Answer
$$ \frac{99\pi}{200\ln(10)}$$
Work Step by Step
Given
$$ y=10^{-x},\ \ \ \ x=0,\ \ x=1 $$
Then the volume of the resulting solid when the bounded region rotate abou $x-$ axis given by
\begin{aligned}
V&= \pi\int_a^bf^2(x)dx\\
&= \pi\int_0^1 10^{-2x}dx\\
&= \frac{-\pi}{2}\frac{10^{-2x}}{\ln (10)}\bigg|_0^1\\
&= \frac{\pi}{2\ln(10)}[1-\frac{1}{100}]\\
&= \frac{99\pi}{200\ln(10)}
\end{aligned}
