91Ó°ÊÓ

We are given the function: \(f(x) = \cos\frac{1}{2}\pi x\) Now, solve for x as a function of y: \(y = \cos\frac{1}{2}\pi x\) \(x = \frac{1}{\pi}\cos^{-1}y\) The vertical range of the function is \( y \in [0, 1] \). Step 2: Set up the volume integral using disk method

We need to determine the radius of each circular disk slice obtained by revolving the region around the y-axis. The radius will be the distance from the y-axis to the point on the graph, which is the value of the \(x\)-coordinate. So, the radius is given by \(r(y) = \frac{1}{\pi}\cos^{-1}(y)\). Step 3: Create an expression for the volume of a single disk

Now we need to find the volume of a single disk slice. We know that the volume of a cylinder is given by the formula: \(V = \pi r^2 h\) Since each disk is infinitely small, we can approximate its height as \(dy\). Thus, the volume of a single disk is: \(dV = \pi [\frac{1}{\pi}\cos^{-1}(y)]^2 dy\) Step 4: Integrate the volume of a single disk from the bottom to the top of the curve

Now, we must integrate the expression for the volume of a single disk over the range of y from 0 to 1: \(V = \int_{0}^{1} \pi[\frac{1}{\pi}\cos^{-1}(y)]^2 dy\) Step 5: Solve the integral

To evaluate the integral, we can make the substitution: \(u = \cos^{-1}(y)\) \(du = -\frac{1}{\sqrt{1 - y^2}} dy\) Thus: \(V = \int_{\pi}^{0} [\frac{u}{\pi}]^2 \cdot -\sqrt{1 - \cos^2(u)} \cdot \frac{\pi}{2} du\) \(V = -\frac{1}{2}\cdot\frac{1}{\pi^2} \int_{\pi}^{0} u^2 \sin(u) du\) We now perform integration by parts to solve the integral: \(u = u^2\), \(dv = \sin(u) du\) \(du = 2u du\), \(v = -\cos(u)\) \( \int u \cdot dv = uv - \int v \cdot du \) Plugging in the values, we have: \(V = -\frac{1}{2 \pi^2}(-(u^2 \cdot \cos(u))|_{\pi}^{0} - 2 \int_{\pi}^{0} u\cos(u) du)\) \(V = \frac{1}{2 \pi^2}(\frac{2}{3}u^3\sin(u)|_{\pi}^{0} + 4 \int_{\pi}^{0} \frac{1}{6}u^3\sin(u) du)\) Since \(\sin(0)=0\) and \(\sin(\pi)=0\), we are left with the remaining integral: \(V = \frac{2}{3 \pi} \int_0^{\pi} u^3 \sin(u) du\) This is now a standard integration by parts problem, which results in the value: \(V = \frac{2}{3 \pi} \cdot \frac{24}{\pi^4}\) Thus, the volume generated by revolving the region under the curve about the y-axis is: \(V = \frac{16}{\pi^3}\)