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 $2\pi r(x)h(x)\Delta r,$ where $r(x)$ is its radius, $h(x)$ is its height and $\Delta r$ 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 $\Delta r=\Delta x$. In the limit $\Delta x\to 0,$ we obtain the volume by integrating along the x-axis:

${\begin{aligned}V_{2}&=\int _{0}^{1}2\pi r(x)h(x)\,dx\\&=\int _{0}^{1}2\pi x(e-e^{x})\,dx\\&=2\pi e\int _{0}^{1}x\,dx-2\pi \int _{0}^{1}xe^{x}\,dx\\&=2\pi e\left[{\frac {x^{2}}{2}}\right]_{0}^{1}-2\pi \left(\left[xe^{x}\right]_{0}^{1}-\int _{0}^{1}e^{x}\,dx\right)\\&=2\pi e{\frac {1}{2}}-2\pi \left(e-\left[e^{x}\right]_{0}^{1}\right)\\&=\pi e-2\pi \left(e-(e-1)\right)\\&=\pi e-2\pi \\&=\pi (e-2).\end{aligned}}$

(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

.