# Having Trouble With an Intergration Problem

• Mikry
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).

#### Mikry

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!

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. 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. 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.