Solid of revolution (barrel): coordinates to use?

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A barrel has the shape of a solid of revolution obtained by rotating about its major axis the part of an ellipse lying between lines through its foci perpendicular to that axis. The barrel is 120cm high and 60cm in radius at its middle. Find its volume

My working...

Do I use the standard cartesian equation of
x^2 /a^2 + y^2/b^2 = 1?

I am not sure where to start...

thank you in advance
 
You know \(\displaystyle b=60\) and \(\displaystyle c=60\)

Then a would be \(\displaystyle \sqrt{b^{2}+c^{2}\)

Use your values and solve the ellipse equation for y.

You then have:

\(\displaystyle \L\\V=2{\pi}\int_{0}^{c}y^{2}dx\)
 
Re: Solid of revolution

Hello, americo74!

Did you make a sketch?

Here's a baby-step rehash of what galactus said . . .

A barrel has the shape of a solid of revolution obtained by rotating about its major axis
the part of an ellipse lying between lines through its foci perpendicular to that axis.
The barrel is 120cm high and 60cm in radius at its middle.
Find its volume.
Turning the barrel on its side, the graph looks like this:
Code:
                  |60
                * * *
            *     |     * 
         *  |     |     |  *
       *    |     |     |    *
      *     |     |     |     *
    - * - - o - - + - - o - - * -
      *     |     |     |60   *
       *    |     |     |    *
         *  |     |     |  *
            *     |     *
                * * *
                  |

As galactus pointed out: \(\displaystyle \,b\,=\,60,\;c\,=\,60.\)

From \(\displaystyle a^2\:=\:b^2\,+\,c^2,\,\) we have: \(\displaystyle \,a^2\:=\:60^2\,+\,60^2\:=\:7200\)

Hence, the ellipse is: \(\displaystyle \L\,\frac{x^2}{7200}\,+\,\frac{y^2}{3600}\;=\;1\;\;\Rightarrow\;\;y^2\:=\:\frac{7200\,-\,x^2}{2}\)

Since: \(\displaystyle \L\,V\;=\;\pi\int^{\;\;\;b}_ay^2\,dx\)
\(\displaystyle \;\;\)we have: \(\displaystyle \L\,V\;=\;2\,\times\,\pi\int^{\;\;\;60}_0\frac{7200\,-\,x^2}{2}\,dx\)
 
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