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Changing limits of integration in double integral

  • Thread starter Tosh5457
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  • #1
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Homework Statement



Invert the limits of integration of the following integrals:

[tex]
1 ) \int_{0}^{4} dx \int_{0}^{x} f(x,y)dy
\int_{0}^{2} dx \int_{0}^{\surd (4 - x^2)} f(x,y)dy
\int_{0}^{1} dy \int_{y}^{2-y} f(x,y)dx
[/tex]

These are 3 different integrals in 3 separate exercises, they're not multiplying. I don't know how to put new lines in latex... Tried \linebreak, \\, nothing worked...

Homework Equations




The Attempt at a Solution



I tried to understand it and I even asked my professor but I didn't get it how to do this...
 
Last edited:

Answers and Replies

  • #2
tiny-tim
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Hi Tosh5457! :smile:

(just close and open the tex to start a new line :wink:)

Let's start with the first one …

always find the shape first …

the shape is a triangle: 0 ≤ x ≤ 4 and y ≤ x …

the other way round it's 0 ≤ y ≤ 4 and … ? :smile:

(and what shapes are the other two?)​
 
  • #3
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28
Hi Tosh5457! :smile:

(just close and open the tex to start a new line :wink:)

Let's start with the first one …

always find the shape first …

the shape is a triangle: 0 ≤ x ≤ 4 and y ≤ x …

the other way round it's 0 ≤ y ≤ 4 and … ? :smile:

(and what shapes are the other two?)​
The first one I know it's 0 ≤ y ≤ 4 and 4 ≤ x ≤ y because I saw the solution, but why can't it be 0 ≤ x ≤ y?
I know this shape is a triangle, but for example if 0 ≤ x ≤ 4 and y ≤ x, when x = 3 y ≤ 3 which is a square. What am I doing wrong?

I can't do the others yet, sorry... I'm really missing something here and I don't know what it is...
 
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  • #4
tiny-tim
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The first one I know it's 0 ≤ y ≤ 4 and 4 ≤ x ≤ y because I saw the solution, but why can't it be 0 ≤ x ≤ y?
No, you've copied this wrong, it's 4 ≥ x ≥ y, isn't it? :redface:

And it can't be 0 ≤ x ≤ y because that contradicts the original y ≤ x. :wink:
And shouldn't this shape be a square? If 0 ≤ x ≤ 4 and y ≤ x, when for example x = 3, y ≤ 3, which is a square.
no …

for x = 3, y ≤ 3,

for x = 4, y ≤ 4,

for x = 2, y ≤ 2,

for x = 1, y ≤ 1 …​

if you draw all those vertical lines you get a triangle.

To put it another way, this is the intersection of two regions …

0 ≤ x ≤ 4 is a vertical strip

0 ≤ y ≤ x is the wedge shape between the positive x-axis and the diagonal line y = x

so the intersection is where the vertical strip cuts the wedge, ie a triangle.

I suggest (in all three cases) you pick a few x-values and draw vertical lines for those x=values to show the maximum extent of y … that'll give you a shading from which you can see the outline of the whole shape.

Alternatively, rewrite the y limit (with an = not a ≤) so that it looks like an equation that you recognise (like y = x is a line).

What shape do you get for the second one? :smile:
 
  • #5
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No, you've copied this wrong, it's 4 ≥ x ≥ y, isn't it?
Yes you're right, I copied it wrong :redface:

if you draw all those vertical lines you get a triangle.
I see, I'd have to draw infinite vertical lines, right?

I usually have no problem seeing the shapes, like you said I draw the functions' graphs and then see the inequations to see what's the area.

So for the others:

2) [tex]\int_{0}^{2} dx \int_{0}^{\surd (4 - x^2)} f(x,y)dy[/tex]

0 ≤ y ≤ 2
0 ≤ x ≤ [tex]\surd (4 - x^2)[/tex]

3) [tex]\int_{0}^{1} dy \int_{y}^{2-y} f(x,y)dx[/tex]

I divided it into 2 regions. For the 1st one:

0 ≤ x ≤ 1
0 ≤ y ≤ x

and the 2nd one:

1 ≤ x ≤ 2
0 ≤ y ≤ x -2

which means I'd have to do 2 double integrals for each domain.

I think I found the "trick" to do this, but isn't it possible to do it just using the initial inequations, without even looking to the functions graph?
 
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  • #6
tiny-tim
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Hi Tosh5457! :smile:
I see, I'd have to draw infinite vertical lines, right?
No, they're short vertical lines from the x-axis to the line x = y. :wink:
So for the others:

I divided it into 2 regions. For the 1st one:

which means I'd have to do 2 double integrals for each domain.
Yes, those are all fine :smile:

(except where you mistyped x2 for y2)
I think I found the "trick" to do this, but isn't it possible to do it just using the initial inequations, without even looking to the functions graph?
Yes it is, but it's so easy to make a mistake (particularly in an exam) that it's best to draw the graph anyway. :wink:
 
  • #7
Ray Vickson
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Yes you're right, I copied it wrong :redface:



I see, I'd have to draw infinite vertical lines, right?

I usually have no problem seeing the shapes, like you said I draw the functions' graphs and then see the inequations to see what's the area.

So for the others:

2) [tex]\int_{0}^{2} dx \int_{0}^{\surd (4 - x^2)} f(x,y)dy[/tex]

0 ≤ y ≤ 2
0 ≤ x ≤ [tex]\surd (4 - x^2)[/tex]

3) [tex]\int_{0}^{1} dy \int_{y}^{2-y} f(x,y)dx[/tex]

I divided it into 2 regions. For the 1st one:

0 ≤ x ≤ 1
0 ≤ y ≤ x

and the 2nd one:

1 ≤ x ≤ 2
0 ≤ y ≤ x -2

which means I'd have to do 2 double integrals for each domain.

I think I found the "trick" to do this, but isn't it possible to do it just using the initial inequations, without even looking to the functions graph?
Perhaps your notation is confusing you. In 2), the dx-integral is "outer" (done last) while the dy-integral is "inner" (done first). So, the dy integration is done before we even get around to integrating with respect to x. What are the limits on the dy-integration (they may depend on x). If you now imagine that you have done the dy-integral, you will now have a function of x alone. What are its integration limits?

Note: you could write, instead:
[tex] \displaystyle
\int_{0}^{2} \{ \int_{0}^{\surd (4 - x^2)} f(x,y)dy \: \} \, dx,
[/tex]
which would make the integration order clearer. It is OK to write it as you did, but only if you understand the ordering conventions involved.

RGV
 

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