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Math Help - Another solid of revolution, thanks

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    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
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    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.
    Last edited by HallsofIvy; October 31st 2013 at 05:16 PM.
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    Re: Another solid of revolution, thanks

    Quote Originally Posted by HallsofIvy View Post
    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...
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