# Calculating regions of double integrals

1. May 2, 2012

### NewtonianAlch

1. The problem statement, all variables and given/known data
Evaluate the follow by first changing the order of integration

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

3. The attempt at a solution

This is the region we're concerned with:

http://www.wolframalpha.com/input/?i=plot%28y%3Dx^2%2C+y+%3D+2+-+x^2%2C+x%3D+1%2C+x%3D+-1%29

The new inequalities would be:

$0 ≤ y ≤ 2$

$\sqrt {-y+2} ≤ x ≤ √y$

$$\int_{y=0}^{2}\int_{x=sqrt(-y + 2)}^{sqrt(y)}dxdy$$

Doing this double integral gives me a final answer of zero which I checked on MAPLE. The correct answer is 8/3

I'm guessing the bounds are incorrect, but I can't figure out why. On the plot y clearly has a lower limit of 0 and an upper limit of 2. The limits of x I just manipulated through the previous inequality.

2. May 2, 2012

### sweet springs

Hi. Calculate

$$\int_{y=x^2}^{2-x^2}dy$$

first. Then integrate in x.

Regards.

3. May 2, 2012

### NewtonianAlch

Yea, but I think they want me to do it the other way, it's a bit futile, since changing the order complicates things here, but I think it's just for practice to get an understanding of both ways.

4. May 2, 2012

### The Gringo

Look at your integral again - particularly the y limits of integration. You cannot do a change of variables with just one integral in this case.

5. May 2, 2012

### sharks

To change the order of integration, basically you need this integral:
$$\iint dxdy$$
Now, to find the limits of this new double integral, split the region vertically in half and describe each half to determine the limits:

For the upper quarter region:
For y fixed, x varies from $x=\sqrt{2-y}$ to x=0
y varies from y=1 to y=2

For the lower quarter region:
For y fixed, x varies from $x=\sqrt{y}$ to x=0
y varies from y= 0 to y=1

Hence the entire area of the region can be found by this double integral:
$$I=2 \left( \int^2_1 \int^0_{\sqrt{2-y}} dxdy + \int^1_0 \int^0_{\sqrt{y}} dxdy \right)=2\int^2_1 \int^0_{\sqrt{2-y}} dxdy + 2\int^1_0 \int^0_{\sqrt{y}} dxdy$$

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Last edited: May 2, 2012
6. May 2, 2012

### NewtonianAlch

That was very insightful thanks for that.

Although, for the lower quarter region do you mean y varies from y = 1 to y = 0?

7. May 2, 2012

### sharks

You should always take the integral limits in the positive direction of the axes.

If you calculate from y=1 to y=0 and then add up, you will get zero as the sum of the area of the regions, since the area of the upper quarter region will cancel out with the (negative) area of the lower quarter region.

8. May 2, 2012

### SammyS

Staff Emeritus
If you solve $\displaystyle y=x^2$ for x you get $\displaystyle x=\pm\sqrt{y\ }\ .$

If you solve $\displaystyle y=2-x^2$ for x you get $\displaystyle x=\pm\sqrt{2-y\ }\ .$

9. May 3, 2012

### sharks

Indeed, the integral could also be solved more directly, without splitting the upper and lower regions into quarters, but the limits can be a little confusing if you don't know which half of the curve corresponds to +ve square root and which half corresponds to the -ve square root.

For the upper half region:
For y fixed, x varies from x=√(2-y) to x=-√(2-y)
y varies from y=1 to y=2

For the lower half region:
For y fixed, x varies from x=-√y to x=√y
y varies from y=0 to y=1

Hence the entire area of the region can be found by this double integral:
$$I= \int^2_1 \int^{-\sqrt{2-y}}_{\sqrt{2-y}} dxdy + \int^1_0 \int^{\sqrt y}_{-\sqrt y} dxdy$$

Maybe someone can confirm this? I based the limits on the negative signs affecting the gradients of each half of the individual curves.

Last edited: May 3, 2012
10. May 3, 2012

### SammyS

Staff Emeritus
To find an area (which in this case should be positive), the order of the limits of integration should be from a lesser value to a greater value,

11. May 3, 2012

### sharks

So, this is the correct integral for the total area:
$$I= \int^2_1 \int^{\sqrt{2-y}}_{-\sqrt{2-y}} dxdy + \int^1_0 \int^{\sqrt y}_{-\sqrt y} dxdy$$

Last edited: May 3, 2012
12. May 3, 2012

### HallsofIvy

Staff Emeritus
Yes, that is correct.