Question

Find the centroid of the region bounded by the given curves. y = 8 sin(2x), y = 8 cos(2x), x = 0, x = 8

Answer 1

The coordinates of the centroid are the average values of the $x$- and $y$-coordinates of the points $(x,y)$ that belong to the region. Let $R$ denote the bounded region. These averages are given by the integral expressions

$\dfrac{\displaystyle \iint_R x \, dA}{\displaystyle \iint_R dA} \text{ and } \dfrac{\displaystyle \iint_R y \, dA}{\displaystyle \iint_R dA}$

The denominator is just the area of $R$, given by

$\displaystyle \iint_R dA = \int_0^8 \int_{\min(8\sin(2x), 8\cos(2x))}^{\max(8\sin(2x),8\cos(2x))} dy \, dx \\\\ ~~~~~~~~ = \int_0^8 |8\sin(2x) - 8\cos(2x)| \, dx \\\\ ~~~~~~~~ = 8\sqrt2 \int_0^8 \left|\sin\left(2x-\frac\pi4\right)\right| \, dx$

where we rewrite the integrand using the identities

$\sin(\alpha + \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$

Now, if

$8(\cos(2x) - \sin(2x)) = R \sin(2x + \alpha) = R \sin(2x) \cos(\alpha) + R \cos(2x) \sin(\alpha)$

with $R>0$, then

$\begin{cases} R\cos(\alpha) = 8 \\ R\sin(\alpha) = -8 \end{cases} \implies \begin{cases}R^2 = 128 \\ \tan(\alpha) = -1\end{cases} \implies R=8\sqrt2\text{ and } \alpha = -\dfrac\pi4$

Find where this simpler sine curve crosses the $x$-axis.

$\sin\left(2x - \dfrac\pi4\right) = 0$

$2x - \dfrac\pi4 = n\pi$

$2x = \dfrac\pi4 + n\pi$

$x = \dfrac\pi8 + \dfrac{n\pi}2$

In the interval [0, 8], this happens a total of 5 times at

$x \in \left\{\dfrac\pi8, \dfrac{5\pi}8, \dfrac{9\pi}8, \dfrac{13\pi}8, \dfrac{17\pi}8\right\}$

See the attached plots, which demonstrates the area between the two curves is the same as the area between the simpler sine wave and the $x$-axis.

By symmetry, the areas of the middle four regions (the completely filled "lobes") are the same, so the area integral reduces to

$\displaystyle \iint_R dA \\\\ ~~~~ = 8\sqrt2 \left(-\int_0^{\pi/8} \sin\left(2x-\frac\pi4\right) \, dx + 4 \int_{\pi/8}^{5\pi/8} \sin\left(2x-\frac\pi4\right) \, dx \right. \\\\ ~~~~~~~~~~~~~~~~~~~~ \left. - \int_{17\pi/8}^8 \sin\left(2x-\frac\pi4\right) \, dx\right)$

The signs of each integral are decided by whether $\sin\left(2x-\frac\pi4\right)$ lies above or below axis over each interval. These integrals are totally doable, but rather tedious. You should end up with

$\displaystyle \iint_R dA = 40\sqrt2 - 4 (1 + \cos(16) + \sin(16)) \\\\ ~~~~~~~~ \approx 57.5508$

Similarly, we compute the slightly more complicated $x$-integral to be

$\displaystyle \iint_R x dA = \int_0^8 \int_{\min(8\sin(2x), 8\cos(2x))}^{\max(8\sin(2x),8\cos(2x))} x \, dy \, dx \\\\ ~~~~~~~~ = 8\sqrt2 \int_0^8 x \left|\sin\left(2x-\frac\pi4\right)\right| \, dx \\\\ ~~~~~~~~ \approx 239.127$

and the even more complicated $y$-integral to be

$\displaystyle \iint_R y dA = \int_0^8 \int_{\min(8\sin(2x), 8\cos(2x))}^{\max(8\sin(2x),8\cos(2x))} y \, dy \, dx \\\\ ~~~~~~~~ = \frac12 \int_0^8 \left(\max(8\sin(2x),8\cos(2x))^2 - \min(8\sin(2x),8\cos(2x))^2\right) \, dx \\\\ ~~~~~~~~ \approx 11.5886$

Then the centroid of $R$ is

$(x,y) = \left(\dfrac{239.127}{57.5508}, \dfrac{11.5886}{57.5508}\right) \approx \boxed{(4.15518, 0.200064)}$