Volume of a solid with rectangular cross-sections

Text

A solid has as its base the region between $y=x^2+2$ and the $x$-axis on $[0,2]$. The cross-section perpendicular to the $x$-axis at abscissa $x$ is a rectangle of height $x+1$. (a) Write the cross-section area $A(x)$; (b) compute the volume.

Solution

(a) The base of the rectangle at height $x$ is $y=x^2+2$ (the distance between the curve and the $x$-axis), the height is $x+1$, so $A(x)=(x^2+2)(x+1)=x^3+x^2+2x+2.$

(b) Volume: $V=\int_0^2(x^3+x^2+2x+2)\,dx=\left[\frac{x^4}{4}+\frac{x^3}{3}+x^2+2x\right]_0^2=4+\frac83+4+4=\frac{44}{3}\approx14.667.$

$\boxed{V=\frac{44}{3}}$