# Having Trouble With an Intergration Problem

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. Please help!

## Answers and Replies

Dick
Science Advisor
Homework Helper
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. 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)???

SteamKing
Staff Emeritus
Science Advisor
Homework Helper
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. Please help!

Your method of obtaining the volume is incorrect.

Review the Second Theorem of Pappus:
http://mathworld.wolfram.com/PappussCentroidTheorem.html

haruspex
Science Advisor
Homework Helper
Gold Member
2020 Award
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.