Find the centroid of the homogeneous figure bounded
Find the centroid of the homogeneous figure bounded by the curve $y=\sin x$ and the lines $y=0$, $x=\dfrac\pi4$.
Explanation of Method
With the region described as $0\le x\le\pi/4,\,0\le y\le\sin x$, the area and static moments reduce to elementary and integration-by-parts computations.
- Compute the area $A$.
- Compute the static moment $M_y$ (integration by parts).
- Compute the static moment $M_x$.
- Divide to get the centroid.
Step 1 · Area
$$ A = \int_0^{\pi/4}\sin x\,dx = 1-\frac{\sqrt2}2 = \frac{2-\sqrt2}2. $$
Step 2 · Static moment $M_y$
By parts, $u=x,\,dv=\sin x\,dx$:
$$ M_y = \int_0^{\pi/4}x\sin x\,dx = \left[-x\cos x\right]_0^{\pi/4}+\int_0^{\pi/4}\cos x\,dx = -\frac{\pi\sqrt2}8+\frac{\sqrt2}2 = \frac{\sqrt2}8(4-\pi). $$
Step 3 · Static moment $M_x$
$$ M_x = \frac12\int_0^{\pi/4}\sin^2x\,dx = \frac14\int_0^{\pi/4}(1-\cos2x)\,dx = \frac14\left(\frac\pi4-\frac12\right) = \frac{\pi-2}{16}. $$
Step 4 · Centroid
Dividing and rationalizing (multiplying by $\dfrac{2+\sqrt2}{2+\sqrt2}$):
$$ \bar x = \frac{M_y}A = \frac{\sqrt2(4-\pi)/8}{(2-\sqrt2)/2} = \frac{(4-\pi)(1+\sqrt2)}4, \qquad \bar y = \frac{M_x}A = \frac{(\pi-2)/16}{(2-\sqrt2)/2} = \frac{(\pi-2)(2+\sqrt2)}{16}. $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically: $A\approx0.292893$, $\bar x\approx0.518095$, $\bar y\approx0.243603$, matching the boxed closed forms exactly.
Result
The centroid of the region is: