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Simplifying help !

  1. Dec 3, 2005 #1
    Simplifying help plz!

    Hi, I was given a double integral question and I managed to do the x integration. After placing the limits I get the following:

    ∫{ (2y²)(√2+y²) - (2y²)(√2y²) } dy

    I know the integrand can be simplified but I don't have a clue. Can anyone help? :confused: Thank you!!!
     
  2. jcsd
  3. Dec 3, 2005 #2
    Is the first square root supposed to cover both 2 and y^2 or is it the square root of 2, plus y^2 (outside the root)?
     
  4. Dec 3, 2005 #3
    oops sorry! Yes the square root is suppose to cover all of 2+y²
     
  5. Dec 3, 2005 #4
    In that case I'm guessing it's the same for the second square root term..
    So you have:
    ∫{ (2y²)√(2+y²) - (2y²)√(2y²) } dy
    Splitting it up into two integrals...
    ∫(2y²)√(2+y²) dy - ∫(2y²)√(2y²) dy
    On the right side, you can pull the y^2 out of the square root:
    ∫(2y²)√(2+y²) dy - ∫(2y²)y√(2) dy
    And then:
    ∫(2y²)√(2+y²) dy - 2√2 ∫y^3 dy
    = ∫(2y²)√(2+y²) dy - (y^4)/(√2) + C
    Now, for the left integral it'll take a bit more work.... let's try a hyperbolic substitution:
    sinh^2 x - cosh^2 x = 1
    sinh^2 x = 1 + cosh^2 x
    So, let's say y = √2 * cosh x
    dy = √2 * sinh x dx; from there:
    ∫(2y²)√(2+y²) dy - (y^4)/(√2) + C
    = ∫(2(2cosh²x))√(2)*sinh x * √2 sinh x dx - (y^4)/(√2) + C
    = 8∫cosh² x sinh² x dx - (y^4)/(√2) + C
    And you can proceed from there.. :\ Or you could have used a trig substitution...
     
  6. Dec 3, 2005 #5
    ooo ok thanks!

    I've also got another quick question. I've been asked to draw the region of integration for the following integral. I'm not sure if I've drawn it right :confused: can someone help? thank you!

    http://tinypic.com/i53khk.jpg
     
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