91Ó°ÊÓ

The region is bounded by the function \(f(x)=e^{-x}\), the vertical line \(x=\ln 2\), and the coordinate axes. First, we find the intersection of \(f(x)\) with the \(x\)-axis by setting \(f(x)\) equal to zero: $$e^{-x}=0$$ However, exponential functions never equal zero, so the function does not intersect the \(x\)-axis. Thus, the region is bounded by the \(y\)-axis.

We need to find the point of intersection between \(f(x)=e^{-x}\) and \(x=\ln 2\). To find the intersection point, we need to substitute \(x=\ln 2\) into the function: $$f(\ln 2)=e^{-\ln 2}$$ Using properties of logarithms and exponentials, we have: $$e^{-\ln 2} = \frac{1}{e^{\ln 2}} = \frac{1}{2}$$ So, the point of intersection is \((\ln 2, \frac{1}{2})\).

We'll use the disk method to find the volume of the solid. In this case, we will integrate with respect to \(y\), as we are revolving the area around the \(y\)-axis. First, we need to find the inverse function of \(f(x)=e^{-x}\), which is \(g(y)= -\ln y\). The radius of each disk is given by \(R(y) = g(y)\) and its thickness is \(dy\). The volume of each disk is given by \(dV = \pi R(y)^2 dy\). We integrate the volume of the disks over the interval \([0, \frac{1}{2}]\) to find the total volume: $$V = \int_{0}^{\frac{1}{2}} \pi g(y)^2 dy$$

Now, we substitute \(g(y)=-\ln y\) into the integral and simplify: $$V = \int_{0}^{\frac{1}{2}} \pi (-\ln y)^2 dy$$ We can make a substitution: let \(u = -\ln y\), so \(du = -\frac{1}{y} dy\) and \(y = e^{-u}\). The limits of integration change to \([$$\ln 2, +\infty$$]\) when \(y\) runs from \(0\) to \(\frac{1}{2}\). The integral becomes: $$V = \int_{\ln 2}^{+\infty} \pi u^2 e^u du$$ To evaluate the integral, we will use integration by parts. Let \(v = u^2\) and \(dw = e^u du\). So, \(dv = 2u du\) and \(w = e^u\). Applying integration by parts twice, we have: $$V = \pi \left[ u^2 e^{u} - 2\int u e^u du\right]_{\ln 2}^{+\infty}$$ $$V = \pi \left[ u^2 e^{u} - 2(u e^{u} - \int e^u du)\right]_{\ln 2}^{+\infty}$$ $$V = \pi \left[ u^2 e^{u} - 2u e^{u} + 2e^u\right]_{\ln 2}^{+\infty}$$ Now we evaluate the integral at the limits, and subtract: $$V = \pi \left[ \lim_{u\to\infty} (u^2e^u - 2u e^u + 2e^u) - \left((\ln2)^2e^{\ln2} - 2\ln2 e^{\ln2} + 2e^{\ln2}\right) \right]$$ $$V = \pi \left[ 0 - \left( 2 - 4 + 4 \right) \right]$$ $$V = 4\pi$$

The volume of the solid that is generated when the given region is revolved about the \(y\)-axis is \(4\pi\).