Having Trouble With an Intergration Problem

  • Thread starter Thread starter Mikry
  • Start date Start date
  • Tags Tags
    Intergration
Mikry
Messages
6
Reaction score
0
I am given that the equation of a curve is y = 9 / (2 - x). They then ask me to find the volume obtained by the region bounded by the curve, the coordinate axes and the line x = 1 when the region is rotated through 360° about the x-axis.

My attempt:

To calculate this I must use the format of V = ∏ ∫ y2 dx. Thus I square the equation, giving me 81 / (2 - x)2.

Now I need to integrate the expression. Here is where I think I'm going wrong...expand the bottom term to get x2 - 4x + 4.
I then try to integrate this expansion, getting 81 / (2x - 4) ln |x2 - 4x + 4|.

This gives me the final equation of V = ∏ [81 / (2x - 4) ln |x2 - 4x + 4|] from x = 0 to x = 1

Long story short I don't come to the right answer after all of that. I'm pretty sure I'm missing something easy and obvious. :frown: Please help!
 
on Phys.org
Mikry said:
I am given that the equation of a curve is y = 9 / (2 - x). They then ask me to find the volume obtained by the region bounded by the curve, the coordinate axes and the line x = 1 when the region is rotated through 360° about the x-axis.

My attempt:

To calculate this I must use the format of V = ∏ ∫ y2 dx. Thus I square the equation, giving me 81 / (2 - x)2.

Now I need to integrate the expression. Here is where I think I'm going wrong...expand the bottom term to get x2 - 4x + 4.
I then try to integrate this expansion, getting 81 / (2x - 4) ln |x2 - 4x + 4|.

This gives me the final equation of V = ∏ [81 / (2x - 4) ln |x2 - 4x + 4|] from x = 0 to x = 1

Long story short I don't come to the right answer after all of that. I'm pretty sure I'm missing something easy and obvious. :frown: Please help!

You didn't integrate that correctly at all. Use the substitution u=(2-x). What's the integral of 1/u^2=u^(-2)?
 
Mikry said:
I am given that the equation of a curve is y = 9 / (2 - x). They then ask me to find the volume obtained by the region bounded by the curve, the coordinate axes and the line x = 1 when the region is rotated through 360° about the x-axis.

My attempt:

To calculate this I must use the format of V = ∏ ∫ y2 dx. Thus I square the equation, giving me 81 / (2 - x)2.

Now I need to integrate the expression. Here is where I think I'm going wrong...expand the bottom term to get x2 - 4x + 4.
I then try to integrate this expansion, getting 81 / (2x - 4) ln |x2 - 4x + 4|.

This gives me the final equation of V = ∏ [81 / (2x - 4) ln |x2 - 4x + 4|] from x = 0 to x = 1

Long story short I don't come to the right answer after all of that. I'm pretty sure I'm missing something easy and obvious. :frown: Please help!

Your method of obtaining the volume is incorrect.

Review the Second Theorem of Pappus:
http://mathworld.wolfram.com/PappussCentroidTheorem.html
 
SteamKing said:
Your method of obtaining the volume is incorrect.
I see no problem with the formulation of the integral, but as Dick notes, the algebra went awry from there.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
2K