# A Few Small Integration Problems

1. May 24, 2010

### Lancelot59

I just have a few small questions.

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

$$y=tan^{3}(x)$$, $$y=1$$, and $$x=0$$ by rotating it about y=1.

$$\pi\int_{0}^{\frac{\pi}{4}} {(tan^{3}(x) - 1)^{2}dx}$$

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:
$$\pi\int_{0}^{\frac{\pi}{4}} {(1 - tan^{3}(x))^{2}dx}$$

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: May 24, 2010
2. May 24, 2010

### Staff: Mentor

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 $\Delta V$ is pi * R2$\Delta x$, where R is the radius from the line y = 1 to the curve.
Yes, that doesn't look right at all, based on the information you have given.

3. May 24, 2010

### Lancelot59

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:

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

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

$$V = \int_{b}^{a} [f(x)+3]^{2} dx$$

Last edited: May 24, 2010
4. May 24, 2010

### Staff: Mentor

I think you have it now. To recap, your integral should look like this:
$$V = \pi \int_{0}^{\pi/4} [tan^3(x) - 1]^{2} dx$$

5. May 24, 2010

### Lancelot59

That's what I got, but the book has the 1 and the cubic tangent reversed.

6. May 24, 2010

### Staff: Mentor

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. May 24, 2010

### Staff: Mentor

I had to look it up in my Halliday & Resnick, which says
$$W = \Delta K = - \Delta U$$
where W = work, $\Delta K$ = change in kinetic energy, and $\Delta U$ = change in potential energy

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

$$W = \int_{x_0}^x F(x) dx$$

8. May 24, 2010

### Lancelot59

Right, I see now.

Great. Thanks for the help!