Integrating with Double Limits: X or Y? Calculating the Correct Integral

MisterP
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Homework Statement
Hello. I have this double integral I have to calculate.
Relevant Equations
Shown below
242015

So i drew sketch.

242016


And I do not understand, how to write integral for calculation, which I should use, X or Y on limit?
242017

Is one of them right?
First answer gives me 65,7
Second 383,4
 
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Hi P,

For which variable are the limits you write ?
What is the first variable you integrate over ?

As long as interchanging integration variables does not give you the same result, please post a bit more of your work...:rolleyes:
 
I wrote that integral for area I need to calculate.. the are between functions x=y^2-4; y = 0 and x=5

First, I integrate (x+2y)
 
How ?
BvU said:
As long as interchanging integration variables does not give you the same result, please post a bit more of your work...:rolleyes:
 
X as constant
 
BvU said:
What is the first variable you integrate over ?
MisterP said:
First, I integrate (x+2y)
x+2y is the integrand (the function to integrate)

You have to integate over an area. The shaded area in your picture
You can choose to integrate first over x and then over y, or vice versa.
 
MisterP said:
X as constant
Ok, so you first integrate over y. What are the bounds ?
 
Bounds are limits? English is not my native..
 
I know how to integrate, I just got confused on which should be as a limit, x = or y = . I have test closing in 15min
 
  • #10
Yes limits
 
  • #11
X limits are from -4 to 5 and Y limits are from 0 to sqrt(4+x)?
 
  • #12
Yes. To answer your original question: the second is correct. For an integration over y you need a limit that is a function of x
 
  • #13
and 65,7 is not the right answer :D
 
  • #14
The result of integrating over y is a function of x only. Then you integrate that over x
 
  • #15
I do not get it, why is 65,7 wrong?
I got the same result as calculator and it says it is wrong (online test)
 
  • #16
Beats me. I find ##\ (xy+y^2) \Big |_0^\sqrt{x+4} = x\sqrt{x+4}+x+4 \ ## Integrate that from -4 to 5 (I use a calculator too) gets same answer as you.

Now the hammer: If we do the integration over x first, the limits are ##y^2-4## and 5

I find ##\ ({1\over 2}x^2+2xy) \Big |_{y^2-4}^5 = 12.5 + 10 y - {1\over 2} (y^2-4)^2 - 2y (y^2-4) \ ##
Integrate that from 0 to 3 using the tool again gets me the same answer again.

(after I found and fixed my own mistake o:) ) :smile:

All I can conclude is: there is something wrong with the test answer
 
  • #17
By the way: #16 is an example of what I mean when I ask
BvU said:
please post a bit more of your work...:rolleyes:
 
  • #18
MisterP said:
X limits are from -4 to 5 and Y limits are from 0 to sqrt(4+x)?
There is no indication in post #1 that ##y \ge 0##. Based on that post, the region you are integrating over should extend down below the x-axis.
 
  • Like
Likes BvU
  • #19
o:)o:)o:)
 
  • #20
Mark44 said:
There is no indication in post #1 that ##y \ge 0##. Based on that post, the region you are integrating over should extend down below the x-axis.
Like this.
242024


242025
 
  • #21
Yes, so it looks like it was just all times 2 :)
 
  • #22
MisterP said:
Yes, so it looks like it was just all times 2 :)
Be Careful about such a leap.

Added in Edit:
I should have said:
No. It's definitely not as simple as multiplying by 2.
 
Last edited:
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