Double integral problem

MisterP

Problem Statement
Hello. I have this double integral I have to calculate.
Relevant Equations
Shown below

So i drew sketch.

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

Is one of them right?
Second 383,4

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BvU

Homework Helper
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...

MisterP

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)

BvU

Homework Helper
How ?
As long as interchanging integration variables does not give you the same result, please post a bit more of your work...

X as constant

BvU

Homework Helper
What is the first variable you integrate over ?
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.

BvU

Homework Helper
X as constant
Ok, so you first integrate over y. What are the bounds ?

MisterP

Bounds are limits? English is not my native..

MisterP

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

Homework Helper
Yes limits

MisterP

X limits are from -4 to 5 and Y limits are from 0 to sqrt(4+x)?

BvU

Homework Helper
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

MisterP

and 65,7 is not the right answer :D

BvU

Homework Helper
The result of integrating over y is a function of x only. Then you integrate that over x

MisterP

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)

BvU

Homework Helper
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 )

All I can conclude is: there is something wrong with the test answer

BvU

Homework Helper
By the way: #16 is an example of what I mean when I ask

Mark44

Mentor
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.

BvU

Homework Helper

SammyS

Staff Emeritus
Homework Helper
Gold Member
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.

MisterP

Yes, so it looks like it was just all times 2 :)

SammyS

Staff Emeritus
Homework Helper
Gold Member
Yes, so it looks like it was just all times 2 :)
Be Careful about such a leap.

I should have said:
No. It's definitely not as simple as multiplying by 2.

Last edited:

"Double integral problem"

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