Determining the Surface Area of a Solid of Revolution

Key Questions

How do you find the surface area of the solid obtained by rotating about the $y$ $y$ -axis the region bounded by $y = x^{2}$ $y=x^2$ on the interval $1 \leq x \leq 2$ $1<=x<=2$ ?

First of all, you are missing a bound. We will assume that the other bound is $y = 0$ $y=0$ or the $x$ $x$ -axis. The answer is $\frac{15 π}{2}$ $(15pi)/2$ .

The first step is to determine whether you are rotating along an axis that is parallel to the independent axis or the axis of the parameter ( $x$ $x$ in this case). And we are not, so this integration should be done with cylindrical shells.

Always draw a diagram to verify what is the parameter and what is the function.

You should note that $y$ $y$ is not always a parameter of $x$ $x$ . For instance, $x = y^{2}$ $x=y^2$ , $x$ $x$ is now a parameter of $y$ $y$ .

The formula for cylindrical shells is:

$V = \int_{a}^{b} 2 π r h d r$ $V=int_a^b2pirhdr$
$h$ $h$ is represented by $y$ $y$ , we have $y = x^{2}$ $y=x^2$ and $y = 0$ $y=0$
$r$ $r$ is represented by $x$ $x$
$V = \int_{1}^{2} 2 π x (x^{2} - 0) d x$ $V=int_1^2 2 pi x (x^2-0) dx$

Now that the substitutions are done, we can solve:

$V = 2 π \int_{1}^{2} x^{3} d x$ $V=2 pi int_1^2x^3dx$
$= 2 π \frac{x^{4}}{4} {∣ ∣ ∣}_{1}^{2}$ $=2pi (x^4)/4|_1^2$
$= 2 π \frac{[2^{4} - 1^{4}]}{4}$ $=2pi([2^4-1^4])/4$
$= \frac{15 π}{2}$ $=(15pi)/2$

Amory W. · · Sep 5 2014
How do you find the surface area of the solid obtained by rotating about the $x$ $x$ -axis the region bounded by $y = e^{x}$ $y=e^x$ on the interval $0 \leq x \leq 1$ $0<=x<=1$ ?

The answer is $\frac{π}{2} [e^{2} - 1]$ $pi/2[e^2-1]$ .

Since you are only given a single function and we are rotating about the axis of the parameter, this requires the disk method. The disk method is:

$V = \int_{a}^{b} A d x$ $V=int_a^b Adx$
$= \int_{a}^{b} π r^{2} d x$ $=int_a^b pi r^2dx$
$= \int_{a}^{b} π {[f (x)]}^{2} d x$ $=int_a^b pi [f(x)]^2dx$

We have the known values:

$f (x) = e^{x}$ $f(x)=e^x$
$a = 0$ $a=0$
$b = 1$ $b=1$

And now we can substitute:

$V = \int_{0}^{1} π {(e^{x})}^{2} d x$ $V=int_0^1 pi (e^x)^2dx$
$= π \int_{0}^{1} e^{2 x} d x$ $=pi int_0^1 e^(2x)dx$
$= π \frac{e^{2 x}}{2} {∣ ∣ ∣}_{0}^{1}$ $=pi (e^(2x))/2|_0^1$
$= \frac{π}{2} [e^{2} - e^{0}]$ $=pi/2[e^2-e^0]$
$= \frac{π}{2} [e^{2} - 1]$ $=pi/2[e^2-1]$

Amory W. · · Sep 16 2014
How do you find the surface area of a solid of revolution?

If the solid is obtained by rotating the graph of $y = f (x)$ $y=f(x)$ from $x = a$ $x=a$ to $x = b$ $x=b$ , then the surface area $S$ $S$ can be found by the integral

$S=2pi int_a^b f(x)sqrt{1+[f'(x)]^2}dx$

Wataru · · Sep 21 2014

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Applications of Definite Integrals

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Determining the Surface Area of a Solid of Revolution

Key Questions

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Applications of Definite Integrals