Dear God help me.

[tinyboy21]

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.

[timo]

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.

[Dave]

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.

[tinyboy21]

is this right?

 

and the final answer is 20/3

[Dave]

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.