91Ó°ÊÓ

Solve the function \(f(x) = e^{-x}\) for x in terms of y. \(f(y) = e^{-x} \Rightarrow e^{-x} = y \Rightarrow -x = \ln y \Rightarrow x = - \ln y\) Now we have the function \(x(y) = -\ln y\).

Using the disk method, the volume of the solid is given by: \(V = \pi \int_{a}^{b} [x(y)]^2 dy\) Here, the radius function will be \(x(y) = - \ln y\), and we need to find the limits of integration in terms of \(y\). At \(x=0\), \(y = f(0) = e^{-0} = 1\). At \(x=\ln 2\), \(y = f(\ln 2) = e^{-\ln 2} = \frac{1}{2}\). So, the limits of integration are from \(y=1\) to \(y=\frac{1}{2}\).

Now, we can evaluate the volume integral: \(V = \pi \int_{1}^{\frac{1}{2}} [\ln y]^2 dy\) To find the antiderivative of \([\ln y]^2\), we can use integration by parts. Let \(u=\ln y\) and \(dv= y^2 dy\). Then, \(du= \frac{1}{y}dy\) and \(v= \frac{-y^3}{3}\). Now use integration by parts: \(\int [\ln y]^2 dy = uv - \int vu dy = -\frac{y^3 \ln y}{3} + \int \frac{y^2}{3} dy\) Now, integrate the remaining term and plug in the limits of integration: \(V = \pi \left[-\frac{y^3 \ln y}{3} + \int \frac{y^2}{3} dy \right]_{1}^\frac{1}{2}\) Evaluate the definite integral: \(V = \pi \left[-\frac{(1/8) \ln (1/2)}{3} + \frac{(1/8)^3}{9} - (-\frac{\ln 1}{3} + \frac{1}{9})\right]\) \(V = \pi \left[ \frac{\ln (2)}{24} + \frac{1}{72} + \frac{1}{9} \right]\) Finally, simplify the expression for the volume: \(V = \frac{\pi}{24}(\ln 2 + 8)\) The volume of the solid generated when the region is revolved about the y-axis is \(V = \frac{\pi}{24}(\ln 2 + 8)\).