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: Iterated Integral

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate the following double integral over the closed region B:

    [tex]\int\int_B x^2 y^3 dx dy[/tex] where B is the region bounded by y = x^2 and y = x

    2. Relevant equations

    3. The attempt at a solution

    I think I set up the paramaters wrong:

    [tex]\int^{1}_{0}\int^{y}_{\sqrt{y}} x^2 y^3 dx dy[/tex]
    Last edited: Feb 9, 2009
  2. jcsd
  3. Feb 9, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    If y is in [0,1], as it looks like you've figured out for your integral, which is larger y or sqrt(y)? Is that what's confusing you?
  4. Feb 9, 2009 #3

    The answer is 1/77, but I get -1/77, but taking it's absolute value should take care of that.
  5. Feb 10, 2009 #4
    What's sqrt(1/4)?
  6. Feb 10, 2009 #5


    User Avatar
    Science Advisor

    Can you explain why "taking its absolute value" is justified?

    Once again, which is larger, y or [itex]\sqrt{y}[/itex] for y between 0 and 1?

    Which is larger 1/4 or [itex]\sqrt{1/4}[/itex]?
  7. Feb 10, 2009 #6
    I get it. Because y goes from 0 - 1, sqrt(y)>y in this case.

    but what if y went from 0 - 2, then would it be this:

    [tex]\int^{1}_{0}\int^{\sqrt{y}} }_{y} x^2 y^3 dx dy + \int^{2}_{1}\int^{y}_{\sqrt{y}} x^2 y^3 dx dy[/tex]
  8. Feb 10, 2009 #7


    User Avatar
    Science Advisor
    Homework Helper

  9. Feb 10, 2009 #8
    Thank you all for your help.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook