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.
Solution:
(a)
Step 1:

First, we sketch the region bounded by the given functions.


Step 2:

Setting the equations equal, we have $x^{2}=2x.$

Solving for $x,$ we get

${\begin{array}{rcl}\displaystyle {0}&=&\displaystyle {x^{2}2x}\\&&\\&=&\displaystyle {x(x2).}\end{array}}$

So, $x=0$ and $x=2.$

If we plug these values into our functions, we get the intersection points

$(0,0)$ and $(2,4).$

This intersection points 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 $h=2xx^{2}.$

Step 2:

So, the volume of the solid is

$\int 2\pi rh~dx~=~\int _{0}^{2}2\pi x(2xx^{2})~dx.$

(c)
Step 1:

We need to integrate

$\int _{0}^{2}2\pi x(2xx^{2})~dx~=~2\pi \int _{0}^{2}2x^{2}x^{3}~dx.$

Step 2:

We have

${\begin{array}{rcl}\displaystyle {\int _{0}^{2}2\pi x(2xx^{2})~dx}&=&\displaystyle {2\pi \int _{0}^{2}2x^{2}x^{3}~dx}\\&&\\&=&\displaystyle {2\pi {\bigg (}{\frac {2x^{3}}{3}}{\frac {x^{4}}{4}}{\bigg )}{\bigg }_{0}^{2}}\\&&\\&=&\displaystyle {2\pi {\bigg (}{\frac {2^{4}}{3}}{\frac {2^{4}}{4}}{\bigg )}2\pi (0)}\\&&\\&=&\displaystyle {{\frac {8\pi }{3}}.}\\\end{array}}$

Final Answer:

(a) $(0,0),(2,4)$ (See Step 1 for the graph)

(b) $\int _{0}^{2}2\pi x(2xx^{2})~dx$

(c) ${\frac {8\pi }{3}}$

Return to Sample Exam