# Find the volume of the solid that results when the region enclosed by the given curve: $(11$$to14)$$y=1+x^{2},x=1,x=2,y=0$ is revolved around the $x$- axis.

Toolbox:
• Area bounded by the curve $t=f(x),$ the x-axis and the ordinates $x=a,x=b$ is $\int \limits_a^b f(x) dx$ or $\int \limits _a^b y dx$
• If the curve lies below the x-axis for $a \leq x \leq b,$ then the area is $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
Step 1:
$y=1+x=>x^2=y-1$ is a parabola with its vertex at (0,1) opening upward
The volume of the solid of revolution about x-axis is $v=\pi \int \limits_1^2 y^2 dx=\pi \int \limits_1^2 (x^2+1)^2$$dx Step 2: \quad= \pi \int \limits_1^2 (x^4 +2x^2 +1)dx \quad=\pi \bigg[ \large\frac{x^5}{5}+\frac{2x^3}{3}$$+x\bigg]_1^2$
$\quad=\pi \bigg[(\large\frac{32}{5}+\frac{11}{3}$$+2)-(\large\frac{1}{5}+\frac{2}{3}$$+1)\bigg]$
$\quad= \pi \bigg[ \large\frac{31}{5}+\large \frac{14}{3}$$+1\bigg] \quad=\pi \large\frac{93+70+15}{15} \quad=\large\frac{178}{15} \pi$$\;units$