can someone how to do this problem please?

Find the surface area of an ornamental light bulb designed by revolving the graph of y=(1/3)x^(1/2)-x^(3/2), on the interval 0 to 1/3, about the x axis.

Printable View

- Dec 2nd 2008, 10:29 PMBigPapasurface area of a light bulb
can someone how to do this problem please?

Find the surface area of an ornamental light bulb designed by revolving the graph of y=(1/3)x^(1/2)-x^(3/2), on the interval 0 to 1/3, about the x axis. - Dec 2nd 2008, 11:05 PMmr fantastic
- Dec 3rd 2008, 01:58 AMshawsend
I recommend we use a better looking lightbulb:

$\displaystyle f(x)=\begin{cases}0.211818 x^2 - 0.042646 x + 0.531034 & 0 \leq x \leq 1.88983 \\ \sqrt{1.35^2-(x-2.4913)^2} & 1.88983\leq x\leq 3.93\end{cases}$