Single Variable Calculus, Chapter 7, 7.2-1, Section 7.2-1, Problem 86

Determine the volume of the solid obtained by rotating about the $y$-axis the region bounded by the curves $y = e^{-x^2}, y = 0, x = 0$ and $x = 1$.

If you use vertical strips, using shell method, notice that these strips have distance to the $y$-axis as $x$. If you rotate these lengths about $y$-axis, you'll get a circumference of $C = 2 \pi x$. Also, the height of the strips resembles the height of the cylinder as $H = y_{\text{upper}} - y_{\text{lower}} = e^{-x^2} - 0$. Thus..


$
\begin{equation}
\begin{aligned}

V =& \int^1_0 C(x) H(x) dx
\\
\\
V =& \int^1_0 2 \pi x (e^{-x^2}) dx
\\
\\
\text{Let } u =& -x^2
\\
\\
du =& -2x dx

\end{aligned}
\end{equation}
$


Make sure that the upper and lower limits are also in terms of $u$.


$
\begin{equation}
\begin{aligned}

V =& 2 \pi \left( \frac{-1}{2} \right) \int^{-(1)^2}_{-(0)^2} e^u du
\\
\\
V =& - \pi \int^{-1}_0 e^u du
\\
\\
V =& - \pi [e^u]^{-1}_0
\\
\\
V =& - \pi [e^{-1} - e^0]
\\
\\
V =& \pi [1 - e^{-1}] \text{ cubic units }
\end{aligned}
\end{equation}
$