009B Sample Final 1, Problem 5

The region bounded by the parabola $y=x^{2}$ and the line $y=2x$ in the first quadrant is revolved about the $y$ -axis to generate a solid.

(a) Sketch the region bounded by the given functions and find their points of intersection.

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

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