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Shell method about the line x=5

alane1994

Active member
Oct 16, 2012
126
I have been going over previous tests in an attempt to better prepare myself for the final that is coming tomorrow. I was posed a question.

Use the shell method to find the volume of the solid generated by revolving the plane region about the given line.

\(\displaystyle y = 4x - x^2\)
, y = 0, about the line x = 5

I know this.
SHELL METHOD
\(\displaystyle V = \int_{a}^{b} 2\pi x (f(x) - g(x))dx\)

I know this is fairly rudimentary, but assistance would be appreciated!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The volume of an arbitrary shell is:

\(\displaystyle dV=2\pi rh\,dx\)

alane1994.jpg

What are $r$ and $h$?
 

alane1994

Active member
Oct 16, 2012
126
Ok, is "h" the highest point of the graph?
And if so, r would be 5-x, where x is the corresponding coordinate for the highest point?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Ok, is "h" the highest point of the graph?
And if so, r would be 5-x, where x is the corresponding coordinate for the highest point?
No, $h$ is the distance from the top curve to the bottom curve at the value of $x$ for the arbitrary shell. The arbitrary shell can be anywhere for $0\le x\le4$. I just drew one such shell.

So we have:

\(\displaystyle r=5-x\)

\(\displaystyle h=\left(4x-x^2 \right)-0=4x-x^2\)

and thus the volume of the shell is:

\(\displaystyle dV=2\pi(5-x)\left(4x-x^2 \right)\,dx\)

Next, you want to sum all the shells:

\(\displaystyle V=2\pi\int_0^4 (5-x)\left(4x-x^2 \right)\,dx\)
 

chisigma

Well-known member
Feb 13, 2012
1,704
I have been going over previous tests in an attempt to better prepare myself for the final that is coming tomorrow. I was posed a question.

Use the shell method to find the volume of the solid generated by revolving the plane region about the given line.

\(\displaystyle y = 4x - x^2\)
, y = 0, about the line x = 5

I know this.
SHELL METHOD
\(\displaystyle V = \int_{a}^{b} 2\pi x (f(x) - g(x))dx\)

I know this is fairly rudimentary, but assistance would be appreciated!
With the substitution $\xi= 5 - x$ You have to compute the volume of the rotation solid about $\xi=0$ of the function $\displaystyle f(\xi) = - 5 + 6 \xi - \xi^{2}$ obtaining...


$\displaystyle V = 2\ \pi\ \int_{1}^{5} \xi\ (- 5 + 6 \xi - \xi^{2})\ d \xi = 64\ \pi\ (1)$

Kind regards

$\chi$ $\sigma$
 

alane1994

Active member
Oct 16, 2012
126
THAT'S MY PROBLEM!!!!!!
I had
\(\displaystyle V = \int_{0}^{4} 2\pi x ((x-5) - (4x-x^2))dx\)

That is where I got off. You need to change the x into 5-x.
\(\displaystyle V = \int_{0}^{4} 2\pi (5-x)(4x-x^2)dx\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
THAT'S MY PROBLEM!!!!!!
I had
\(\displaystyle V = \int_{0}^{4} 2\pi x ((x-5) - (4x-x^2))dx\)

That is where I got off. You need to change the x into 5-x.
\(\displaystyle V = \int_{0}^{4} 2\pi (5-x)(4x-x^2)dx\)
You can't get from the first formula to the second simply by replacing $x$ with $5-x$.

The formula you cited in your original post looks like it was meant for revolution about the $y$-axis. I find it easier to not try to use such a formula, but to just look at one element of the entire volume, whether it be a shell, disk or washer. Once you have the elemental volume, then you can add all the elements by integrating.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Another picture:

 

alane1994

Active member
Oct 16, 2012
126
Ok, I have arrived at an answer of \(\displaystyle 64\pi\).