A cross$-$section $in$ the plane $Px$ has the shape $of$ a washer $(an$ annular ring$)$ with inner radius $x2$ and $an$ outer radius $3x,sowe$ find the cross$-$sectional area $by$ subtracting the area $of$ the inner circle from the area $of$ the outer circle. $A(x)=9x2-(x2)2=9x2-x4.$ Therefore, $we$ have $V=3A(x)dx0=3(9x2-x4)dx0=81-243530=1625.$

$A(x)=9x^2-(x^2{)}^2=(9x^2-x^4,)$

Therefore, $we$ have

$V={\displaystyle \underset{0}{\overset{3}{}}}A(x)dx={\displaystyle \underset{0}{\overset{3}{}}}(9x^2-x^4)dx$

$=\left[\begin{array}{ccc}& & \end{array}\right]$

${}_0^3=$

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