Need help on finding the surface area generated by revolving the curve

y=square root of 25-x^2 around the X-axis for -3 ≤ x ≤ 4

Thank you

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- Feb 16th 2010, 11:08 AMmacadamianchicaFind the surface area generated by revolving the curve
Need help on finding the surface area generated by revolving the curve

y=square root of 25-x^2 around the X-axis for -3 ≤ x ≤ 4

Thank you - Feb 16th 2010, 12:13 PMblertta
- Feb 16th 2010, 12:18 PMArchie Meade
As the curve revolves around the x-axis, each point on graph of the function traces out a circle, with f(x) as radius.

To calculate the surface area, the circle circumferences are integrated.

$\displaystyle R=f(x),\ -3\le{x}\le{4}$

Hence, the curved surface area is.... $\displaystyle 2\pi\int_{-3}^4f(x)dx\ =\ 2\pi\int_{-3}^4\sqrt{25-x^2}dx$

You may or may not want to include the circular side faces. - Feb 17th 2010, 01:34 AMblertta
- Feb 17th 2010, 01:45 AMArchie Meade
This is not the "integration of a function problem".

It's a 3-dimensional "surface of revolution".