1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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


    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


    User Avatar
    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


    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


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted