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Deriving the first moment of area of semicircle

by Elbobo
Tags: deriving, moment, semicircle
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Elbobo
#1
Apr25-11, 11:47 PM
P: 145
1. The problem statement, all variables and given/known data
Derive via integration the first moment of area Q of a semicircle with radius r.

2. Relevant equations
[tex]Q = \int_{A} y dA[/tex]

[tex] A_{semicircle} = \frac{\pi r^{2} }{2}[/tex]

3. The attempt at a solution
[tex] A = \frac{\pi r^{2} }{2}[/tex]
[tex] A(y) = \frac{\pi y^{2} }{2}[/tex]
[tex] dA = \pi y dy[/tex]

[tex]Q = \int^{y=r}_{y=0} y dA[/tex]
[tex] = \int^{r}_{0} \pi y^{2} dy[/tex]
[tex] = \frac{\pi}{3} [y^{3}]^{r}_{0}[/tex]

[tex] Q = \frac{\pi r^{3}}{3}[/tex]


But the answer is [tex]\frac{2 r^{3} }{3}[/tex], which my textbook derived from the equation [tex]Q = (area) \times (centroidal height) [/tex]. I want to know how to derive the Q for any shape without knowing its centroidal height beforehand. Can someone help me out with why I got a different and wrong answer?
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nvn
#2
Apr26-11, 12:47 AM
Sci Advisor
HW Helper
P: 2,124
Elbobo: dA is not pi*y*dy. Hint: Shouldn't dA instead be, dA = 2[(r^2 - y^2)^0.5]*dy? Try again.
Elbobo
#3
Apr26-11, 12:55 AM
P: 145
Sorry, I really don't understand why dA equals that. My A(y) must be wrong then? What should it be and why?

nvn
#4
Apr26-11, 08:35 AM
Sci Advisor
HW Helper
P: 2,124
Deriving the first moment of area of semicircle

Elbobo: A(y) = integral(dA), integrated from y = y1 to y = r. In your particular case, y1 = 0.


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