side 10ft under the ground. The radius is 4ft and the length is 10ft. How much work is required to empty the tank up to ground level if it is initially full of water?

Problem Statement

We have a cylindrical tank that is laying on its side 10ft under the ground. The radius is 4ft and the length is 10ft. How much work is required to empty the tank up to ground level if it is initially full of water?

Solution

2510 video

video by Dr Phil Clark

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A conical water tank that is 18ft high and has a radius of 12ft at the top is filled to a depth of 15ft. Find the work done in pumping out all the water over the top edge of the tank.

Problem Statement

A conical water tank that is 18ft high and has a radius of 12ft at the top is filled to a depth of 15ft. Find the work done in pumping out all the water over the top edge of the tank.

Solution

2512 video

video by Dr Phil Clark

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A cylindrical container with a depth of 3 feet and a radius of 1 foot is filled with water to a depth of 2 feet. How much work is done to pump all the water out over the top edge of the container? Water weighs 62.4 lbs/ft.

Problem Statement

A cylindrical container with a depth of 3 feet and a radius of 1 foot is filled with water to a depth of 2 feet. How much work is done to pump all the water out over the top edge of the container? Water weighs 62.4 lbs/ft.

Solution

2514 video

video by Dr Phil Clark

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A cylindrical water tank that is 18 feet high and has a radius of 12 feet at the top is filled to a depth of 15 feet. Find the work done in pumping out all of the water over the top edge of the tank.

Problem Statement

A cylindrical water tank that is 18 feet high and has a radius of 12 feet at the top is filled to a depth of 15 feet. Find the work done in pumping out all of the water over the top edge of the tank.

Solution

2515 video

video by Dr Phil Clark

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A trough that has a triangular cross section that is 5m high, 3m wide at the top and 8m long is filled up to 3 meters with water. Given that the density of water is 1000 kg/m3, how much work is required in order to empty the trough?

Problem Statement

A trough that has a triangular cross section that is 5m high, 3m wide at the top and 8m long is filled up to 3 meters with water. Given that the density of water is 1000 kg/m3, how much work is required in order to empty the trough?

Solution

2518 video

video by Dr Phil Clark

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You CAN Ace Calculus

Topics You Need To Understand For This Page

Related Topics and Links

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Practice Instructions

Unless otherwise instructed, solve these problems giving your answers in exact terms.
Note - Although we recommend in the article on successfully solving word problems to organize your problems so that you work several similar problems one right after another, we have not arranged them like that here. We let you do it, if you choose to.

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