Answer
$ \pi (e-1)^2 \text{cubic units}$
Work Step by Step
Our aim is to compute the volume of the given curve when it is revolved around $y$-axis by using the shell method.
Shell method for computing the volume of the curve: Let us consider two functions $f(x)$ and $g(x)$ (both are continuous functions) with $f(x) \geq g(x)$ on the interval $[m, n]$ and $R$ defines the region which revolves about the y-axis and is known as the region bounded by the curves $y=f(x)$ and $y=g(x)$ between the lines $x=m$ and $x=n$. Then, the volume of the solid can be expressed as:
$Volume, V=\int_m^n 2 \pi x [f(x)-g(x)] \ dx$
Here, $2\pi x=\text{Shell Circumference}$ and $[f(x)-g(x)] =\text {Height of the shell}$
We are given that $y=e^{x^2}$ and $y=e^{2-x^2}$ on $[0,1]$
Thus, $V=\int_m^n 2 \pi x [f(x)-g(x)] \ dx=2 \pi \int_0^1 x(e^{2-x^2}-e^{x^2}) \ dx$
or, $=2 \pi \times [- \dfrac{e^{2-x^2}}{2}-\dfrac{e^{x^2}}{2}]_0^1$
or, $=2 \pi \times [- \dfrac{e^{2-(1)^2}}{2}-\dfrac{e^{(1)^2}}{2}]$
or, $= \pi(-e-e+e^2+1)$
or, $= \pi (e-1)^2 \text{cubic units}$
