Deriving the first moment of area of semicircle

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Discussion Overview

The discussion revolves around deriving the first moment of area \( Q \) of a semicircle with radius \( r \) using integration. Participants are exploring the mathematical formulation and attempting to clarify the correct expressions for area and differential area elements.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • The initial attempt at deriving \( Q \) uses the equation \( Q = \int_{A} y dA \) and proposes \( dA = \pi y dy \), leading to a calculated result of \( \frac{\pi r^{3}}{3} \).
  • Some participants propose that \( dA \) should instead be expressed as \( dA = 2\sqrt{r^2 - y^2} dy \), suggesting a different approach to finding the area element.
  • There is a request for clarification on the correct formulation of \( A(y) \) and how it relates to \( dA \), indicating uncertainty about the integration process.
  • Participants discuss the need to derive \( Q \) without prior knowledge of the centroidal height, highlighting a desire for a more general method applicable to various shapes.
  • Areas of Agreement / Disagreement

    Participants do not appear to reach a consensus on the correct expression for \( dA \) or the approach to deriving \( Q \). Multiple competing views remain regarding the formulation of the area element and the integration process.

    Contextual Notes

    Limitations include potential misunderstandings of the geometric relationships involved in the semicircle and the assumptions made in the integration process. The discussion does not resolve the mathematical steps leading to the differing results.

    Who May Find This Useful

    This discussion may be useful for students and practitioners interested in the mathematical derivation of moments of area, particularly in the context of semicircular shapes and integration techniques.

Elbobo
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Homework Statement


Derive via integration the first moment of area Q of a semicircle with radius r.

Homework Equations


Q = \int_{A} y dA

A_{semicircle} = \frac{\pi r^{2} }{2}

The Attempt at a Solution


A = \frac{\pi r^{2} }{2}
A(y) = \frac{\pi y^{2} }{2}
dA = \pi y dy

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

Q = \frac{\pi r^{3}}{3}But the answer is \frac{2 r^{3} }{3}, which my textbook derived from the equation Q = (area) \times (centroidal height). 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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Elbobo: dA is not pi*y*dy. Hint: Shouldn't dA instead be, dA = 2[(r^2 - y^2)^0.5]*dy? Try again.
 
Sorry, I really don't understand why dA equals that. My A(y) must be wrong then? What should it be and why?
 
Elbobo: A(y) = integral(dA), integrated from y = y1 to y = r. In your particular case, y1 = 0.
 

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