Single Variable Calculus, Chapter 7, 7.4-2, Section 7.4-2, Problem 52

Find the volume of the region under the curve $y = 10^{-x}$ from $x = 0$ to $x = 1$ that is rotated about the $x$-axis.

By using vertical strips, notice that if you slice the figure, you'll get a cross section with radius $10^{-x}$, thus, the cross sectional area is

$A = \pi (10^{-x})^2$. Hence, the volume is


$
\begin{equation}
\begin{aligned}

V =& \int^1_0 A(x) dx
\\
\\
V =& \int^1_0 \pi (10^{-x})^2 dx
\\
\\
V =& \int^1_0 \pi \left( \frac{1}{10 x} \right)^2 dx
\\
\\
V =& \pi \int^1_0 10^{-2x} dx


\end{aligned}
\end{equation}
$


Let $u = -2x$, then $du = -2 dx$, so $\displaystyle dx = \frac{-du}{2}$. When $x = 0, u = 0$ and when $x = 1, u = -2$.

Therefore,


$
\begin{equation}
\begin{aligned}

V =& \pi \int^{-2}_0 10^u \cdot \frac{-du}{2}
\\
\\
V =& \frac{- \pi}{2} \int^{-2}_0 10^u du
\\
\\
V =& \frac{- \pi}{2} \left[ \frac{10^u}{\ln 10} \right]^{-2}_0
\\
\\
V =& \frac{- \pi}{2} \left[ \frac{10^{-2}}{\ln 10} - \frac{10^0}{\ln 10} \right]
\\
\\
V =& 0.6754 \text{ cubic units }

\end{aligned}
\end{equation}
$