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Double Integral

  1. Nov 15, 2005 #1
    I need to solve a double integral and I have no idea what to change the variables to:
    [tex] \iint_{R} \cos ( \frac{y-x}{y+x}) \ dA [/tex]
    [tex] R=\{(x,y) \mid \ -x+1 \leq y \leq -x+2, 1 \leq x \leq 2 \}
    [/tex]
    I tried to set [itex] u=y-x [/itex] and [itex]v=y+x[/itex], but I still can't solve the resulting integral. I also tried setting [itex] v + u= \frac{y-x}{y+x} [/itex] and differentiating implicitly when finding the Jacobian, but the Jacobian turns out to be zero.
     
    Last edited: Nov 15, 2005
  2. jcsd
  3. Nov 15, 2005 #2
    [tex] \iint_{R} \cos ( \frac{y-x}{y+x}) \ dA = \iint_{R} \cos ( 1 - \frac{2x}{y+x}) \ dA [/tex]
    [tex] R=\{(x,y) \mid \ -x+1 \leq y \leq -x+2, 1 \leq x \leq 2 \} \\[/tex]
    [tex] u = 2x [/tex]
    [tex] v = x+y \\ [/tex]
    [tex] R=\{(u,v) \mid \ 1 \leq v \leq 2, \ 2 \leq u \leq 4 \}[/tex]
    [tex] \frac{\partial (x,y)}{\partial (u,v)} = \frac{1}{2} [/tex]
    [tex] \frac{1}{2} \int_{1}^{2} \int_{2}^{4} \cos ( 1- \frac{u}{v}) \ dudv [/tex]

    After putting in these variables, I can't solve the integral. If I change what u and v are, then either the Jacobian is zero or I still can't solve the intergral.
     
    Last edited: Nov 15, 2005
  4. Nov 15, 2005 #3
    [tex] u=2x[/tex]
    [tex] v= \frac{1}{y+x} [/tex]

    [tex] R=\{(u,v) \mid \ \frac{1}{2} \leq v \leq 1, \ 2 \leq u \leq 4 \}[/tex]

    [tex] \frac{\partial (x,y)}{\partial (u,v)} = \frac{1}{2v^2} [/tex]

    [tex] \frac{1}{2} \int_{ \frac{1}{2} }^{1} \int_{2}^{4} \frac{1}{2v^2} \cos ( 1- uv) \ dudv [/tex]

    I can't do this one either.
     
    Last edited: Nov 16, 2005
  5. Nov 16, 2005 #4

    benorin

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    Homework Helper

    There's a good reason for that!

    The value of that integral is:

    [tex] \frac{1}{4} \left[ -\sin 1\cos 4 + 4\sin 1 \cos ^{2} 2 -4\cos 1 \cos 4 -20\sin 1 \mbox{Ci}(2)+16\sin 1 \mbox{Ci}(4)-16\cos 1 \mbox{Si}(4) \right.[/tex]
    [tex] -4\cos 1\cos 2 \sin 2 -4\sin 1\sin 4 +10\sin 1\sin 2+20\cos 1\mbox{Si}(2)-4\cos ^{2} 1 +3\cos 1\sin 2 [/tex]
    [tex]\left. +10\cos 1\cos 2-3\sin 1\cos 2 -4\cos 1 \mbox{Si}(1) + 4\sin 1\mbox{Ci}(1)-4\sin^{2} 1 + \cos 1 \sin 4 \right] [/tex] :bugeye:

    so says maple, where Ci and Si are the cosine and sine integral functions :cry: .
     
    Last edited: Nov 16, 2005
  6. Nov 16, 2005 #5
    Well, the answer should be [itex] \frac{3}{2} \sin 1 [/itex], so I think that this problem is probably an error in my book.

    I'll see what my professor says about it. Thanks.
     
  7. Nov 16, 2005 #6

    benorin

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    Homework Helper

    I get that problem from Stewart, Calculus 4th ed. ch15.9 promblem # 21 (I had that text for clalc I->III). The given answer is [itex] \frac{3}{2} \sin 1 [/itex] how ever the given region of integration is the trapezoidal region with verticies (1,0), (2,0), (0,2), and (0,1). Your bounds should then be:

    [tex] R=\{(x,y) \mid \ -x+1 \leq y \leq -x+2, 0 \leq x \leq 2 \}[/tex]
     
  8. Nov 17, 2005 #7
    yeah, i figured it out. I was just being dumb. Thanks.
     
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