Discussion

The centroid of the quarter disc \( x^{2}+y^{2} \le a^{2} \) with \( x, y \ge 0 \) has \( \bar{x} = \)

Answer: D

Divide the first moment by the area, working in polar coordinates with \( x = r\cos\theta \):

\[ \iint x\,dA \;=\; \int_{0}^{\pi/2}\!\!\int_{0}^{a} r\cos\theta \cdot r\,dr\,d\theta \;=\; \frac{a^{3}}{3}. \]

The area is \( \pi a^{2}/4 \), so

\[ \bar{x} \;=\; \frac{a^{3}/3}{\pi a^{2}/4} \;=\; \frac{4a}{3\pi} \;\approx\; 0.424a. \]

By symmetry \( \bar{y} \) takes the same value

No comment is present. Be the first to comment.

Loading…