# Math Help - Another solid of revolution, thanks

1. ## Another solid of revolution, thanks

I have now another problem, alike to my previous posted, but I'm confused since I get a negative results, wich is impossible, that means I have a mistake somewhere:

I have to find the solid of revolution by rotating about x axis the area bounded by y=sec(x) , y=1, x = -1 and x = 1;

the area would be pi(sec^2(x) - 1) and the volume is the integral from -1 to 1; how do I evaluate these x-values to integrate the function?

thanks

2. ## Re: Another solid of revolution, thanks

The region bounded by those curves is rather small! And the simplest way to find the volume rotated around the x-axis is to do it as two integrals. The volumes bounded above by y= sec(x), below by y= 0, on the left by x= -1, and on the right by x= 1, rotated around the x-axis, is $\pi\int_{-1}^1 sec^2(x) dx$. The volume bounded above by y= 1, below by y= 0, on the left by x= -1, and on the right by x= 1, rotated around the x-axis, is $\pi\int_{-1}^1dx= 2\pi$ (actually the volume of a cylinder with radius 1 and height 2). The volume you want is the difference between those two. That is the same as $\pi \int_{-1}^1 (sec^2(x)- 1) dx$.

That is based on the fact that the volume of a cylindrical "donut" with inner radius r, outer radius R, and height h, is $\pi(R^2- r^2)$ which is the same as the volume of the whole cylinder, $\pi R^2h$ minus the volume of the inner cylinder, $\pi r^2h$.

3. ## Re: Another solid of revolution, thanks

Originally Posted by HallsofIvy
The region bounded by those curves is rather small! And the simplest way to find the volume rotated around the x-axis is to do it as two integrals. The volumes bounded above by y= sec(x), below by y= 0, on the left by x= -1, and on the right by x= 1, rotated around the x-axis, is $\pi\int_{-1}^1 sec^2(x) dx$. The volume bounded above by y= 1, below by y= 0, on the left by x= -1, and on the right by x= 1, rotated around the x-axis, is $\pi\int_{-1}^1dx= 2\pi$ (actually the volume of a cylinder with radius 1 and height 2). The volume you want is the difference between those two. That is the same as $\pi \int_{-1}^1 (sec^2(x)- 1) dx$.

That is based on the fact that the volume of a cylindrical "donut" with inner radius r, outer radius R, and height h, is $\pi(R^2- r^2)$ which is the same as the volume of the whole cylinder, $\pi R^2h$ minus the volume of the inner cylinder, $\pi r^2h$.
and by solving this integral I did the following $\pi \int_{-1}^1 (sec^2(x)- 1) dx =$

$2\pi \int_{0}^{1} (sec^2(x)- 1) dx =$ by symmetry

$2\pi (tan(x) - x)_{0}^{1} =$ which gives me

$2\pi ( \frac{\pi}{4} - 1) =$ , which is negative... what am I doing wrong,

even if I evaluate the integral otherwise $\pi \int_{-1}^1 (sec^2(x)- 1) dx = \pi \int_{-1}^1 (tan^2(x)) dx$

which is $2\pi (tan(x) - x)_{0}^{1} =$ gives me the same results...