A Few Small Integration Problems

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In summary, the conversation involves setting up integrals to find the volume defined by a function and rotating it about a specific axis, as well as a conceptual question about work and potential energy. The correct setup for the integral is discussed and the relation between work and potential energy is explained.
  • #1
Lancelot59
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I just have a few small questions.

For this first question I need to setup the integral to find the volume defined by.

[tex]y=tan^{3}(x)[/tex], [tex]y=1[/tex], and [tex]x=0[/tex] by rotating it about y=1.

My result was this:
[tex]\pi\int_{0}^{\frac{\pi}{4}} {(tan^{3}(x) - 1)^{2}dx}[/tex]

From my understanding all you need to do is offset the function by the axis in order to make it work. But the actual answer is:
[tex]\pi\int_{0}^{\frac{\pi}{4}} {(1 - tan^{3}(x))^{2}dx}[/tex]

Which looks like it would do something else entirely. I had a similar issue when doing the same thing for sin(x) about y=1. What's going on here?

My second issue is a conceptual question involving work. Just to make sure I have the order of everything correct, how does say, the work done to move a spring relate to it's potential energy? Is it the integral of work = potential energy?
 
Last edited:
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  • #2
Lancelot59 said:
I just have a few small questions.

For this first question I need to setup the integral to find the volume defined by.

[tex]y=tan^{3}(x)[/tex], [tex]y=1[/tex], and [tex]x=0[/tex] by rotating it about y=1.
The tangent function (and its cube) are not continuous in the interval [0, pi], which you show below. The region being rotated is bounded by the y-axis, the line y = 1, and the graph of y = tan^3(x). For another, you are apparently using disks, but haven't set up the integral correctly. Using disks, the typical volume element [itex]\Delta V[/itex] is pi * R2[itex]\Delta x[/itex], where R is the radius from the line y = 1 to the curve.
Lancelot59 said:
My result was this:
[tex]\pi\int_{0}^{\pi} {(tan^{3}(x) - 1)dx}[/tex]

From my understanding all you need to do is offset the function by the axis in order to make it work. But the actual answer is:
[tex]\pi\int_{0}^{\pi} {(1 - tan^{3}(x))dx}[/tex]


Which looks like it would do something else entirely. I had a similar issue when doing the same thing for sin(x) about y=1. What's going on here?
Yes, that doesn't look right at all, based on the information you have given.
Lancelot59 said:
My second issue is a conceptual question involving work. Just to make sure I have the order of everything correct, how does say, the work done to move a spring relate to it's potential energy? Is it the integral of work = potential energy?
 
  • #3
Sorry, I forgot to square the function when I entered it in...and I meant to put pi/4. I've corrected it now.

From what I was told in class though, all I need to do to get the volume using the disk method is this:

[tex]V = \int_{b}^{a} [f(x)]^{2} dx[/tex]

and if I was revolving around a differnet axis, say y=-3 then I would just do this:

[tex]V = \int_{b}^{a} [f(x)+3]^{2} dx[/tex]
 
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  • #4
I think you have it now. To recap, your integral should look like this:
[tex]V = \pi \int_{0}^{\pi/4} [tan^3(x) - 1]^{2} dx[/tex]
 
  • #5
That's what I got, but the book has the 1 and the cubic tangent reversed.
 
  • #6
When you calculate the overall radius of the shell you should have R - r, with R being the larger y value and r being the smaller y value. In this case, the overall radius is 1 - tan^3(x). Doesn't matter in this case, because it's squared.
 
  • #7
Lancelot59 said:
My second issue is a conceptual question involving work. Just to make sure I have the order of everything correct, how does say, the work done to move a spring relate to it's potential energy? Is it the integral of work = potential energy?
I had to look it up in my Halliday & Resnick, which says
[tex]W = \Delta K = - \Delta U[/tex]
where W = work, [itex]\Delta K[/itex] = change in kinetic energy, and [itex]\Delta U[/itex] = change in potential energy

If a force F is applied in a straight line, work W is defined as

[tex]W = \int_{x_0}^x F(x) dx[/tex]
 
  • #8
Mark44 said:
When you calculate the overall radius of the shell you should have R - r, with R being the larger y value and r being the smaller y value. In this case, the overall radius is 1 - tan^3(x). Doesn't matter in this case, because it's squared.
Right, I see now.

Mark44 said:
I had to look it up in my Halliday & Resnick, which says
[tex]W = \Delta K = - \Delta U[/tex]
where W = work, [itex]\Delta K[/itex] = change in kinetic energy, and [itex]\Delta U[/itex] = change in potential energy

If a force F is applied in a straight line, work W is defined as

[tex]W = \int_{x_0}^x F(x) dx[/tex]

Great. Thanks for the help!
 

1. What is integration?

Integration is a mathematical process that involves finding the area under a curve or the accumulation of a quantity over a certain range. It is often used to solve problems in calculus, physics, and engineering.

2. What are some common integration techniques?

Some common integration techniques include substitution, integration by parts, partial fractions, and trigonometric substitution. These techniques are used to simplify the integrand and make it easier to solve.

3. What are "a few small integration problems"?

"A few small integration problems" refers to a set of simple integration problems that can be solved using basic integration techniques. These problems typically involve straightforward integrands and can be used to practice and improve integration skills.

4. How can I improve my integration skills?

The best way to improve your integration skills is to practice solving a variety of integration problems. Start with simple problems and gradually work your way up to more complex ones. It's also helpful to review the fundamental rules and techniques of integration regularly.

5. Why is integration important in science?

Integration is important in science because it allows us to model and analyze real-world phenomena. It is used in many scientific fields, such as physics, chemistry, and biology, to solve problems and make predictions. Without integration, many scientific concepts and theories would not be possible.

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