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Surface area
Hi!
Problem is to find the area. (See picture below, along with description of problem).
The book says, $\displaystyle A=2\pi\cdot \bar{x}\cdot L $ , where $\displaystyle \bar{x}$ is the ycoordinate of the centroid of the line. And L is the length of the line.
I donīt get correct results :/
I did for the A circular arc: $\displaystyle 2\pi\cdot \left(\underbrace{100\frac{50\cdot sin(\frac{\pi}{4})}{\frac{\pi}{4}}\cdot cos(\frac{\pi}{4})}_{\text{xcoordinate}}\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 $\displaystyle 9.87(10^{4}) $
Thx!


I'd first calculate it the regular way using the formula:
$\displaystyle S=2\pi \int_a^b x\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy }{dt}\right)^2}dt$
Right?
Me, I'd use complex coordinates. The equation for the top curve is $\displaystyle 4+4i +2 e^{it}$. You can do the bottom one. Thus, the parametric form of the top curve is $\displaystyle x=4+2\cos(t), y=4+2\sin(t)$. 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 ycoordinate of the centroid and see if I get agreement.