Surface area

**Twig** · July 11th 2009, 02:45 AM

Hi!

Problem is to find the area. (See picture below, along with description of problem).

The book says, $A=2\pi\cdot \bar{x}\cdot L$ , where $\bar{x}$ is the y-coordinate of the centroid of the line. And L is the length of the line.

I don´t get correct results :/

I did for the A circular arc: $2\pi\cdot \left(\underbrace{100-\frac{50\cdot sin(\frac{\pi}{4})}{\frac{\pi}{4}}\cdot cos(\frac{\pi}{4})}_{\text{x-coordinate}}\right)\cdot \underbrace{\frac{2\pi\cdot 50}{4}}_{\text{Length of arc}}$

Then I did the same for the B arc. And I add them.
The result is incorrect. Even if I try to add the top area as well, I get the wrong result.

Note: Instead of 2 inch use 50mm , and the answer should be $9.87(10^{4})$

Thx!

**malaygoel** · July 11th 2009, 06:16 AM

These articles may help you:

Surface of Revolution -- from Wolfram MathWorld
Pappus's Centroid Theorem -- from Wolfram MathWorld

However, answer to your problem is $16\pi^2$

**shawsend** · July 11th 2009, 06:17 AM

I'd first calculate it the regular way using the formula:

$S=2\pi \int_a^b x\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy }{dt}\right)^2}dt$

Right?

Me, I'd use complex coordinates. The equation for the top curve is $4+4i +2 e^{it}$ . You can do the bottom one. Thus, the parametric form of the top curve is $x=4+2\cos(t), y=4+2\sin(t)$ . Now, just substitute both expressions, making sure you get the starting and ending t times correctly, into the formula above.

After I get the right answer, then I'd go on to calculate the moment of the figure and then the y-coordinate of the centroid and see if I get agreement.

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