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Some double integrals

  1. Sep 26, 2009 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data

    1) By evaluating a double integral find the volume of the tetrahedron bounded by planes x+2y+z=2, x= 2y ,z=0 and y=0 (answer = 1/3)

    2)

    [tex]\int_{0} ^{1} \int_{x} ^{1} sin(y^2) dydx[/tex]

    i) explain why this can be evaluated by integrating wrt y first.
    ii) sketch the region of integration R of the double integration
    iii)Evaluate the integral by expressing it as a Type II region first.


    2. Relevant equations



    3. The attempt at a solution

    My query is with 1 and 2ii)

    for 1

    my region I integrated over was x/2≤y≤(2-x)/2 and 0≤x≤1

    found when z=0 to get y=(2-x)/2 and this intersects y=x/2 at (1,1/2)

    yet when I compute

    [tex]\int_{0} ^{1} \int_{\frac{x}{2}} ^{\frac{2-x}{2}} (2-2y-x) dydx[/tex]

    I don't get 1/3. If it's a problem with my algebra I'll fix it. But are my boundaries correct?

    2) I believe it can't be done with dy first since the integral of sin(y2) can't be expressed in terms of elementary functions. Is the reason or is there some other factor that prohibits this?
     
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  3. Sep 26, 2009 #2

    LCKurtz

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    Your boundaries look correct in your integral for 1.

    And your answer to 2 looks correct too. And that problem will go away when you reverse the integration order.
     
    Last edited: Sep 26, 2009
  4. Sep 26, 2009 #3

    rock.freak667

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    ok for some reason I can't seem to get 1/3. This is what I have


    [tex]\int_{0} ^{1} \int_{\frac{x}{2}} ^{\frac{2-x}{2}} (2-2y-x)dydx[/tex]

    [tex] = \int_{0} ^{1} \left[ 2y-y^2 -xy \right] _{\frac{x}{2}} ^{\frac{2-x}{2}} dx[/tex]

    [tex]= \int_{0} ^{1} \left( (2-x)- \frac{1}{4}(2-x)^2 -\frac{x}{2}(2-x) -x + \frac{x^2}{4} + \frac{x^2}{2} \right)[/tex]


    [tex]= \int_{0} ^{1} (2-x-1+x-\frac{x^2}{4}-x+\frac{x^2}{2}-x+\frac{x^2}{4}+\frac{x^2}{2}) dx[/tex]

    [tex]= \int_{0} ^{1} (1-2x+\frac{x^2}{2}) dx[/tex]

    [tex]= \left[ x-x^2 +\frac{x^3}{6} \right]_{0} ^{1} = 1/6[/tex]

    :confused: where am I going wrong? Because I clearly can't see it

    EDIT: for the second one as well, I can't get the answer, I must be doing something wrong

    [tex]\int_{0} ^{1} \int_{x} ^{1} siny^2 dxdy[/tex]

    [tex]=\int_{0} ^{1} \left[ xsiny^2 \right] _{x} ^{1} dy[/tex]

    [tex]=\int_{0} ^{1} (xsin1-xsinx^2) dx[/tex]

    [tex]= \left[ \frac{x^2}{2}sin1 + \frac{1}{2}cos(x^2) \right] _{0} ^{1} [/tex]

    [tex]= \frac{1}{2}sin1 + \frac{1}{2}cos1 - \frac{1}{2}[/tex]


    answer is (1-cos1)/2 :confused: :confused:
     
    Last edited: Sep 27, 2009
  5. Sep 27, 2009 #4

    LCKurtz

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    It's late here so I will be brief. Your limits are correct on the first one and Maple checks it as 1/3. Recheck your algebra.

    On the second one your reversed limits are wrong. You need to draw a picture of the region from the first integral and use it to get the dx dy limits.

    Sack time here.
     
  6. Sep 27, 2009 #5

    rock.freak667

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    Thanks for your help, I just need a bit of understanding with the 2nd one still. I found the new limits as y≤x≤0 and 0≤y≤1, but to get the answer the y range must be 1≤y≤0 and this makes no sense to me.
     
  7. Sep 27, 2009 #6

    LCKurtz

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    Agreed, 1≤y≤0 makes no sense. Again, to work the problem you must draw the region described in the original integral. You should see a triangular region in the first quadrant above the line y = x and below the line y = 1. Then simply use that picture to get the limits integrating in the x direction first. You need to look at the picture to get the new limits.
     
  8. Sep 27, 2009 #7

    rock.freak667

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    I got out this one, but in my integration working above can you see any error I am making because I keep getting answers which are not 1/3 . I just tried it again and got an answer of -1/2
     
  9. Sep 27, 2009 #8

    LCKurtz

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    The x2/2 term in the second line from the bottom is wrong.
     
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