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Homework Help: Multivariable calculus -double int

  1. Jul 29, 2006 #1
    R is the interior of the region, in the x,y plane, bounded by the parabola [tex]y = 4 - (x - 3)^2[/tex] and for which [tex]x \leq 3\,\mbox{and}\,y \geq 0[/tex].
    Sketch the region R, and evaluate the double integral [tex]\iint_R 2xy\,dx\,dy[/tex]


    I've drawn the region, but I am unsure as to what to do with the integral and how R links into it. Do I simply make 3 and 0 the upper and lower limits for x, and make 4 and 0 the upper and lower limits for y when integrating? ie. [tex]\int_{0}^{4}\int_{0}^{3} 2xy\,dx\,dy[/tex]?
     
    Last edited: Jul 29, 2006
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  3. Jul 29, 2006 #2

    benorin

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    No, the domain of integration of [tex]\int_{0}^{4}\int_{0}^{3} 2xy\,dx\,dy[/tex] is a rectangle, having (0,0) and (3,4) as opposite corners. The given region R is parabolic, and easily described in the form [tex]f_1(y) \leq x\leq f_2(y), a\leq y\leq b[/tex]. To determine the two functions bounding x, look at the picture of R: for each fixed y value, x ranges between what curve (as function of y) and 3? after that, the bounds on y are constants: [tex]0\leq y\leq 4[/tex]
     
  4. Jul 30, 2006 #3
    I still don't really understand how I should start off the question. x is bounded by [tex]y = 4 - (x - 3)^2\,\mbox{and}\,y = 3[/tex]. So my equation would be [tex]\int_{0}^{4}\int_{\sqrt{4 - y}}^{3} 2xy\,dx\,dy[/tex]?
     
  5. Jul 30, 2006 #4

    HallsofIvy

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    You said you'd drawn the graph- that should tell you all you need to know. You should see that the parabola is symmetric about x= 3 and crosses the x-axis at x= 1 and x= 5. Those are important because you are requiring that [itex]y\ge 0[/itex]. Also, because you are requiring that [itex]x\le 3[/itex], the region over which you are integrating is the left half of that parabola- only up to the axis of symmetry.

    The way you have it set up, you are doing the "inner" integral with respect to x and the "outer" integral with respect to y. That's perfectly fine. Since the limits on the "outer" integral must be constants, what are the smallest and largest possible values of y in that region? Yes, they are 0 and 4, exactly as you have. For each y, then, what are the limits on x? Imagine drawing a horizontal line across the parabola. The lower value for x is on the parabola while the higher value is on the axis of symmetry, x= 3. Clearly the upper limit of integration is 3, as you have. What about the lower limit? That's on the parabola given by y= 4- (x-3)2. Solve that for x (it is not [itex]\sqrt{4- x}[/itex]!). That quadratic equation will have two roots and you will have to decide which to choose.

    As I said, that choice of order of integration is perfectly valid but I suspect most people would have chosen the other way: integrate with respect to y first and then x. What is the smallest value x takes on in that region? What is the largest? Now draw a vertical line at some representative value of x. What are the lowest and highest values of y in terms of x?
     
  6. Jul 30, 2006 #5
    [tex]x = \sqrt{4 - y} + 3[/tex]
    [tex]x = \pm2 + 3 = 1\,\mbox{or}\,5[/tex]
    As [tex]x \leq 3[/tex] choose x = 1.

    After solving the integral, I get:
    [tex]\int_{1}^{3}\int_{0}^{4} 2xy\,dy\,dx = 64[/tex]
     
  7. Jul 30, 2006 #6

    HallsofIvy

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    In spite of the fact that you were just told that was wrong??
    Integrating x from 1 to 3 and y from 0 to 4 integrates over the rectangle bounded by x=1, x=3, y= 0, y= 4. You were told that in benorin's first response. Unless the region of integration is a rectangle, the inner integral must depend on the second variable.
    To repeat (this time with emphasis!)
    "Now draw a vertical line at some representative value of x. What are the lowest and highest values of y in terms of x?"
     
  8. Jul 31, 2006 #7
    Ok, I think I got you. I chose to integrate with respect to x first.

    [tex]\int_{0}^{4}\int_{-\sqrt{4-y}+3}^{3} 2xy\,dx\,dy[/tex]
    [tex]\int_{0}^{4}[x^2 y]_{-\sqrt{4-y}+3}^{3}\,dy[/tex]
    [tex][5y^2 - 7y - 4(4 - y)^\frac{3}{2}]_{0}^{4}[/tex]
    Answer is 52 units cubed.
     
  9. Aug 1, 2006 #8

    HallsofIvy

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    [tex]\int_{0}^{4}[x^2 y]_{-\sqrt{4-y}+3}^{3}\,dy[/tex]

    [tex]\int_0^49y- y(3- \sqrt{4-y})^2 dy= \int_0^4y^2-4y- 6y\sqrt{4-y}dy[/tex]
    which doesn't seem to give what you have.
     
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