Hi,

I am solving a problem which involves calculating the surface are of a solid revolution about the x – axis. I have found that the surface area can be given by:

$\displaystyle A = 2\Pi \int_{-70}^{110}r+\frac{rd}{\sqrt{r^{2}-x^{2}}}\: dx$

I have done this using the trapezium rule as I know the values of r and d, so I have a rough estimate of a figure I am aiming for.

The question I am trying to answer says to use the standard integral:

$\displaystyle \int \frac{dx}{\sqrt{k^{2}-x^{2}}}= \tan^{-1}\left ( \frac{x}{\sqrt{k^{2}-x^{2}}} \right )+C$

I am struggling to interpret the r+rd/√(r^2-x^2) into the form dx/√(k^2-x^2) in order to evaluate the integral. Any help or guidance would be appreciated as I have spent too much time trying to do this.

Thanks