Deriving the first moment of area of semicircle

Click For Summary
SUMMARY

The discussion focuses on deriving the first moment of area (Q) for a semicircle with radius r using integration. The initial attempt incorrectly defined the differential area (dA) as πy dy, leading to an erroneous result of Q = πr³/3. The correct approach involves redefining dA as 2√(r² - y²) dy, which aligns with the formula Q = (area) × (centroidal height), ultimately yielding Q = 2r³/3. Participants emphasized the importance of accurately determining dA for successful integration.

PREREQUISITES
  • Understanding of integral calculus
  • Familiarity with the concept of moments of area
  • Knowledge of semicircular geometry
  • Ability to perform variable substitutions in integrals
NEXT STEPS
  • Study the derivation of moments of area for various geometric shapes
  • Learn about the centroidal height and its application in area calculations
  • Explore integration techniques involving polar coordinates
  • Investigate the relationship between area and moment of inertia
USEFUL FOR

Students studying structural engineering, civil engineering, or mechanical engineering, particularly those focusing on mechanics of materials and geometric properties of shapes.

Elbobo
Messages
145
Reaction score
0

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?
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
2K
Replies
6
Views
1K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
561
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K