Having Trouble With an Intergration Problem

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The discussion centers on calculating the volume of a region bounded by the curve y = 9 / (2 - x), the coordinate axes, and the line x = 1 when rotated around the x-axis. The initial approach involves squaring the function and attempting to integrate it, but the user struggles with the integration process. Suggestions include using substitution to simplify the integral and correcting algebraic mistakes made during integration. The conversation highlights the importance of proper integration techniques and algebraic manipulation to arrive at the correct volume. Overall, the user seeks clarification on the integration process to resolve their calculation errors.
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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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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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