surface area of a light bulb

**BigPapa** · December 2nd 2008, 10:29 PM

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.

**mr fantastic** · December 2nd 2008, 11:05 PM

Originally Posted by BigPapa

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.

I assume you're familiar with the formula and can calculate dy/dx.

Start by showing that $\sqrt{\left(\frac{dy}{dx} \right)^2 + 1} = \frac{1}{6} \left( x^{-1/2} + 9 x^{1/2}\right)$ .

The rest should by routine.

**shawsend** · December 3rd 2008, 01:58 AM

I recommend we use a better looking lightbulb:

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

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