# Double integral

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• Jan 31st 2011, 10:15 AM
dervast
Double integral
Dear all some naive question.

If I have a double integral over a x,y area.

Usually this process is like this
I calculate the integral for x and plugging in the two limits of x and then I feed this result
and I calculate the integral for y and I plug in the two limits of y.

This is the process explained here:
YouTube - Calculating a Double Integral (5th minute).

IS it possible to calculate the integral of x.. the integral of y and then plug in together the x and y limits to find the number I am looking for?

IF this is not can you show me a simple example that depicts why this is not possible?

Best Regards
Alex
• Jan 31st 2011, 10:24 AM
mr fantastic
Quote:

Originally Posted by dervast
Dear all some naive question.

If I have a double integral over a x,y area.

Usually this process is like this
I calculate the integral for x and plugging in the two limits of x and then I feed this result
and I calculate the integral for y and I plug in the two limits of y.

This is the process explained here:
YouTube - Calculating a Double Integral (5th minute).

IS it possible to calculate the integral of x.. the integral of y and then plug in together the x and y limits to find the number I am looking for? Mr F says: No.

IF this is not can you show me a simple example that depicts why this is not possible? Mr F says: Pick a question form your textbook and try doing it your way. Post your working here for review.

Best Regards
Alex

..
• Jan 31st 2011, 12:42 PM
AllanCuz
Quote:

Originally Posted by dervast
Dear all some naive question.

If I have a double integral over a x,y area.

Usually this process is like this
I calculate the integral for x and plugging in the two limits of x and then I feed this result
and I calculate the integral for y and I plug in the two limits of y.

This is the process explained here:
YouTube - Calculating a Double Integral (5th minute).

IS it possible to calculate the integral of x.. the integral of y and then plug in together the x and y limits to find the number I am looking for?

IF this is not can you show me a simple example that depicts why this is not possible?

Best Regards
Alex

You are asking if we can take,

$A = \int_a^b dx \int_{ g(x) }^{ h(x) } f(x,y) dy$

And make it,

$\int_a^b dx = P$ , $\int_{ g(x) }^{ h(x) } f(x,y) dy = G$

$\to A = P + G$

??

I can see a few problems with this. First, the dx integral depends on the function produced from the dy integral. Second, you would appear to be missing the whole reason for the double integral! We are essentially sweeping some function across a finite interval [a,b] where the finite interval is on the perpendicular axis to our function. But that function or the height of the sweeping arm depends on where it takes place on the interval [a,b]. We of course represent this via integration to obtain an area (or volume). We cannot treat these by themselves. They are necessarily dependent.
• Jan 31st 2011, 01:14 PM
TriKri
Sure it is possible! Usually (although not always) when integrating a function f(x), it is possible to find a primitive function F(x). Then the integral of f from $x_1$ to $x_2$ will be the difference $F(x_2) - F(x_1)$.

Say that you instead have a function $f(x,\,y)$ that you want to integrate two times, one time with respect to x and then a second time with respect to y. After integrating with x you will get a primitive function, let's call that one $F_x,$ which is a function of x and y. The integral of f between $x_1$ and $x_2$ will be $F_x(x_2,\,y) - F_x(x_1,\,y)$. Now you integrate this difference with respect to y. The function $F_x$ appears two times in the expression, using two different x values, but integrate it once for an arbitrary value of x to get a primitive function $F$ of $F_x$, which only depends on x and y (instead of a function depending on $x_1$, $x_2$ and y). Since F is the primitive function of $F_x$, the primitive function of $F_x(x_2,\,y) - F_x(x_1,\,y)$ will be $F(x_2,\,y) - F(x_1,\,y)$. Now, this is the moment when you insert your limits for y, to get the integral between $y_1$ and $y_2$. After doing that, your expression will look like

$(F(x_2,\,y_2) - F(x_1,\,y_2)) - (F(x_2,\,y_1) - F(x_1,\,y_1)) =$

$= F(x_2,\,y_2) - F(x_1,\,y_2) - F(x_2,\,y_1) + F(x_1,\,y_1)$

I guess this is the expression you are looking for. Note however that this only works when you have a rectangular surface in the xy-plane with sides parallel to the x and y axes, over which you want to integrate. For other shapes, you will have to find other expressions.
• Feb 1st 2011, 12:37 AM
dervast
Quote:

Originally Posted by mr fantastic
..

Thanks a lot I will pick an example and I ll try to find out... My questions is more to build understanding why this is not possible... So a simple example my let me figure out why.
• Feb 1st 2011, 12:53 AM
TriKri
I just showed you that it is possible, or is there anything I have misunderstodd?
• Feb 1st 2011, 02:02 AM
mr fantastic
Quote:

Originally Posted by TriKri
I just showed you that it is possible, or is there anything I have misunderstodd?

I think the OP might have been wondering if it's always possible. Clearly there will be some special cases where it is, but in general it won't be. I'm sure the OP will discover this by working through a simple example.
• Feb 1st 2011, 02:09 AM
dervast
Quote:

Originally Posted by TriKri
I just showed you that it is possible, or is there anything I have misunderstodd?

So it works but only for rectangular areas so If I have some polynomials of order greater than 1 (2,3,4 or more) it does not work as my function does not create any more a rectangular surface.

Best Regards
Alex
• Feb 1st 2011, 02:28 AM
dervast
Quote:

Originally Posted by TriKri
Sure it is possible! Usually (although not always) when integrating a function f(x), it is possible to find a primitive function F(x). Then the integral of f from $x_1$ to $x_2$ will be the difference $F(x_2) - F(x_1)$.

Say that you instead have a function $f(x,\,y)$ that you want to integrate two times, one time with respect to x and then a second time with respect to y. After integrating with x you will get a primitive function, let's call that one $F_x,$ which is a function of x and y. The integral of f between $x_1$ and $x_2$ will be $F_x(x_2,\,y) - F_x(x_1,\,y)$. Now you integrate this difference with respect to y. The function $F_x$ appears two times in the expression, using two different x values, but integrate it once for an arbitrary value of x to get a primitive function $F$ of $F_x$, which only depends on x and y (instead of a function depending on $x_1$, $x_2$ and y). Since F is the primitive function of $F_x$, the primitive function of $F_x(x_2,\,y) - F_x(x_1,\,y)$ will be $F(x_2,\,y) - F(x_1,\,y)$. Now, this is the moment when you insert your limits for y, to get the integral between $y_1$ and $y_2$. After doing that, your expression will look like

$(F(x_2,\,y_2) - F(x_1,\,y_2)) - (F(x_2,\,y_1) - F(x_1,\,y_1)) =$

$= F(x_2,\,y_2) - F(x_1,\,y_2) - F(x_2,\,y_1) + F(x_1,\,y_1)$

I guess this is the expression you are looking for. Note however that this only works when you have a rectangular surface in the xy-plane with sides parallel to the x and y axes, over which you want to integrate. For other shapes, you will have to find other expressions.

I just read it carefully and I think I found my mistake... I was thinking that at the end I could something like

$= F(x_2,\,y_2) - F(x_1,\,y_1)$ converting directly from the one-dimension case $= F(x_2)- F(x_1)$ but your example clearly depicts that my reasoning was totaly wrong.

So it seems that it might be easier to stick to the simple rule as the video I posted above depicts

Thank you very much both

Best Regards
Alex
• Feb 1st 2011, 06:12 AM
dervast
One more question. How do I calculate an indefinite double integral dxdy? I first find dx and then I feed it to dy?

Best Regards
Alex
• Feb 1st 2011, 06:38 AM
Prove It
There's no such thing as an indefinite double integral...
• Feb 1st 2011, 07:06 AM
dervast
Quote:

Originally Posted by Prove It
There's no such thing as an indefinite double integral...

But there are:
indifite one-dimension integral
definitie one-dimension integral
definite two-dimension integral
right

and there is not indefinite two-dimensional itengral
• Feb 1st 2011, 11:26 AM
TriKri
Okay, don't worry about missing my post.

I think that an indefinite integral is one that has no given limits, and therefore results in a primitive function rather than in a value of the integral, right? In that case, I guess the function F of x and y in my example is the closest you will get to an indefinite two dimensional integral, if there is such a thing. I have never seen it in use myself what I can remember.
• Feb 1st 2011, 10:12 PM
dervast
But as you also commented before your F x and y equation only applies in a rectangle area... I think that there can be no definitie multi-dimensional integral as there is no possiblity to find the antiderivative of such a function. Just a guess
• Feb 2nd 2011, 05:06 AM
TriKri
I guess you mean indefinite multi-dimensional integral. Yes, this was only for the rectangular case, you are right. An arbitrary surface can be varied in an infinite numbers of ways (it can be extended in an infinite number of edge points), hence I believe that it's difficult to find a primitive function from which an integral over an arbitrary surface easily can be expressed, withouth using at least one more integral. And of course it doesn't become easier to find a primitive function which makes any sense when the dimension of the integral increases even more.
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