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

`(\displaystyle \iint_R x \, dA)/(\displaystyle \iint_R dA) \text{ and } (\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 = -\frac\pi4`

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

`\sin\left(2x - \frac\pi4\right) = 0`

`2x - \frac\pi4 = n\pi`

`2x = \frac\pi4 + n\pi`

`x = \frac\pi8 + \frac{n\pi}2`

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

`x \in \left\{\frac\pi8, \frac{5\pi}8, \frac{9\pi}8, \frac{13\pi}8, \frac{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((239.127)/(57.5508), (11.5886)/(57.5508)\right) \approx \boxed{(4.15518, 0.200064)}`