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Reflecting About A Line- Solids

  1. May 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid generated by revolving the region bounded by the graphs of [itex] y= x^2 - 4x + 5 [/itex] and [itex] y= 5- x [/itex] about the line [itex] y= -1 [/itex]


    2. Relevant equations
    That is what I'm trying to figure out here, which one of the methods to use.


    3. The attempt at a solution

    I have these steps so far...

    x^2 - 4x + 5= 5 - x
    x^2 - 3x = 0
    x(x-3)= 0
    x= 0 and x= 3
    These are my intersection points. From 0 to 3 on this graph, the function 5- x has greater y-values than those of x^2- 4x + 5, so I get (5-x)- (x^2-4x+5)= 3x-x^2. And now I am stumped. What method would I use? And how am I supposed to know if it's a disk, shell, or washer method looking at the graph? I am clueless on how to do that. Thank you for your help.
     
  2. jcsd
  3. May 22, 2013 #2

    CAF123

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    Gold Member

    Generally it will be easier to do the question via the method of shells or via disks/washers. Boths methods work. I think using disks is easier in this case,
     
  4. May 22, 2013 #3

    Mark44

    Staff: Mentor

    You can always use either method (shells vs. disks/washers). The question really is which of the two methods results in an integral that is easier to evaluate.

    I find it helpful in these problems to draw two graphs. The first shows the region that is being revolved; the second shows a cross-section of the solid of revolution, with some detail on the typical volume element (a shell or disk/washer), including its dimensions.
     
  5. May 22, 2013 #4
    So using the disk method which if I'm not mistaken is (f(x)^2) dx, I should get:

    (3x-x^2)^2= 9x^2 - 6x^3 + x^4 with a and b being 0 and 3 respectively.

    108- [0]= 108?

    But when I draw the graph.. I don't see how the y= -1 would affect this at all then?
     
  6. May 22, 2013 #5

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    Which means it's wrong, eh? The formula$$
    \int_a^b \pi f^2(x)dx$$is for the area under ##y=f(x)## rotated about the ##x## axis. That isn't what you have, and you forgot the ##\pi## anyway. You want the formula$$
    V=\int_a^b \pi (r_{upper}^2 - r_{lower}^2)\, dx$$where ##r## is the radius of revolution for the appropriate curve. The axis being ##y=-1## will affect that.
     
  7. May 22, 2013 #6
    So by using the correct formula, I get R = (5-x)+1 and r = (x^2-4x+5)+1

    Then after I plug in and solve, V= 162pi/5. Is this correct?
     
  8. May 22, 2013 #7

    Mark44

    Staff: Mentor

    It would be helpful if you showed the work you did to get that result. I'm not saying that it's wrong, but I can't tell without duplicating your effort.
     
  9. May 22, 2013 #8
    Well I did this:

    pi * integral { (5-x)+1)^2) with a=0 and b=2 - ((x^2-4x+6)^2)} dx

    Then I just simplified.
     
  10. May 22, 2013 #9

    Mark44

    Staff: Mentor

    This looks good so far.

    Here's your integral in LaTeX.
    $$ \pi \int_0^3 [(6 - x)^2 - (x^2 - 4x + 6)^2]dx$$

    There are lots of opportunities for algebra errors to creep in, so show us the rest of your work.

    Click the Quote button to see what I did
     
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