Volume by Shells, continually getting wrong answers

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Homework Help Overview

The discussion revolves around finding the volume of a solid of revolution using the Method of Shells for the function y = x^(1/3) - 2, specifically across the x-axis for the interval 8 <= x <= 27. Participants are attempting to understand the setup and calculations involved in applying the method correctly.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the setup of the radius and height for the shells, with some questioning whether they have defined these parameters correctly. There are attempts to integrate in terms of y, and some participants express frustration over consistently arriving at incorrect answers despite multiple attempts and seeking help.

Discussion Status

Some participants have provided guidance on the correct interpretation of the shell width and height, suggesting that drawing the graph may help clarify the relationships between the variables. There is an acknowledgment of confusion regarding when to subtract values in the setup, indicating that multiple interpretations are being explored.

Contextual Notes

Participants mention the pressure of an upcoming exam and the challenges faced in understanding the concepts, highlighting the emotional aspect of the learning process. There is also a reference to previous similar problems that led to different outcomes, emphasizing the need for clarity in problem setup.

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Homework Statement


y = x^(1/3) - 2

Find volume across the x-axis on 8 <= x <= 27 using the Method of Shells


Homework Equations


V = 2pi * Integral(r * h) from a -> b


The Attempt at a Solution



Let x = (y + 2)^3

Let h = (y +2)^3 // Set height

Let r = y // Set radius

Attempt:

2pi * Integral(y(y+3)^3 dy) from 0 to 1 // Integrate in terms of y

I multiplied (y+3)^3 out, and combined with y for:

2pi * Integral(y^4 + 6y^3 + 12y^2 + 8y dy) from 0 to 1

=

2pi ((y^5)/5 + ((3/2)*y^4) + (4y^3) + (4y^2)) from 0 to 1

I then calculate F(1) - F(0) and get (194/10)pi = (97/5)pi.

However, the correct answer in the book is (38/5)pi.

I also tried u-substitution and still got (97/5)pi.

Please help... this is extremely frustrating for me. I have been working really hard at finding volumes for the last two weeks and still continue to get most of my answers wrong. I've been to tutoring, my professor's office hours multiple times, and now I'm here. I just don't see what I'm doing wrong. I have an exam tomorrow morning and feel like I'm going to flunk it even though I've put in more than my fair share of studying.

I did set the radius and height correctly, right? If so, what is going on?
 
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sschmiggles said:

Homework Statement


y = x^(1/3) - 2

Find volume across the x-axis on 8 <= x <= 27 using the Method of Shells
What's the exact wording in this problem? Is the graph of your function revolved around the x-axis?
sschmiggles said:

Homework Equations


V = 2pi * Integral(r * h) from a -> b


The Attempt at a Solution



Let x = (y + 2)^3

Let h = (y +2)^3 // Set height

Let r = y // Set radius

Attempt:

2pi * Integral(y(y+3)^3 dy) from 0 to 1 // Integrate in terms of y

I multiplied (y+3)^3 out, and combined with y for:

2pi * Integral(y^4 + 6y^3 + 12y^2 + 8y dy) from 0 to 1

=

2pi ((y^5)/5 + ((3/2)*y^4) + (4y^3) + (4y^2)) from 0 to 1

I then calculate F(1) - F(0) and get (194/10)pi = (97/5)pi.

However, the correct answer in the book is (38/5)pi.

I also tried u-substitution and still got (97/5)pi.

Please help... this is extremely frustrating for me. I have been working really hard at finding volumes for the last two weeks and still continue to get most of my answers wrong. I've been to tutoring, my professor's office hours multiple times, and now I'm here. I just don't see what I'm doing wrong. I have an exam tomorrow morning and feel like I'm going to flunk it even though I've put in more than my fair share of studying.

I did set the radius and height correctly, right? If so, what is going on?
 
sschmiggles said:

Homework Statement


y = x^(1/3) - 2

Find volume across the x-axis on 8 <= x <= 27 using the Method of Shells


Homework Equations


V = 2pi * Integral(r * h) from a -> b


The Attempt at a Solution



Let x = (y + 2)^3

Let h = (y +2)^3 // Set height
This (above) is incorrect. The width of your shell is 27 - (y + 2)3
sschmiggles said:
Let r = y // Set radius

Attempt:

2pi * Integral(y(y+3)^3 dy) from 0 to 1 // Integrate in terms of y

I multiplied (y+3)^3 out, and combined with y for:

2pi * Integral(y^4 + 6y^3 + 12y^2 + 8y dy) from 0 to 1

=

2pi ((y^5)/5 + ((3/2)*y^4) + (4y^3) + (4y^2)) from 0 to 1

I then calculate F(1) - F(0) and get (194/10)pi = (97/5)pi.

However, the correct answer in the book is (38/5)pi.
I get this answer.
sschmiggles said:
I also tried u-substitution and still got (97/5)pi.

Please help... this is extremely frustrating for me. I have been working really hard at finding volumes for the last two weeks and still continue to get most of my answers wrong. I've been to tutoring, my professor's office hours multiple times, and now I'm here. I just don't see what I'm doing wrong. I have an exam tomorrow morning and feel like I'm going to flunk it even though I've put in more than my fair share of studying.

I did set the radius and height correctly, right? If so, what is going on?
 
Mark44 said:
This (above) is incorrect. The width of your shell is 27 - (y + 2)3
I get this answer.
Thanks, I've got it now too, but how am I supposed to know when to subtract the width/height from x without looking at a graph?

I just did a very similar problem (y = 8 - x^3), subtracted from x for the height, and got the wrong answer. I got the right answer when I didn't subtract h from x.

I'm so confused
 
sschmiggles said:
Thanks, I've got it now too, but how am I supposed to know when to subtract the width/height from x without looking at a graph?
You probably won't know. That's why you should always draw the graph.
sschmiggles said:
I just did a very similar problem (y = 8 - x^3), subtracted from x for the height, and got the wrong answer. I got the right answer when I didn't subtract h from x.

I'm so confused
 
Mark44 said:
You probably won't know. That's why you should always draw the graph.
Thanks a lot. I was making this a lot harder for myself than I needed to.
 
The first thing I did was to sketch a graph of the curve, and then to sketch a drawing of the rotated solid object. From the drawing it was obvious that the width of the shell was 27 - (y + 2)2.

When you have expressions that represent length of some sort, if it's a vertical length, you want yupper - ylower. If it's a horizontal length, you want xright - xleft. Having a drawing helps you figure this out, so that you have the right signs for your lengths.
 

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