1. Not finding help here? Sign up for a free 30min 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!

Another Integral Word Problem

  1. Apr 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Here is the problem:
    WordProblem19.jpg


    3. The attempt at a solution
    I found the intersection point to be (1.10,2.78). Now I don't know what to use for the lower limit.
     
  2. jcsd
  3. Apr 25, 2009 #2

    jhae2.718

    User Avatar
    Gold Member

    We are revolving the curve around the x axis. Since the area of R is limited by the y-axis, this means our limits of integration will be the x-coordinate where the two graphs intersect the y-axis and the point where they intersect each other.

    What method of finding volume are you using? Cross sections or cylindrical shell? If using cross sections this will be a dx problem; if using the shell method this will be a dy problem. For this problem I would use the cross section/disk method and thus use x-coordinates as limits of integration. For a, this will be a washer.
     
  4. Apr 25, 2009 #3
    Ya the upper limit is the point of intersection but what do you mean by when the graphs interse t the y axis? You cant have 3 limits?

    Edit: so from 0 to 1.10?
     
  5. Apr 25, 2009 #4

    jhae2.718

    User Avatar
    Gold Member

    It is zero--since we are bounded by the y-axis and the intersection, our limits of integration are 0,1.10.

    When we find the volume we are going to add cross sections of the curve from 0 until we reach 1.10. Think of a layer cake--the bottom is zero and the top is the intersection of the curve. Turn it sideways and let the axis down the middle be the x-axis, and that is what we are doing.
     
    Last edited: Apr 25, 2009
  6. Apr 25, 2009 #5
    And what about the second part?
     
  7. Apr 25, 2009 #6

    jhae2.718

    User Avatar
    Gold Member

    We'll let f(x)=4-x2 and g(x)=1+2sin(x)
    Use the same limits of integration, but a different cross section. The area of a square is A=s2. In this case s=f(x)-g(x), f(x)>g(x) on the interval of the limits of integration. Take Sab[f(x)-g(x)]2dx.

    For a, remember to treat it as a washer. V=pi*Sab[f(x)]2-[g(x)]2dx.
     
  8. Apr 25, 2009 #7
    just a general question, for both parts if I use the limits 0 to 1.10, won't that also include the area under the curve for 1+2sin(x) curve until x=1.10?
     
  9. Apr 25, 2009 #8

    jhae2.718

    User Avatar
    Gold Member

    If you treat it as a washer for a, you will have subtracted the volume of 1+2sin(x). For part b, you also subtract the area of 1+2sin(x). 4-x2 includes the area/volume resulting from 1+2sin(x), so by subtracting it we get the area of R.

    Earlier I accidently used 2.78 instead of 1.10...fixed my above post.
     
  10. Apr 26, 2009 #9
    thank you very much for all the help you have provided.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Another Integral Word Problem
Loading...