Double Intergration dydx or dxdy

eochu · #1 $Old$ November 2nd, 2009, 08:48 PM

Hi,

I am confused as to why Type 1 double integration is dydx. As in the attached image, it seems that dx strips are taken first between a and b.

However, type 1 integration is dydx thus horizontal strips (dy) are taken first and then vertical dx strips between a and b. This I do not understand.

Jameson · #2 $Old$ November 2nd, 2009, 09:16 PM

The image is correct. Type I regions have fixed x-values for boundaries and varying y-values. A Type II is the opposite, the x-values vary and fixed y-values are boundaries of the region.

Why do you say you first take horizontal strips? It seems that vertical strips are actually taken and then stretched horizontally between the bounds of x=a to x=b.

eochu · #3 $Old$ November 2nd, 2009, 09:56 PM

Quote:

Originally Posted by Jameson $View Post$

Why do you say you first take horizontal strips?

I have updated my fist post to better explain myself. Thanks for your quick reply.

Jameson · #4 $Old$ November 3rd, 2009, 01:58 AM

Quote:

Originally Posted by eochu $View Post$

I have updated my fist post to better explain myself. Thanks for your quick reply.

I think I understand you better now.

The point of an iterated integral like the examples you posted is to find the area of the region. So the end result should be a number clearly. The x-values of all these shapes fit nicely into the boundaries x=a to x=b. But the y-values are all over the place and must use variable expressions to represent them in general.

You can use either order of integration but sometimes only one way is possible and very often one way is much easier. If for these regions you went dxdy, then the inner bounds must be expressions in terms of variable y and the outer bounds must be constants. So $g_{1}(y) \le x \le g_{2}(y)$ and $a \le y \le b$ . See how difficult it is to describe the region like this?

eochu · #5 $Old$ November 3rd, 2009, 04:09 AM

Hi,

I think I understand a little more.

One thing that still confuses me: on the inside of the double integral - when I see dy on the inside I think of horizontal strips - which would make it very hard to evaluate the function because of the varying y values?

I mean, I would expect dx to be on the inside because then the vertical strips would make sense going from the higher function to the lower function.

Do you understand what I mean?

Jameson · #6 $Old$ November 3rd, 2009, 05:34 AM

Quote:

Originally Posted by eochu $View Post$

Hi,

I think I understand a little more.

One thing that still confuses me: on the inside of the double integral - when I see dy on the inside I think of horizontal strips - which would make it very hard to evaluate the function because of the varying y values?

I mean, I would expect dx to be on the inside because then the vertical strips would make sense going from the higher function to the lower function.

Do you understand what I mean?

When you take an integral of the single variable x, you write dx and the bounds pertain to moving across the x-axis.

With a function of two variables though, f(x,y), you either construct the region of integration, A, by dx then dy or dy then dx, like you've said. The point is though that this is taking place in a 3-d plane so you've got to think of terms of that.

If you fix x to be a set value, then imagine a slice in the y-z plane. You can find the area of this slice (it's standing upright though remember) by $\int_{g_1(x)}^{g_2(x)}f(x,y)dy$ Now if you sum up these vertical slices for every x, and keep letting taking thinner and thinner slices of x you will get the real volume of f(x,y) over the region.

If f(x,y)=1 then you can just view the xy-plane for simplicity but I think it's important to think of it like the above description. I don't know if that helps any but I hope this clarifies somethings.

Look at this website for visuals on this process:

Double Integrals and Volume

eochu · #7 $Old$ November 3rd, 2009, 05:34 PM

So the double integral is finding the area of a 3d shape projected onto a surface (defined by z)?

So, is the area of each of these thin slices evaluated by $\int_{g_1(x)}^{g_2(x)}f(x,y)dy$ ? Then the slices of x's are summed up over a fixed interval?

Is that correct?

Jameson · #8 $Old$ November 3rd, 2009, 06:33 PM

The double integral can be used to find the area of a region in the xy-plane or the volume of a solid over the xy-plane on the and below the z-values relating to the region.

The double integral over a region is the area of that region only when f(x,y)=1. When this is the case we just look at the xy-plane for simplicity.

When dy is the inner variable of integration, like you said we are slicing infinitely small pieces of the yz-plane, evaluating the height across the y-boundaries, then summing these areas across the x-boundaries. If you look at the image you attached, notice that these slices are vertical ones in the xy-plane, not horizontal. This is probably what led to your initial confusion.

I hope this clears things up at least somewhat.