# Problem with volume of solid of revolution

I was attempting to solve this problem regarding a volume of a solid of revolution (reference: Apostol, Volume 1, Section 6.26, Question 18 (c)). The question is (paraphrased) as follows:

Let $f(x) = e^{-2x} \forall x$. Denote by $S(t)$ the ordinate set of $f$ over the interval $[0,t]$, where $t>0$. Let $W(t)$ be the volume of the solid obtained by rotating $S(t)$ about the $y$ axis. Compute $W(t)$.

My attempt at this is as follows:

$y = e^{-2x} \Rightarrow \log y = -2x \Rightarrow x = \frac{\log y}{-2}$

Which means $$W(t) = \pi \int_{e^{-2t}}^1 \left(\frac{\log y}{-2}\right)^2 dy$$ $$= \frac{\pi}{4}\int_{e^{-2t}}^1 \left(\log y \right)^2 dy$$ $$= \frac{\pi}{4} \left[y(\log y)^2 - 2y(\log y -1)\right]_{e^{-2t}}^1$$ $$= \frac{\pi}{4} \left( (1(\log 1)^2 -2\cdot 1(\log 1 -1)) - (e^{-2t}(\log e^{-2t})^2-2e^{-2t}(\log e^{-2t}-1))\right)$$ $$= \frac{\pi}{4} \left(2 - (4t^2e^{-2t}+4te^{-2t}+2e^{-2t})\right)$$ $$= \frac{\pi}{4} \left(2 - 2e^{-2t}(2t^2+2t+1)\right)$$ $$= \frac{\pi}{2}\left(1-e^{-2t}(2t^2+2t+1)\right)$$ which is where I have my issue since the answer in the back of the book is $\frac{\pi}{2}\left(1-e^{-2t}(2t+1)\right)$, so either the book is wrong or I am wrong. If anyone can shed some light on this that would be appreciated

I would say assignment deceiving. S(t) = the arc function f(t), $t\in \langle 0,1 \rangle$.
$W(t)=2\pi\int_{0}^{t}xf(x)\,dx=2\pi\int_{0}^{t}xe^{-2x}\,dx=\cdots =\frac{\pi}{2}\left(1-e^{-2t}(2t+1)\right)$.