Why is this a type II double integral?

[Avatrin]

∫∫$_{A}$xy$^{2}$dxdy

A is the area between y = x^2, y = 2-x and x$\geq$0.

I am told that this is a type II double integral and I thus have to:

∫$^{1}_{0}$∫$^{2-y}_{√y}$xy$^{2}$dxdy

But, why can't I do this?

∫$^{2}_{0}$∫$^{2-x}_{x^2}$xy$^{2}$dydx

 

[Millennial]

Did you hear of switching the order of integration?

 

[mathman]

∫∫$_{A}$xy$^{2}$dxdy

A is the area between y = x^2, y = 2-x and x$\geq$0.

I am told that this is a type II double integral and I thus have to:

∫$^{1}_{0}$∫$^{2-y}_{√y}$xy$^{2}$dxdy

But, why can't I do this?

∫$^{2}_{0}$∫$^{2-x}_{x^2}$xy$^{2}$dydx

The second integral should have (0,1) as the range for x, since the curves cross at x = 1.

 

[Avatrin]

It still is not the right answer. The solution to the equation you suggest is 47/120. The correct solution is 17/120 (i.e. the solution to the type II double integral)

 

[HallsofIvy]

∫∫$_{A}$xy$^{2}$dxdy

A is the area between y = x^2, y = 2-x and x$\geq$0.

I am told that this is a type II double integral and I thus have to:

∫$^{1}_{0}$∫$^{2-y}_{√y}$xy$^{2}$dxdy

Who told you this? That is clearly incorrect. For y between 0 and 1, x ranges between 0 and $\sqrt{y}$. For y between 1 and 2, x ranges between 0 and 2- y. Integrating in that order, the area has to be done as two separate integrals:
$$\int_{y=0}^1\int_{x= 0}^{\sqrt{y}} xy^2dxdy+ \int_{y=1}^2\int_{x= 0}^{2-y} xy^2 dxdy$$

Why can't I do this?

∫$^{2}_{0}$∫$^{2-x}_{x^2}$xy$^{2}$dydx

You can. It is the first integral that is incorrect.

 

[Avatrin]

Who told you this? That is clearly incorrect.

Well, it is an example in my book. The text is in Norwegian, but you will get the math:
http://img805.imageshack.us/img805/8316/eksempel2kap6.png

 

[algebrat]

∫∫$_{A}$xy$^{2}$dxdy

A is the area between y = x^2, y = 2-x and x$\geq$0.

I am told that this is a type II double integral and I thus have to:

∫$^{1}_{0}$∫$^{2-y}_{√y}$xy$^{2}$dxdy

But, why can't I do this?

∫$^{2}_{0}$∫$^{2-x}_{x^2}$xy$^{2}$dydx

Mathman and Halls of Ivy are both wrong.

Here's how your second integral needs to be corrected:

$$\int_{x=0}^1\int_{y=0}^{x^2}xy^2\ dydx+\int_{x=0}^1\int_{y=0}^{2-x}xy^2\ dydx$$

Of course that is easier to see with that picture above, but it goes to show you, even experts can make mistake with limits of integration for multidimensional regions. I've seen some that get really tricky in 3 dimensional regions. For instance, after you have the three constraints y=2-x, y=x^2 and x=0, you have to be careful that you pick the region in the plane that is "bounded" by these. What I mean is, there are actually 8 regions, created by the 3 lines (find them). Two of them are "bounded", by that I mean, we can't head infinitely far away. Can you find the two bounded regions (out of the eight total). But only one of them has a side or face of x=0.

My guess, Mathman jumped to the conclusion that we should stop at something corresponding to a point of intersection. But just using algebra to figure out the limits is very tricky, if possible, and one should definitely draw them to get a better and/or faster answer.

As for Halls of Ivy, it looks as if they chose the wrong bounded region, and may have in effect assumed that y=0 was an edge as well.

You may have preferred the type I because you wanted to avoid the root, but now you have to divide your integral into two separate regions. And the type II won in the end because that root never survived long, polynomials are so easy to integrate hooray!