# Thread: Volume of a solid rotated aroun the x axis

1. ## Volume of a solid rotated aroun the x axis

R is the region in the first and second quadrants bounded above by the graph y=(20)/(1 + x^2) and below by the line y= 2. find the volume of the solid when generated by the x axis.

Okay, so I set the hyperbola function as the outer radius and the horizontal line as the inner radius. Then I squared it and multipied by pi and integrated, but I'm getting a really big unreasonable number.

2. Originally Posted by helpplz
R is the region in the first and second quadrants bounded above by the graph y=(20)/(1 + x^2) and below by the line y= 2. find the volume of the solid when generated by the x axis.

Okay, so I set the hyperbola function as the outer radius and the horizontal line as the inner radius. Then I squared it and multipied by pi and integrated, but I'm getting a really big unreasonable number.
I'm guess you did

$\displaystyle V = \pi\int\limits_{-3}^{3} (\frac{20}{1 + x^2} - 2)^2dx$

You're supposed to square each individual one like.

$\displaystyle V = \pi\int\limits_{-3}^{3} (\frac{20}{1 + x^2})^2 - {2}^2dx$

Even if you did that correctly, I got for my final answer

$\displaystyle V = 96\pi + 400arctan(3) \approx 1871.190104$

3. Ah, okay. But you don't think I set my integral up wrong? (As in, used to wrong inner/outer radii?)

4. Originally Posted by helpplz
Ah, okay. But you don't think I set my integral up wrong? (As in, used to wrong inner/outer radii?)
The truth is I don't know for sure how you set it up; you didn't explain too well.

Yes, you are correct about setting the hyperbolic function as the outer radii and the y = 2 as the inner radii.

But in the integrand, did you subtract them then square them?
Or did you square each one individually then subtract them?

Write it out..

5. I subtracted and then squared. Which you explained was incorrect. Thanks for that. I was glad that my radii weren't wrong though, because that's my biggest struggle right now.