91Ó°ÊÓ

To find the volume of the solid, we will integrate the volume element in spherical coordinates over the given region. The volume element in spherical coordinates is given by \(dV = r^2 \sin(\varphi) dr d\varphi d\theta\). The limits of integration will be determined by the radius and angle bounds given by the cylinders and cones.

The radius will range between the two cylindrical bounds, so \(r\) will go from 1 to 2. The angle \(\varphi\) will be bounded by the two cones, so \(\varphi\) will range from \(\pi/6\) to \(\pi/3\). Finally, the angle \(\theta\) will vary between 0 and \(2\pi\) since the solid spans the entire circle in the \(xy\)-plane. Thus, the limits of integration are: - \(r\): 1 to 2 - \(\varphi\): \(\pi/6\) to \(\pi/3\) - \(\theta\): 0 to \(2\pi\)

Now we can set up the triple integral for the volume: $$V = \int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \int_{1}^{2} r^2 \sin(\varphi) dr d\varphi d\theta$$ First, we will integrate with respect to \(r\): $$V = \int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \left[\frac{1}{3}r^3\right]_{1}^{2} \sin(\varphi) d\varphi d\theta$$ Now, we evaluate the expression inside the brackets: $$V = \int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \left( \frac{8}{3} - \frac{1}{3} \right) \sin(\varphi) d\varphi d\theta = \int_{0}^{2\pi} \int_{\pi/6}^{\pi/3} \frac{7}{3} \sin(\varphi) d\varphi d\theta$$ Next, we will integrate with respect to \(\varphi\): $$V = \int_{0}^{2\pi} \left[-\frac{7}{3}\cos(\varphi)\right]_{\pi/6}^{\pi/3} d\theta$$ Now, we evaluate the expression inside the brackets: $$V = \int_{0}^{2\pi} \left( -\frac{7}{6} + \frac{7}{3} \right) d\theta = \int_{0}^{2\pi} \frac{7}{6} d\theta$$ Finally, we will integrate with respect to \(\theta\): $$V = \left[\frac{7}{6}\theta\right]_{0}^{2\pi}$$ Now, we evaluate the expression inside the brackets: $$V = \frac{7}{6}(2\pi) - \frac{7}{6}(0) = \frac{7\pi}{3}$$ Thus, the volume of the solid is \(\frac{7\pi}{3}\) cubic units.