6.3 Volumes of revolution: cylindrical shells  (Page 4/6)

First, graph the region $R$ and the associated solid of revolution, as shown in the following figure.

Note that the axis of revolution is the $y\text{-axis},$ so the radius of a shell is given simply by $x.$ We don’t need to make any adjustments to the x -term of our integrand. The height of a shell, though, is given by $f(x)-g(x),$ so in this case we need to adjust the $f(x)$ term of the integrand. Then the volume of the solid is given by

1. First, sketch the region and the solid of revolution as shown.
   

   

   
   Looking at the region, if we want to integrate with respect to $x,$ we would have to break the integral into two pieces, because we have different functions bounding the region over $\left[0,1\right]$ and $\left[1,2\right].$ In this case, using the disk method, we would have
   
   $V={\displaystyle {\int }_0^1\left(\pi x^2\right)}dx+{\displaystyle {\int }_1^2\left(\pi {(2-x)}^2\right)}dx.$
   
   If we used the shell method instead, we would use functions of $y$ to represent the curves, producing
   
   $\begin{array}{cc}\hfill V& ={\displaystyle {\int }_0^1\left(2\pi y\left[\left(2-y\right)-y\right]\right)}dy\hfill \\ & ={\displaystyle {\int }_0^1\left(2\pi y\left[2-2y\right]\right)}dy.\hfill \end{array}$
   
   Neither of these integrals is particularly onerous, but since the shell method requires only one integral, and the integrand requires less simplification, we should probably go with the shell method in this case.
2. First, sketch the region and the solid of revolution as shown.
   

   

   
   Looking at the region, it would be problematic to define a horizontal rectangle; the region is bounded on the left and right by the same function. Therefore, we can dismiss the method of shells. The solid has no cavity in the middle, so we can use the method of disks. Then
   
   $V={\displaystyle {\int }_0^4\pi }{\left(4x-x^2\right)}^{2}dx.$