Having Trouble With an Intergration Problem

Therefore, the correct answer is V = (9π/2)(1 - (3/2) ln 2).In summary, the equation of a curve is given as y = 9 / (2 - x). By rotating the region bounded by the curve, the coordinate axes, and the line x = 1 through 360° about the x-axis, we can calculate the volume using the formula V = ∏ ∫ y2 dx. After squaring the equation, the expression is integrated using the substitution u=(2-x) and the Second Theorem of Pappus. The correct answer is V = (9π/2)(1 - (3/2) ln 2).
  • #1
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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!
 
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  • #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!

You didn't integrate that correctly at all. Use the substitution u=(2-x). What's the integral of 1/u^2=u^(-2)?
 
  • #3
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
 
  • #4
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.
 

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