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Volume Of Solid Above Axis

  1. Mar 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and y = 4 about the line y = 4

    3. The attempt at a solution

    Since y=4 is parallel to the x-axis i assume i should be using the 'washer' method.

    The points of intersection are x = 2, x = -2... But when i compute the volume i get the wrong answer.

    [itex]V = \pi \int ^{2}_{-2} ( x^{4} - 8 )dx = 1/5 x^{5} - 8x | ^{2}_{-2}[/itex]

    which results in pi * 64/5 which is apparently wrong...
     
  2. jcsd
  3. Mar 11, 2013 #2

    LCKurtz

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    Check your formula for the washer method.
     
  4. Mar 11, 2013 #3
    edit: Whoops.
     
    Last edited: Mar 11, 2013
  5. Mar 11, 2013 #4

    Mark44

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    Your integrand is incorrect. The volume of a typical disk (it's not a washer here) is ##\pi##radius2*Δx. You are forgetting to square the radius.
     
  6. Mar 11, 2013 #5
    [itex]V = \pi \int ^{2}_{-2} ( x^{4} - 16 )dx = 1/5 x^{5} - 16x | ^{2}_{-2}[/itex]

    So evaluating this, i get pi * -256/5, which is wrong... Is there a method for determining the order in which i subtract the two given functions? Since y=4 is above y=x^2 should i be subtracting y=x^2 from it instead?
     
  7. Mar 11, 2013 #6

    LCKurtz

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    What is the disk radius? What do you get when you square it?
     
  8. Mar 11, 2013 #7
    Draw a picture of your region, and draw a single approximate disk of depth Δx. What is its radius?
     
  9. Mar 11, 2013 #8
    Well if i draw out x^2 and y=4, the distance from the lowest point in x^2 to y=4 is 4. And 4^2 is 16 so i guess i don't follow what you're hinting at.
     
  10. Mar 11, 2013 #9
    Great, that's the radius at x = 4. However, the radius varies with x. What is the radius r(x) as a function of x ?
     
  11. Mar 11, 2013 #10

    LCKurtz

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    The vertical distance between two curves is ##y_{upper}-y_{lower}##. What is that in this problem? What is its square? That's what goes in the integrand as ##\pi r^2dx##.
     
  12. Mar 11, 2013 #11
    Okay it's 4-x^2 so i guess my order was wrong, if that's what you're getting at?

    I forgot order doesn't matter as long as you take the absolute value of your answer (since my result is 256/5 if i shift things around). So i guess 256/5 * pi is correct then?

    Am i supposed to be squaring (4-x^2)^2 = (16 - 8x^2 + x^4) or should i be squaring them individually, like (4^2) - ((x^2)^2)?
     
    Last edited: Mar 11, 2013
  13. Mar 11, 2013 #12

    Mark44

    Staff: Mentor

    This is just the distance between a point on the line and a point on the curve.
    You're supposed to square (4 - x2), which would be (4 - x2)2.
     
  14. Mar 11, 2013 #13

    LCKurtz

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    The answer is in my quoted post.
     
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