Volume by Shells, continually getting wrong answers

  • #1

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?
 

Answers and Replies

  • #2
34,178
5,794

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?

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?
 
  • #3
34,178
5,794

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
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.
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?
 
  • #4
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
 
  • #5
34,178
5,794
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.
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
 
  • #6
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.
 
  • #7
34,178
5,794
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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