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:
Review volumes of revolutions

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 $\int _{0}^{1}2\pi x(e^{x}-ex)~dx$.

(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
${\begin{array}{rcl}\displaystyle {\int _{0}^{1}2\pi x(e^{x}-ex)~dx}&=&\displaystyle {2\pi {\bigg (}xe^{x}{\bigg |}_{0}^{1}-\int _{0}^{1}e^{x}dx{\bigg )}-{\frac {2\pi ex^{3}}{3}}{\bigg |}_{0}^{1}}\\&&\\&=&\displaystyle {2\pi {\bigg (}xe^{x}-e^{x}{\bigg )}{\bigg |}_{0}^{1}-{\frac {2\pi e}{3}}}\\&&\\&=&\displaystyle {2\pi (e-e-(-1))-{\frac {2\pi e}{3}}}\\&&\\&=&\displaystyle {2\pi -{\frac {2\pi e}{3}}}\\\end{array}}$

Final Answer:
(a) $(1,e)$ (See (a) Step 1 for the graph)
(b) $\int _{0}^{1}2\pi x(e^{x}-ex)~dx$
(c) $2\pi -{\frac {2\pi e}{3}}$

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