# Thread: Volume using Washer/Disc method

1. ## Volume using Washer/Disc method

I have a problem I'm working on for a study guide, but it's worded a little oddly compared to the other problems.

Set up the integral and find the volume of the bounded region with the washer/disc method.

15. Below by y=3x^2_1 and above by y=4, rotated about the line y=4.

It doesn't seem like it should be that hard, but again the wording is throwing me off. Any help would be much appreciated. The answer is (48pi/5). I just need to figure out how to get there.

2. ## Re: Volume using Washer/Disc method

Originally Posted by rberg1897
I have a problem I'm working on for a study guide, but it's worded a little oddly compared to the other problems.

Set up the integral and find the volume of the bounded region with the washer/disc method.

15. Below by y=3x^2_1 and above by y=4, rotated about the line y=4.

It doesn't seem like it should be that hard, but again the wording is throwing me off. Any help would be much appreciated. The answer is (48pi/5). I just need to figure out how to get there.
I think something is wrong with your question. Here is a graph of the region in question.

It will have infinite volume if it is rotated around y=4

3. ## Re: Volume using Washer/Disc method

I think he means the bounded part between the parabola and the line (bounded below by the parabola and bounded above by the line). If so, the integral goes like:

$\displaystyle \int_{min(x)}^{max(x)} \pi (radius)^2 dx$

where radius is $\displaystyle 4 - (3x^2-1)$, and min(x) and max(x) are the x-coordinates of the intersection points.

- Hollywood