Science:Math Exam Resources/Courses/MATH103/April 2011/Question 04 (b)/Solution 2

Shell method. In this solution, we consider the solid to be a sum of cylindrical shells. Each cylindrical shell has volume

$2\pi r(x)h(x)\mathrm{\Delta }r,$

where

$r(x)$

is its radius,

$h(x)$

is its height and

$\mathrm{\Delta }r$

is its thickness. The bowl will be made up by rotating the blue shaded region in the 2D figure

xy 2D image

. From there we have

$r(x)=x$

,

$h(x)=f(1)-f(x)=e-e^x,$

and

$\mathrm{\Delta }r=\mathrm{\Delta }x$

. In the limit

$\mathrm{\Delta }x\to 0,$

we obtain the volume by integrating along the x-axis:

$\begin{array}{rl}{V}_2& =\underset{0}{\overset{1}{\int }}2\pi r(x)h(x)dx\\ & =\underset{0}{\overset{1}{\int }}2\pi x(e-e^x)dx\\ & =2\pi e\underset{0}{\overset{1}{\int }}xdx-2\pi \underset{0}{\overset{1}{\int }}x{e}^{x}dx\\ & =2\pi e{\left[\frac{x^2}{2}\right]}_0^1-2\pi ({\left[x{e}^x\right]}_0^1-\underset{0}{\overset{1}{\int }}{e}^{x}dx)\\ & =2\pi e\frac{1}{2}-2\pi (e-{\left[e^x\right]}_0^1)\\ & =\pi e-2\pi (e-(e-1))\\ & =\pi e-2\pi \\ & =\pi (e-2).\end{array}$

(We used integration by parts in the second integral to get from the third to the fourth line.)

Reality check: Note that we obtain the same value for

$V_2$

with both methods. To visualize the full bowl we can look at the 3D figure

Full 3D image from rotation

.