Another simple double integral

In summary, the conversation was about the process of integrating a function with respect to x first and then with respect to y. The speaker had some confusion about the limits of integration and the order in which to integrate. The expert clarified that there is no specific order to integrate and it can be done in whichever way is easier. Additionally, the expert suggested trying different methods to learn more about the problem.
  • #1
WrongMan
149
15

Homework Statement


So i think i got this straight since my last question. let's see :)
So my area of integration is: y=4 ; y=x2 and y=(x-2)2
the function is |x-1|
i must integrate with respect to dx first.

The Attempt at a Solution


So i sketched the area (see attatchment graphs should be cross at x=1 sorry for my bad sketch xD) and found that the upper limit for dx is x2 (so y½) and the lower is (x-2)(so y½-2)2
then noticed that since the function is the absolute value of x-1 i can divide it up into 2 functions
-x+1 (for x<1) and x-1 (for x>1) and with that in mind in my sketch i drew the line x=1 so my integrals are:
y½-21(-x+1) dx (this integral is from y½-2 to 1)
and ∫1y½(x-1)dx (this integral is from 1 to y½)
and then integrate all this with respect to y from 0 to 4.
 

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  • #2
On your dx lower limit, try recomputing it. You need to take the negative sqrt of (x-2)^2. (The x-2 for the left half of the parabola is negative.)
 
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  • #3
Charles Link said:
On your dx lower limit, try recomputing it. You need to take the negative sqrt of (x-2)^2. (The x-2 for the left half of the parabola is negative.)
oh right it is (sqrt(y) + 2) damn my distractions
that is what you meant right?
 
  • #4
WrongMan said:
oh right it is (sqrt(y) + 2) damn my distractions
that is what you meant right?
No. -sqrt(y)+2 .i.e. ## -y^{1/2}+2 ## (You also needed a +2, but you need the minus sqrt.)
 
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  • #5
Charles Link said:
No. ## -y^{1/2}+2 ##
say whaaaat? XD

so (x-2)2=y <=> x-2=y½ <=> x= y½ +2
no?
 
  • #6
WrongMan said:
say whaaaat? XD

so (x-2)2=y <=> x-2=y½ <=> x= y½ +2
no?
When you take the sqrt, there are are two solutions. You want the left one because your boundary is on the left side of this parabola. The left one is the negative sqrt. e.g. sqrt(4)=+/- 2. In this case you want "-2". When y=4, x-2=-2 ==>> x=0 is the root you are looking for.
 
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  • #7
Charles Link said:
When you take the sqrt, there are are two solutions. You want the left one because your boundary is on the left side of this parabola. The left one is the negative sqrt. e.g. sqrt(4)=+/- 2. In this case you want "-2". When y=4, x-2=-2 ==>> x=0 is the root you are looking for.
ohhhh i get it. thanks!
 
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  • #8
WrongMan said:

Homework Statement


So i think i got this straight since my last question. let's see :)
So my area of integration is: y=4 ; y=x2 and y=(x-2)2
the function is |x-1|
i must integrate with respect to dx first.

The Attempt at a Solution


So i sketched the area (see attatchment graphs should be cross at x=1 sorry for my bad sketch xD) and found that the upper limit for dx is x2 (so y½) and the lower is (x-2)(so y½-2)2
then noticed that since the function is the absolute value of x-1 i can divide it up into 2 functions
-x+1 (for x<1) and x-1 (for x>1) and with that in mind in my sketch i drew the line x=1 so my integrals are:
y½-21(-x+1) dx (this integral is from y½-2 to 1)
and ∫1y½(x-1)dx (this integral is from 1 to y½)
and then integrate all this with respect to y from 0 to 4.

Why must you integrate with respect to x first? Did somebody order you to do that? The problem is a lot easier if you integrate first with respect to y.
 
  • #9
Ray Vickson said:
Why must you integrate with respect to x first? Did somebody order you to do that? The problem is a lot easier if you integrate first with respect to y.
This is from the same set of problems from yesterday.
here is what it says:
calculate ∫∫Df(x,y) dxdy where:
then i get like 10 different values for f(x,y) and D
 
  • #10
WrongMan said:
This is from the same set of problems from yesterday.
here is what it says:
calculate ∫∫Df(x,y) dxdy where:
then i get like 10 different values for f(x,y) and D

I do not see anywhere a statement to the effect that you must integrate first with respect to x. It is perfectly legitimate to interchange the order of integration (first x, then y, or first y, then x), and the notation ##dx \, dy## has nothing to do with it. I really hope you do not think otherwise.

There may be reasons to do x first, then y, if the point of the exercise is to make you practice setting up integration limits, etc., but if the aim of the exercise is to simply compute an integral, you can use any order you want, and do whichever is easier. You might even learn something by doing the same problem two different ways.
 

1. What is a double integral?

A double integral is a type of mathematical operation that involves integrating a function of two variables over a two-dimensional region. It is often used in calculus to find the area under a curve in two dimensions.

2. How is a double integral calculated?

A double integral is calculated by first setting up limits of integration for both variables, which define the region of integration. Then, the function is evaluated at each point within this region and summed up. This sum represents the value of the double integral.

3. What is the difference between a single and a double integral?

A single integral involves integrating a function of one variable over a one-dimensional interval, while a double integral involves integrating a function of two variables over a two-dimensional region. In other words, a single integral calculates the area under a curve, while a double integral calculates the volume under a surface.

4. What are some real-world applications of double integrals?

Double integrals have many applications in physics, engineering, and other sciences. They can be used to calculate the moment of inertia of a solid object, the center of mass of a system, or the work done by a force on a moving object. They can also be used to solve problems involving fluid flow and heat transfer.

5. Are there any techniques for making double integrals easier to solve?

Yes, there are several techniques for simplifying and solving double integrals. These include changing the order of integration, using symmetry to reduce the number of integrals, and using substitution to convert the integral into a simpler form. It is also important to have a good understanding of basic integration rules and methods.

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