tinyboy21 Posted February 8, 2005 Share Posted February 8, 2005 This is an extra credit problem for my calc class, and I doubt anyone in my class will get it. Please help, as I am also clueless. Ellipse E is centered at the origin, and has a horizontal minor axis of length 4. If you rotate the portion of E which falls only in the first and second quadrants about the x-axis, the resulting rotational solid has volume 800pi/27 . Find the length of E's major axis without using a calculator. Link to comment Share on other sites More sharing options...

timo Posted February 8, 2005 Share Posted February 8, 2005 Is the answer supposed to be a function of the resulting volume V? In that case (and if I understood your question correctly - I´m not a native english speaker) you could calculate the resulting volume V as a function of minor axis lenght a and major axis length b (V=f(a,b)) and solve the resulting equation for the length of the major axis (b=g(V,a)), then. Link to comment Share on other sites More sharing options...

Dave Posted February 9, 2005 Share Posted February 9, 2005 The equation of your ellipse is (in this case) [math]\frac{x^2}{16} + \frac{y^2}{a^2} = 1[/math] where a is the semi-major length. Consider the rotation about the x-axis of the ellipse: [math]V = \pi\int_{-2}^{2} y^2 \, dx = a \int_{-2}^{2} 1-\frac{x^2}{16}\,dx[/math]. If you know that [math]V = \frac{800\pi}{27}[/math] then by evaluating the integral you can calculate a precisely. Link to comment Share on other sites More sharing options...

tinyboy21 Posted February 9, 2005 Author Share Posted February 9, 2005 is this right? and the final answer is 20/3 Link to comment Share on other sites More sharing options...

Dave Posted February 9, 2005 Share Posted February 9, 2005 I think you're missing a factor of 2 there somewhere. Btw, I got my volume of integration formula wrong at the top, changed it now though. Link to comment Share on other sites More sharing options...

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