91Ó°ÊÓ

We will compute the volume by integrating the area of each disk from 0 to π along the x-axis. The disk area is given by π (radius)² where the radius is the function value (y = sin(x)). We have: \(V_x = \int_{0}^{\pi} \pi\big(\sin{x}\big)^2 dx\)

To evaluate the integral, we can use the power reduction formula: \(\sin^2{x} = \frac{1-\cos{2x}}{2}\) \(V_x = \int_{0}^{\pi} \pi\left(\frac{1-\cos{2x}}{2}\right) dx\) Now we can distribute the pi and integrate: \(V_x = \pi\int_{0}^{\pi}\frac{1}{2} - \pi\int_{0}^{\pi} \frac{\cos{2x}}{2}dx\) Then, we have: \(V_x = \frac{\pi}{2}(x - \frac{\sin{2x}}{4}) \Big|_0^\pi\) Thus, \(V_x = \frac{\pi^2}{2}\)

To find the volume when R is revolved about the y-axis, we must first express x in terms of y using the original function y = sin(x). This gives us x = arcsin(y). Now, using the disk method, we will compute the volume by integrating the area of each disk from 0 to 1 along the y-axis: \(V_y = \int_{0}^{1} \pi\left(\arcsin{y}\right)^2 dy\)

This integral is a bit more complicated. We'll need to use substitution: Let \(u = \arcsin{y}\), then \(y = \sin{u}\) and \(dy = \cos{u}du\). Also, the new limits will be \(u=0\) and \(u=\pi/2\). The integral becomes: \(V_y = \int_{0}^{\frac{\pi}{2}} \pi\left(u\right)^2 \cos{u} du\) Now integrate by parts. Let \(v = u^2\) and \(dw = \cos{u} du\) then \(dv= 2udu\) and \(w=\sin{u}\): \(V_y = \pi \left[u^2\sin{u} \Big|_0^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} 2u\sin{u} du\right]\) Evaluating the first term, we have: \(V_y = \pi \left[\frac{\pi^2}{4}\sin{\frac{\pi}{2}}\right] - \pi\left[ \int_{0}^{\frac{\pi}{2}} 2u\sin{u} du\right]\) Now, we must integrate by parts again: Let \(v=2u\) and \(dw=\sin{u}du\) then \(dv=2du\) and \(w=-\cos{u}\): \( \int_{0}^{\frac{\pi}{2}} 2u\sin{u} du = -2u\cos{u}\Big|_0^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}}2\cos{u}du = -2u\cos{u}\Big|_0^{\frac{\pi}{2}} + 2\sin{u}\Big|_0^{\frac{\pi}{2}}\) Thus, we have: \(V_y = \pi\left[\frac{\pi^2}{4} - \left(\left(\frac{\pi}{2}\right)\cos{\frac{\pi}{2}}+ 2\sin{\frac{\pi}{2}}\right)\right]\) Finally, \(V_y = \frac{\pi^3}{4}\)

Comparing the volumes: \(V_x = \frac{\pi^2}{2}\) and \(V_y = \frac{\pi^3}{4}\) Since \(\pi>2\), we can see that \(\frac{\pi^3}{4} > \frac{\pi^2}{2}\) Thus the volume of the solid generated when R is revolved about the y-axis (Vy) is greater than the volume of the solid generated when R is revolved about the x-axis (Vx).