Shell method about the line x=5

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Discussion Overview

The discussion revolves around using the shell method to find the volume of the solid generated by revolving the region bounded by the curve $$y = 4x - x^2$$ and the line $$y = 0$$ about the line $$x = 5$$. Participants explore the setup of the integral and the definitions of the variables involved in the shell method.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant states the shell method formula for volume, $$V = \int_{a}^{b} 2\pi x (f(x) - g(x))dx$$, and seeks assistance in applying it.
  • Another participant introduces the concept of an arbitrary shell, defining the volume element as $$dV=2\pi rh\,dx$$ and asks for clarification on the variables $$r$$ and $$h$$.
  • There is a discussion about the definitions of $$r$$ and $$h$$, with one participant suggesting that $$h$$ represents the highest point of the graph and $$r$$ is the distance from the line of revolution to the point on the curve.
  • One participant provides a corrected expression for $$r$$ as $$r=5-x$$ and for $$h$$ as $$h=4x-x^2$$, leading to the formulation of the volume integral as $$V=2\pi\int_0^4 (5-x)(4x-x^2)\,dx$$.
  • Another participant presents an alternative approach using a substitution $$\xi= 5 - x$$ and computes the volume integral, arriving at a numerical result of $$64\pi$$.
  • One participant expresses confusion about their earlier setup of the integral and acknowledges the need to adjust their expression to match the correct formulation involving $$5-x$$.
  • Another participant critiques the initial formula presented, suggesting it was intended for revolution about the $$y$$-axis and emphasizes the importance of visualizing the elemental volume directly.
  • A participant claims to have arrived at an answer of $$64\pi$$, indicating a potential resolution to their inquiry.

Areas of Agreement / Disagreement

Participants express varying interpretations of the shell method and how to set up the integral correctly. There is no clear consensus on the best approach, and multiple perspectives on the definitions of the variables and the setup of the integral remain evident.

Contextual Notes

Some participants appear to be working from different interpretations of the shell method, leading to confusion regarding the appropriate setup for the integral. There are also indications that some mathematical steps may be unresolved or misapplied.

alane1994
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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.

$$y = 4x - x^2$$
, y = 0, about the line x = 5

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

I know this is fairly rudimentary, but assistance would be appreciated!
 
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The volume of an arbitrary shell is:

$$dV=2\pi rh\,dx$$

View attachment 1091

What are $r$ and $h$?
 

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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?
 
alane1994 said:
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:

$$r=5-x$$

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

and thus the volume of the shell is:

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

Next, you want to sum all the shells:

$$V=2\pi\int_0^4 (5-x)\left(4x-x^2 \right)\,dx$$
 
alane1994 said:
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.

$$y = 4x - x^2$$
, y = 0, about the line x = 5

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

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

$\chi$ $\sigma$
 
THAT'S MY PROBLEM!
I had
$$V = \int_{0}^{4} 2\pi x ((x-5) - (4x-x^2))dx$$

That is where I got off. You need to change the x into 5-x.
$$V = \int_{0}^{4} 2\pi (5-x)(4x-x^2)dx$$
 
alane1994 said:
THAT'S MY PROBLEM!
I had
$$V = \int_{0}^{4} 2\pi x ((x-5) - (4x-x^2))dx$$

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

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.
 
Another picture:

shell.png
 
Ok, I have arrived at an answer of $$64\pi$$.
 

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