Answer
$V\approx 1.98595$
Work Step by Step
Given $$y=e^{ -x^2} ; \quad\left[0,1\right] ; \quad y \text { -axis; } \quad T_{8}$$
Since
$ V= 2\pi \int_{0}^{1} xe^{ -x^2}dx$
Now, we will evaluate the integral using $T_8$, since $\Delta x=\dfrac{b-a}{n}=\dfrac{1}{8}$
$T_{n}=\dfrac{1}{3}[y_0+4y_1+2y_2+..+4y_{N-3}+2y_{N-2}+4y_{N-1}+y_N]\Delta x\\
T_{8}=\dfrac{1}{3}[y_0+4y_1+2y_2+..+4y_{N-3}+2y_{N-2}+y_N]\Delta x$
The volume can be approximated as:
$V= 2\pi \int_{0}^{1} xe^{ -x^2}dx \approx 2 \pi S_8 \\ \approx 2\pi (\dfrac{1}{3}) \dfrac{1}{8} [0+4(\dfrac{1}{8}e^{-1/64}++4(\dfrac{7}{8}e^{-49/64}+e^{-1}]\\ \approx 1.98595$
Hence, $V\approx 1.98595$
