91Ó°ÊÓ

To determine the intersection points, set the two functions equal to each other and solve for \(x\): $$\sin{x} = 1 - \sin{x}.$$ Adding \(\sin{x}\) to both sides of the equation and simplifying gives us: $$2\sin{x} = 1.$$ Now, to solve for \(x\), we can take the inverse sine of both sides: $$x = \arcsin{\frac{1}{2}}.$$ Recall that the interval given is \(\frac{\pi}{6} \le x \le \frac{5\pi}{6}\). Therefore, the intersection points are: $$x = \frac{\pi}{6} \quad \text{and}\quad x = \frac{5\pi}{6}.$$

The Washer Method involves finding the volume of the solid generated by the outer function, then subtracting the volume of the solid generated by the inner function, leaving only the volume of the solid generated by the region \(R\). As \(y=\sin x\) is above \(y= 1-\sin x\), we'll take them as outer and inner functions respectively. The volume formula for the Washer Method is: $$V = \pi\int_{a}^{b}\left[ (R_{\text{outer}}(x))^2 - (R_{\text{inner}}(x))^2 \right] dx,$$ where \(R_{\text{outer}}(x)\) is the distance between the outer function and the line about which we're revolving, and \(R_{\text{inner}}(x)\) is the distance between the inner function and that same line. Both distances are taken as positive. The interval bounds are \(a=\frac{\pi}{6}\) and \(b=\frac{5\pi}{6}\). The distance \(R_{\text{outer}}(x)\) from the outer function to the line \(y=-1\) is: $$R_{\text{outer}}(x) = (-1) - (\sin x) = 1 - \sin x,$$ and the distance \(R_{\text{inner}}(x)\) from the inner function to the line \(y=-1\) is: $$R_{\text{inner}}(x) = (-1) - (1-\sin x) = \sin x - 1.$$ Now substitute those distances into the Washer Method formula and integrate over the given interval.

We substitute the values for \(R_{\text{outer}}(x)\) and \(R_{\text{inner}}(x)\) into the Washer Method formula: $$V = \pi\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \left[(1 - \sin{x})^2 - (\sin{x} - 1)^2\right] dx.$$ This simplifies to: $$V = \pi\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 4\sin{x}(1 - \sin{x})\, dx.$$ Now, evaluate the integral using trigonometric substitution: Let \(u = 1 - \sin{x}\) then, \(-\mathrm{d}u = \mathrm{d}\sin x\) So the integral becomes: $$V = -\pi\int_{\frac{1}{2}}^{1} 4u(1 - u)(-du) = 4\pi\int_{\frac{1}{2}}^{1} u(1 - u) du.$$ Integrate \(u(1-u)\) with respect to \(u\): $$V = 4\pi \left[ \int_{\frac{1}{2}}^{1} u - u^2\, du\right] = 4\pi\left[\frac{u^2}{2} - \frac{u^3}{3}\right]_{\frac{1}{2}}^{1}.$$ Now, evaluate the definite integral: $$V = 4\pi \left[\left(\frac{1}{2} - \frac{1}{3}\right) - \left(\frac{1}{8} - \frac{1}{24}\right)\right] = 4\pi\left(\frac{1}{6}+\frac{1}{24}\right) = \frac{5\pi}{3}.$$ The final volume of the solid generated by revolving the region \(R\) around the line \(y = -1\) is: $$V = \frac{5\pi}{3}.$$