Deriving the first moment of area of semicircle

[Elbobo]

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?

 

[nvn]

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]

Sorry, I really don't understand why dA equals that. My A(y) must be wrong then? What should it be and why?

 

[nvn]

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