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
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.