Hello Forrest,
I would begin by drawing a diagram of the region to be revolved, with the axis of rotation and the lengths of the inner and outer radius ($r$ and $R$ respectively):
View attachment 1315
The volume of an arbitrary washer is:
$$dV=\pi\left(R^2-r^2 \right)\,dx$$
where:
$$R=1+1-e^{-x}=2-e^{-x}$$
$$r=1$$
and so we have:
$$dV=\pi\left(\left(2-e^{-x} \right)^2-(1)^2 \right)\,dx=\pi\left(e^{-2x}-4e^{-x}+3 \right)\,dx$$
Summing the washers through integration, we obtain:
$$V=\pi\int_0^2 e^{-2x}-4e^{-x}+3\,dx$$
Applying the FTOC, we find:
$$V=\pi\left[-\frac{1}{2}e^{-2x}+4e^{-x}+3x \right]_0^2=\pi\left(\left(-\frac{1}{2}e^{-4}+4e^{-2}+6 \right)-\left(-\frac{1}{2}+4 \right) \right)=\frac{\pi}{2e^4}\left(5e^4+8e^2-1 \right)$$
If we wish to check our result by using the shell method, then refer to the following diagram:
View attachment 1316
The volume of an arbitrary shell is:
$$dV=2\pi rh\,dy$$
where:
$$r=2-y$$
$$h=2+\ln(y)$$
And so we have:
$$dV=2\pi (2-y)\left(2+\ln(y) \right)\,dy=2\pi\left(4+2\ln(y)-2y-y\ln(y) \right)\,dy$$
Summing the shells by integration, we have:
$$V=2\pi\int_{e^{-2}}^1 4+2\ln(y)-2y-y\ln(y)\,dy$$
Each of the terms can be readily integrated except for those involving the natural log function, so let's use integration by parts:
i) $$I=\int \ln(y)\,dy$$
$$u=\ln(y)\,\therefore\,du=\frac{1}{y}\,dy$$
$$dv=dy\,\therefore\,v=y$$
Hence:
$$I=y\ln(y)-\int\,dy=y\left(\ln(y)-1 \right)$$
ii) $$I=\int y\ln(y)\,dy$$
$$u=\ln(y)\,\therefore\,du=\frac{1}{y}\,dy$$
$$dv=y\,dy\,\therefore\,v=\frac{1}{2}y^2$$
Hence:
$$I=\frac{1}{2}y^2\ln(y)-\frac{1}{2}\int y\,dy$$
$$I=\frac{1}{2}y^2\ln(y)-\frac{1}{4}y^2=\frac{y^2}{4}\left(2\ln(y)-1 \right)$$
Now we may apply the FTOC to compute the volume:
$$V=2\pi\left[4y+2y\left(\ln(y)-1 \right)-y^2-\frac{y^2}{4}\left(2\ln(y)-1 \right) \right]_{e^{-2}}^1=2\pi\left(\left(\frac{5}{4} \right)-\left(-\frac{2}{e^2}+\frac{1}{4e^4} \right) \right)=$$
$$\frac{\pi}{2e^4}\left(5e^4+8e^2-1 \right)$$
And this checks with the washer method.
