Answer
$45 \pi $
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 $y=-x^2+4x+2; y=x^2-6x+10$ on $[1, 4]$
Thus, $V=\int_m^n 2 \pi x [f(x)-g(x)] \ dx=2 \pi \int_1^4 x[-x^2+4x+2-(x^2-6x+10)] \ dx$
or, $=2 \pi \int_1^4 (-2x^3+10x^2-8x) \ dx$
or, $=2 \pi [ \dfrac{-x^4}{2}+\dfrac{10x^3}{3}-4x^2]_1^4$
or, $=45 \pi $
