009B Sample Final 1, Problem 5

Consider the solid obtained by rotating the area bounded by the following three functions about the $y$-axis:

$x=0$, $y=e^{x}$, and $y=ex$.

(a) Sketch the region bounded by the given three functions. Find the intersection point of the two functions:

$y=e^{x}$ and $y=ex$. (There is only one.)

(b) Set up the integral for the volume of the solid.

(c) Find the volume of the solid by computing the integral.

Foundations:
Recall:
1. You can find the intersection points of two functions, say $f(x),g(x),$
by setting $f(x)=g(x)$ and solving for $x$.
2. The volume of a solid obtained by rotating an area around the $y$-axis using cylindrical shells is given by
$\int 2\pi rh~dx,$ where $r$ is the radius of the shells and $h$ is the height of the shells.

Solution:

(a)

Step 1:
First, we sketch the region bounded by the three functions.
Insert graph here.

Step 2:
Setting the equations equal, we have $e^{x}=ex$.
We get one intersection point, which is $(1,e)$.
This intersection point can be seen in the graph shown in Step 1.

(b)

Step 1:
We proceed using cylindrical shells. The radius of the shells is given by $r=x$.
The height of the shells is given by $h=e^{x}-ex$.

Step 2:
So, the volume of the solid is
$\int 2\pi rh\,dx\,=\,\int _{0}^{1}2\pi x(e^{x}-ex)\,dx.$

(c)

Step 1:
We need to integrate
$\int _{0}^{1}2\pi x(e^{x}-ex)\,dx\,=\,2\pi \int _{0}^{1}xe^{x}\,dx-2\pi \int _{0}^{1}ex^{2}\,dx.$

Step 2:
For the first integral, we need to use integration by parts.
Let $u=x$ and $dv=e^{x}dx$. Then, $du=dx$ and $v=e^{x}$.
So, the integral becomes
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int_0^1 2\pi x(e^x-ex)~dx} & = & \displaystyle{2\pi\bigg(xe^x\bigg|_0^1 -\int_0^1 e^xdx\bigg)-\frac{2\pi ex^3}{3}\bigg|_0^1}\\ &&\\ & = & \displaystyle{2\pi\bigg(xe^x-e^x\bigg)\bigg|_0^1-\frac{2\pi e}{3}}\\ &&\\ & = & \displaystyle{2\pi(e-e-(-1))-\frac{2\pi e}{3}}\\ &&\\ & = & \displaystyle{2\pi-\frac{2\pi e}{3}}.\\ \end{array}}

Final Answer:
(a)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1,e)} (See Step 1 for the graph)
(b)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^1 2\pi x(e^x-ex)~dx}
(c)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\pi-\frac{2\pi e}{3}}

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