# Volumes of Revolution

• Jan 31st 2010, 11:51 AM
dalbir4444
Volumes of Revolution
Calculate the volume of the solid obtained when the triangle with vertices (2,5), (6,1) , (4,4) is rotated about the line x=-3 using either cylindrical shells or disks.

I found the slope for each line, and found an equation fro each. Then, using cylindrical shells with radius x+3, I found the volume generated by each line and then subtracted the results. However, I got a negative value.
• Jan 31st 2010, 03:45 PM
TKHunny
Couple of things wrong with this:

1) Lines don't tend to generate much volume.

2) You have three lines. Which did you subract from which?

3) What were you integration limits? What did your integrals look like?

4) Have you ever heard of Pappas' Theorem?

5) Please show your work. Someone will guide you once you demonstrate where it is that you need guidance.
• Jan 31st 2010, 04:29 PM
dalbir4444
My first integral was 2pi(3+x)(7-x) in the interval [2,6].
My second integral was 2pi(3+x)(-0.5x+6) over the interval [2,4].
My third integral was 2pi(3+x)(-1.5x+10) over the interval [4,6].

I subtracted the second and third integral from the first one. We haven't learned Pappas' Theorem yet, so we can only use cylindrical shells or disks method.
• Jan 31st 2010, 07:38 PM
TKHunny
You're upside down. y = 7-x is less than the other two on [2,6].

That's a lot of nice calculus to go down with an analytic geometry error.
• Feb 1st 2010, 08:16 AM
dalbir4444
That's why I was getting a negative answer, but I can't seem to visualize the volume of revolution. Am I on the right path, and if so, which ones am I supposed to subtract?
• Feb 1st 2010, 09:21 AM
TKHunny
1) Visualization should become unnecessary. If you learn to rely on it with so few variables, what will you do when the number increases?

2) Part of your difficulty visualizing, I think, is your unusual construction. You probably cannot visualise subtraction of volumes. If you were to define the two pieces that you want as single entities, it may help. For one thing, it will get rid of all the unnecessary distractions down to the x-axis.

In this example, using just the domains of your integrals, rather than [2,4]+[4,6]-[2,6], as you have it, try ([2,4]-[2,4])+([4,6]-[4,6]) - as in, also split up the subtraction piece into rational chunks. It's much more palatable for your eyes.
• Feb 1st 2010, 04:26 PM
dalbir4444
I don't completely understand what you mean. I know that the limits of my integrals are incorrect, but what I'm not sure of is why they are incorrect. You said " try ([2,4]-[2,4])+([4,6]-[4,6]) - as in, also split up the subtraction piece into rational chunks". Which integrals am I supposed to take over these limits?
• Feb 1st 2010, 06:23 PM
TKHunny
I knew that description with just the limits would not be clear. I regretted it when I hit the [submit] button.

You have these

$\displaystyle 2\pi\cdot\int_{2}^{4}(x+3)\left(6-\frac{x}{2}\right)\;dx$

$\displaystyle 2\pi\cdot\int_{4}^{6}(x+3)\left(10-\frac{3}{2}x\right)\;dx$

$\displaystyle 2\pi\cdot\int_{2}^{6}(x+3)(7-x)\;dx$

These three chunks are "right" and the do solve the problem. They do not contribute to your visualization. This next version should contribute to your visualization.

$\displaystyle 2\pi\cdot\int_{2}^{4}(x+3)\left[\left(6-\frac{x}{2}\right)-(7-x)\right]\;dx$

$\displaystyle 2\pi\cdot\int_{4}^{6}(x+3)\left[\left(10-\frac{3}{2}x\right)-(7-x)\right]\;dx$

It's more a picture of the pieces you want. Same thing, just a different presentation.