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A solid of revolution is defined as a 3-dimensional object that is formed by rotating a 2-dimensional plane curve around a straight line (the axis) in the same plane. The basic characteristic of a solid of revolution is that they possess rotational symmetry. A pyramid, on the other hand, does not possess rotational symmetry. It cannot be obtained by rotating a 2-dimensional figure around a straight line. Therefore, a pyramid is not a solid of revolution. This statement is false.

The disk method is a technique used to find the volume of a solid of revolution. It involves dividing the region into infinitely small disks, and summing the volume of these disks to find the volume of the solid. A hemisphere is a solid of revolution, as it can be obtained by rotating a semi-circle around its diameter. Therefore, its volume can be computed using the disk method. Let's compute the volume of a hemisphere of radius r using the disk method: 1. Consider a semi-circle with radius r in the xy-plane with its center at the origin and its diameter on the x-axis. 2. The equation of this semi-circle is \(y=\sqrt{r^2-x^2}\). 3. The thickness of each disk is \(dx\) and its volume is given by \(dV = \pi y^2 dx\). 4. To find the volume of the hemisphere, integrate dV over the interval \([-r, r]\): $$V = \int_{-r}^{r} \pi (\sqrt{r^2-x^2})^2 dx$$. 5. Solving the integral: $$V = \pi \int_{-r}^{r} (r^2-x^2) dx$$ $$V = \pi \left( \left[r^2x - \frac{x^3}{3}\right]_{-r}^r\right)$$ $$V = \pi \left(\left[r^3 - \frac{r^3}{3}\right] - \left[-r^3 + \frac{r^3}{3}\right]\right)$$ $$V = \frac{4}{3}\pi r^3$$. Hence, the volume of a hemisphere can be computed using the disk method. This statement is true.

Let's first compute the volumes of both solids separately and then compare the results: Solid 1: The region \(R_{1}\) is bounded by \(y=\cos x\) and the x-axis on \([-\frac{\pi}{2}, \frac{\pi}{2}]\). To find the volume of the solid generated when \(R_1\) is revolved about the x-axis, we can use the disk method: $$V_1 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \pi (\cos{x})^2 dx$$ $$V_1 = \pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\cos (2x)}{2} dx$$ $$V_1 = \pi \left[\frac{1}{2}x + \frac{1}{4}\sin (2x)\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$$ $$V_1 = \pi \cdot \frac{1}{2}\pi = \frac{\pi^2}{2}$$ Solid 2: The region \(R_{2}\) is bounded by \(y=\sin x\) and the x-axis on \([0, \pi]\). To find the volume of the solid generated when \(R_2\) is revolved about the x-axis, again we use the disk method: $$V_2 = \int_{0}^{\pi} \pi (\sin{x})^2 dx$$ $$V_2 = \pi \int_{0}^{\pi} \frac{1- \cos(2x)}{2} dx$$ $$V_2 = \pi \left[\frac{1}{2}x - \frac{1}{4}\sin (2x)\right]_{0}^{\pi}$$ $$V_2 = \pi \cdot \frac{1}{2}\pi = \frac{\pi^2}{2}$$ The volumes of the solids generated when \(R_1\) and \(R_2\) are revolved about the x-axis are equal, as \(V_1 = V_2 = \frac{\pi^2}{2}\). This statement is true.