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Calc 3, double integrals, how do they find the bounds of the graph to integrate?

  1. Jan 29, 2007 #1
    Hello everyone, i'm having issues figuring out how you find the bounds of each of these graphs, I have the solution manual but I still don't see how they did it. For example:

    The directions say to evaluate the double integral:


    [​IMG]

    On the first image, for #15.
    THe directions say: doulbe integral D y^3 dA. D is the triangluar region with vertices (0,2) (1,1) (3,2)

    from (0,2) to (1,1) they got the equation of the line to be:
    x = 2-y;
    and from
    (1,1) to (3,2) they got
    x = 2y-1;

    So I used the line formula, y = mx + b;
    b is the intercept of the y-axis:
    m = (y1-y2)/(x1-x2);
    m = (1-2)/(1-3) = -1/-2 = 1/2;
    y = 1/2x + 2
    y-2 = 1/2x
    x = 2y-4

    They also got bounds 1 to 2 for the dy, and i'm also not sure how they came up with that either.

    But this isn't the answer, and also for 17, 19 and 21 i don't see how they are getting these bounds...any help would be great!

    For #17, i see how they get y = sqrt(4-x^2) and y = -sqrt(4-x^2) is it dy bounds becuase its in the y -axis and the dx bounds is -2 to 2 becuase in the x direction the radius is 2?


    but for #19 i don't see how they are getting x^4, or x, i see from (0,0) to (1,1) thats the distance of 1, so i'm guessing thats y the dx is 0 to 1?
     
    Last edited: Jan 30, 2007
  2. jcsd
  3. Jan 30, 2007 #2

    AlephZero

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    Does this help explain what they did in #15? If not, you need to probably review what double integrals are before trying to do the questions.

    #17, the equation of the circle is x^4 + y^2 = 4
     

    Attached Files:

  4. Jan 30, 2007 #3

    HallsofIvy

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    This is to be the line from (1, 1) to (3,2)? Isn't it obvious that if you put y= 1 into your equation, x= 2-4= -2, not 1? Yes, the slope is 1/2 but where did you get that "2"? As you said, "b is the intercept of the y-axis" and you know that y= 2 when x= 3, not 0!! Since you are not given the y-intercept, use instead y= m(x-x0)+ y0. y= (1/2)(x-1)+ 1 or y= (1/2)(x-3)+ 2. Those both give y= (1/2)x+ 1/2 or 2y= x+1 so x= 2y-1 just as they said.

    Look at the graph! The whole point of the limits of integration is to cover the region. The lowest value of y in that region is at the lower vertex, (1, 1). The highest value is on the upper boundary line, where y= 2.

    Now, for each possible y value, imagine a horizontal line crossing the region. The left side is the line x= 2-y and the right side is the line x= 2y-1. For each y, x must go from 2-y on the left to 2y- 1 on the right. That's why they solved for x!
    (You could integrate in the other order, first y then x, but it is harder because of that "break" in the lower boundary. Clearly the lowest possible value for x is 0 and the highest is 3. For every x, the upper bound is that horizontal line y= 2, but the formula for the lower boundary changes at x= 1. It would be best to break it into two integrals. For x= 0 to x= 1, the lower boundary is given by the line y= 2- x (I've switch back to "y= " since I need the lower and upper values of y now.) The upper limit on y is, of course, 2. For x= 1 to 3, however, the lower boundary is the line x= 2y- 1 or y= (1/2)x+ 1/2. The integral is
    [tex]\int_{x=0}^1\int_{y=2-x}^2 y^3 dydx+\int_{x=1}^3\int_{y=(1/2)x+1/2}^2 y^3dydx[/tex]


    More correctly, the x bounds are -2 to 2 because the lowest value of x in that region is -2 and the highest value is 2. (If the circle were centered at some point other than (0,0) but still had radius 2, the x bounds would not be -2 and 2.) Now, for each x, imagine a vertical line at that value of x. The point where that vertical line crosses the lower arc of the circle is [itex]y= -\sqrt{4-x^2}[/itex] and the point where it crosses the upper arc is [itex]\sqrt{4-x^2}[/itex]. For each value of x, y ranges from one to the other.


    (You keep saying "the dx" or "the dx bounds" and "the dy bounds". It would be better to say "the x bounds" or "y bounds"- the "dx" has nothing to do with it.) For #19, again, look at the picture. It is obvious that the smallest value of x in that region occurs where y= x and y= x4 intersect on the left, (0,0), while the highest value of x occurs where they cross on the right, (1, 1). The smallest value of x in that region is 0 and the largest is 1 and you must cover the entire region- x must go from 0 to 1.
    Now, for each x, imagine a vertical line at that x. The lowest point on that line is where it crosses the curve y= x4 and the highest is where it crosses the line y= x. To cover that region, for each x y must range from y= x4 up to y= x.

    To be sure you understand, how would you reverse the order of integration?
    That is suppose you had
    [tex]\int_{y=?}^? \int_{x=?}^? x+2y dxdy[/tex]
    What would the ?s be? Since the "outer" integral is in y, those limits must be the lowest and highest values for y in the region. What are they? (You should be able to read them off the graph.) Now, for each y, imagine a horizontal line across the graph at that y. The left boundary is the line y= x or x= ?. The right boundary is the line y= x4 or x= ? (remember that x is positive here).
     
  5. Jan 30, 2007 #4

    Thanks for the help guys, once I figured out how to graph functions without a calculator like the good old days I did fine, i got all the problems correct except i'm having issues on what you just brought up Ivey...

    The directions are: Sketch the region of integration and change the order of integration.

    Now I can sketch the integration just fine, for example:
    [​IMG]

    But i'm confused on how they went from

    0 <= y <= x^(1/2), 0 <= x <= 4

    to

    y^2 <= x <= 4, 0 <= y <= 2



    I know y is going from 0 (the lower bound) and up to x^(1/2) which is y's upper bound.

    I also know x is going from 0 to 4.

    So they want me to convert a type I region to a type II region and thats where i'm having difficulties...

    I see if you let y = x^(1/2) then if you solve for x, you'll get x = y^2 which is the lower bound of x but how did they get 4 to be the upper bound of x?


    If i can sketch the orginal region of integration is there a quick way to look at the graph and see how it would be a change of integration, any tips?

    Thanks!
     
  6. Jan 31, 2007 #5

    HallsofIvy

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    Look at your graph! (One error- the point is (4, 2), not (2, 4) but since you say 0<x<4, 0<y<2, I'm sure that's what you meant.)

    After you have the graph drawn, ignore the previous limits of integration.

    You "outside" integral is now with respect to y. What is the smallest numeric value of y in that region? What is the largest value of y in that region? Those are the lower and upper limits on the y-integral.

    Now imagine a horzontal line, y= constant, in that region. Heck, you've got the graph- go ahead and draw a horizontal line! What curve marks the left end of that line? x= what there? What "curve" marks the right end of the line? x= what there? Those x values- which may depend on y are the limits for the x-integral.
     
  7. Feb 1, 2007 #6
    Thanks for the help Ivy, I got a 100% on the quiz! w00t!
     
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