# Word problems (volume &amp; work by integrals)

• silicon_hobo
In summary, the conversation is about solving two word problems involving volume and work by integrals. The first problem involves finding the volume of a pie-dish with specific dimensions, while the second problem involves finding the work required to pump all of the water from a cylindrical tank with given dimensions. The conversation includes some calculations and corrections, and ends with a discussion about the center of gravity and its relation to the work required to lift the fluid.
silicon_hobo
[SOLVED] Word problems (volume &amp; work by integrals)

## Homework Statement

a) A pie-dish with circular cross-section is 24cm across the top, 16cm across the bottom and 6cm deep. Find the volume of the dish.

b) A water tank in the form of a cylinder with height 20 ft and radius 6 ft is half full of water, find the work required to pump all of the water over the top rim (water weighs 62.5lb/ft^3.

## The Attempt at a Solution

a)I'm pretty sure I've got this one, just looking for some confirmation.
http://www.mcp-server.com/~lush/shillmud/int1.8.JPG

b)I found this one tougher. I used 32 ft/s for gravity and 1.36 J/lbft. Am I on the right track? Thanks for reading.
http://www.mcp-server.com/~lush/shillmud/int1.9.JPG

I agree with your first result.

For the other problem, I guess we are assuming that the cylindrical tank is upright (symmetry axis oriented vertically). Tank-draining problems also are given using horizontally-placed cylinders, so I'm presuming you had a picture of this.

Three problems on your second result --

1) The 2250(pi)(delta_x) is already in pounds. Keep in mind that, in the English system, pounds are units of weight and thus force. You do not need to multiply by 32 ft/(sec^2) because the unit already contains the value of g implicitly. (The unit of mass in the English system is the slug, which has a weight of 32.2 pounds. The value you are given for water is a weight density.) Just as well, because you dropped a decimal in the next line: 2250 times 32 is 72,000; but you aren't going to use this anyway.

2) So you now multiply 2250(pi)(delta_x) by the distance you will lift each infinitesimal layer of water and integrate the infinitesimal work from x = 10 to x = 20. Fine, but the integral of x dx is (1/2)(x^2). You would now have 2250(pi)(1/2)(400-100).

3) On the last line, when you took your result over to convert it to Joules, you dropped the (pi). That would be 2,160,000(pi) x 1.36 , but must now be corrected further in any event.

Here's a quick way to check your answer on a problem with a simple symmetry such as this one. You are asked to lift a total cylindrical volume of water (pi)(6^2)(10) ft^3 with a weight of (pi)(6^2)(10)(62.5) lb vertically by an average distance of 15 ft (which is the location of its center of gravity below the lip of the tank). Thus the total work to be done is

(pi)(6^2)(10)(62.5)(15) ft-lb .

Last edited:
Hey. Thanks for your your help. Just one more thing, where did you get 15 for the center of gravity? Is it simply halfway between the surface of the water and the top of the tank? Cheers.

-SH

silicon_hobo said:
Hey. Thanks for your your help. Just one more thing, where did you get 15 for the center of gravity? Is it simply halfway between the surface of the water and the top of the tank?

The surface of the water is 10 feet down from the rim of the tank and the bottom is 20 feet down. So the center of the cylinder of water is 15 feet down from the rim, so this represents the average distance that the infinitesimal layers of water will be lifted.

In fact, there is a theorem stating that the total amount of work required to lift the fluid is equal to the weight of the fluid times the distance its center of mass or gravity needs to be raised.

## 1. What are volume and work by integrals?

Volume and work by integrals are concepts used in mathematics and physics to calculate the volume of an object or the work done by a force over a given distance. These concepts involve using integrals, which are mathematical tools used to find the area under a curve.

## 2. How do you solve word problems involving volume and work by integrals?

To solve word problems involving volume and work by integrals, you first need to identify the given information and the unknown variables. Then, you can set up an integral using the appropriate formula for the problem and solve for the unknown variable using integration techniques. It is important to carefully read the problem and understand the physical meaning of the variables involved.

## 3. What are some common formulas used to find volume and work by integrals?

Some common formulas used to find volume and work by integrals include the formula for calculating the volume of a solid of revolution, which involves using the cross-sectional area of the object and integrating it over the given interval. Another formula is the work formula which involves multiplying the force and distance and integrating it over the given interval.

## 4. How do volume and work by integrals relate to real-world scenarios?

Volume and work by integrals are used in various real-world scenarios, such as calculating the volume of a water tank, finding the work done by a force to lift an object, or determining the amount of fluid flowing through a pipe. These concepts are also used in engineering, physics, and other fields to solve practical problems.

## 5. Are there any common mistakes to avoid when solving word problems involving volume and work by integrals?

One common mistake to avoid when solving word problems involving volume and work by integrals is not carefully considering the units of measurement. It is important to ensure that all variables and formulas are using the same units to avoid errors in the final answer. Another mistake is not carefully setting up the integral or using the wrong formula for the given problem. It is important to double-check the setup before solving the integral.

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