Double/triple intergral bounds

  • Thread starter Jacob87411
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Hello,

I am having trouble understanding how to determine what is the lower and what is the upper bound in some calculus problems. For example:

Evaluate the double integral xydA, where D is the region bounded by the line y=x-1 and the parabola y^2=2x+6. Now you set it up to take the integral in terms of x first and the upper bound is y+1 and the lower is 1/2y^2+3...I guess im just confused on why it isnt the other way around not just in this problem but in general i have problems recognizing which is which. thanks
 

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  • #2
arildno
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You cannot express the relation [itex]y^{2}=2x+6[/itex] in such a way that y is a SINGLE function of the x!
Thus, you cannot really, use x as your independent variable (or rather, it becomes very hard to do so!).

However, you may use y as your independent variable.

I strongly advise you to actually DRAW the region of integration to get a feel for this, which obviously is something you didn't bother about doing.
 
  • #3
arildno
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Just adding to Halls' post:
For slightly "nastier", but still "nice" regions, we often manage to make a full change of variables, so that in the new set of variables, the domain is VERY nice (preferably a rectangle, since that's the simplest type of domain).

For still nastier regions, we get head-aches.
 
  • #4
HallsofIvy
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Jacob87411 said:
Hello,

I am having trouble understanding how to determine what is the lower and what is the upper bound in some calculus problems. For example:

Evaluate the double integral xydA, where D is the region bounded by the line y=x-1 and the parabola y^2=2x+6. Now you set it up to take the integral in terms of x first and the upper bound is y+1 and the lower is 1/2y^2+3...I guess im just confused on why it isnt the other way around not just in this problem but in general i have problems recognizing which is which. thanks
Your objective is to "cover" the region. I recommend that you first graph the given curves. The next thing you have to do is decide in what order you want to do the integrations. You want to do the x- integration first (a good choice- I'll explain why in a moment). That means that the limits for the "outer", y, integration must be numbers while the limits for the "inner", x, integration may depend on y. In order to "cover" the region, the limits of integeration on y must go from the least possible value of y to the largest possible value of y. Obviously, from the graph, that occurs where the two curves cross. Solving y2= 2x+ 6 and y= x- 1 simultaneously, we find that the curves cross when y= -2 and y= 4. The limits on the outer integral are y= -2 and y= 1.
Now you have to decide on the limits for each y. Draw (or imagine) a horizontal line across your graph (representing a specific value of y). you see that the left endpoint of that line is on the parabola, the right endpoint on the line. Of course, the left x-value is the lower value, the right x-value is the higher value. On the parabola y2= 2x+ 6 so x= (1/2)y2- 3. That will be the lower limit of integration.
On the line y= x- 1 so x= y+ 1. That will be the upper limit of integration.
[tex]\int_{y= -2}^4\int_{x=\frac{1}{2}y^2- 3}^{y+1} xy dx dy[/tex]
I recommend, by the way, actually writing the "x= " and "y= " in the limits as I did here as a reminder.

Suppose you had decided to integrate with respect to y first- in other words to do the integrals in the reverse order. Now we have to identify the lowest and highest values of x. We can see that the lowest value of x occurs at the vertex of the parabola, x= -3. The highest value is where the line and parabola cross at (5, 4). So the limits of integration on the outer integral will be x= -3, x= 5. Now imagine a vertical line across the graphs (representing a specific value of x). The upper end is on the parabola y2=- 2x+ 6 or [itex]y= \sqrt{2x+6}[/itex]. But the parabola crosses the line at (-1, -2) so the lower end has two different expressions depending on whether x is less than or larger than -1. For x< -1, [itex]y= -\sqrt{2x+6}[/itex]. For x> -1, y= x-1. That's why "x first
" was a good choice! However, we can break the region into two parts and say that the integral is
[tex]\int_{x=-3}^{-1}\int_{y= -\sqrt{2x+6}}^{\sqrt{2x+6}}xy dydx+ \int_{x=-1}^5\int_{y= x+1}^{\sqrt{2x+6}}xy dydx[/tex]
You might try doing it both ways and see if they give the same answer.

In 2 dimensions, we have only those 2 orders: dxdy and dydx. In 3 dimension we have 3!= 6 possible orders: dxdydz, dxdzdy, dydxdz, dydzdx, dzdxdy, and dzdydx. Determining the limits of integration can be quite complicated!
 
  • #5
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so x= (1/2)y2- 3. That will be the lower limit of integration.
On the line y= x- 1 so x= y+ 1. That will be the upper limit of integration.
Thats the part I don't understand. How do you know which is which, which is upper and which is lower.
 
  • #6
arildno
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The difference function between the limit functions is continuous, hence, between two zeros, the difference function must necesssarily be single-signed. Pick a y in between, and find your sign.
 
  • #7
HallsofIvy
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Better yet, actually draw the graph!
 

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