# Multiple integral

1. Aug 10, 2008

### asi123

1. The problem statement, all variables and given/known data

Hey.
Which one is correct, the right one or the left one?
I think it's the right one, but I'm not sure...

2. Relevant equations

3. The attempt at a solution

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2. Aug 11, 2008

### dynamicsolo

When you choose to change the order of integration in this way, turn the graph (or your head) 90º to see which curve is now "on top" and "on bottom" of the region. The y = x^3 [ now x = y^(1/3) ] becomes the "upper" curve for integration along the y-axis.

However, I believe you do not have these integrals set up right. Shouldn't they be

$$\int^{1}_{0} \int^{x^2}_{x^3} F(x,y) dx \, dy$$

and

$$\int^{1}_{0} \int^{y^{1/3} }_{y^{1/2} } F(x,y) dy \, dx$$ ?

If you were only integrating infinitesimal two-dimensional elements of the area within the bounded region to evaluate the total area, that is the order the differentials would be written in (with F(x,y) = 1).

Last edited: Aug 11, 2008
3. Aug 11, 2008

### kidmode01

So I'm guessing F(x,y) is your density function? I know in statistics it is common to use a lowercase f(x,y) to represent your density function and a capital F(x,y) to represent your distribution function but I'll keep your notation.

In your top integral you are doing:
$$\int \int F(x,y)dydx$$
where you will be fixing a arbitrary value of x.

So from the x-axis up you will be hitting your x^3 function first and then your x^2 function so your bounds are correct there.

In your bottom integral you are doing:
$$\int \int F(x,y)dxdy$$
where you are fixing an arbitrary value of y

and from the y-axis and to the right you will hit your x^2 function first and then your x^3 function.
So your thinking is correct I believe, someone will come along and correct me if I'm wrong hopefully haha.