# describe the region of integration (double integrals)

• Feb 14th 2009, 06:06 AM
James0502
describe the region of integration (double integrals)
describe the region of integration of the folowing integrals

int(between 0 and 2) int (between sqrt(4-x^2) and 0) (x^2 + y^2) dy dx

int(2 and -1) int(x+2 and x^2) (x+y) dy dx

how would I go about describing integrals like this?

sorry if it appears I havent thought about this - I have - I just have no idea how to think about it!

many thanks
• Feb 14th 2009, 06:22 AM
Mush
Quote:

Originally Posted by James0502
describe the region of integration of the folowing integrals

int(between 0 and 2) int (between sqrt(4-x^2) and 0) (x^2 + y^2) dy dx

int(2 and -1) int(x+2 and x^2) (x+y) dy dx

how would I go about describing integrals like this?

sorry if it appears I havent thought about this - I have - I just have no idea how to think about it!

many thanks

The region is defined by the limits of the integral.

The first integral is :

$\displaystyle \int_0^2 \int_{\sqrt{4-x^2}}^{0} (x^2+y^2)dydx$

The y limits state that $\displaystyle \sqrt{4-x^2} < y < 0$

Now, the region $\displaystyle y = \sqrt{4-x^2}$ can be expressed $\displaystyle y^2+x^2 = 2^2$, which is a circle of radius 2 with its center at the original, and the region $\displaystyle y = 0$ is a straight line in line with the x axis! So in the y axis, the region is bounded between the x axis, and the circle described.

And the x axis limits are x = 0 and x = 2. So I'm sure you can understand what this means!
• Feb 14th 2009, 08:29 AM
HallsofIvy
Quote:

Originally Posted by James0502
describe the region of integration of the folowing integrals

int(between 0 and 2) int (between sqrt(4-x^2) and 0) (x^2 + y^2) dy dx

int(2 and -1) int(x+2 and x^2) (x+y) dy dx

how would I go about describing integrals like this?

sorry if it appears I havent thought about this - I have - I just have no idea how to think about it!

many thanks

That says that x must lie between 0 and 2. On a graph, draw vertical lines x= 0 and x= 2 to bound that region. For every x, y must be between $\displaystyle \sqrt{4- x^2}$ and 0 so you want to draw y= 0 and $\displaystyle \sqrt{4- x^2}$. The region is that bounded by those three lines and one curve. If you recognize that $\displaystyle x^2+ y^2= 4$ as a circle of radius
2, you should see that $\displaystyle y= \sqrt{4- x^2}$ is, since the square root is positive, the upper half of that circle. Since x lies between 0 and 2, this is the fourth of the circle in the first quadrant.
• Feb 15th 2009, 07:49 AM
James0502
yes, thankyou all for the help - I have the left half of a circle - is this correct?

I'm struggling with the second one though.. I have drawn the limits but I cant see how the x + y relates?

many thanks
• Feb 15th 2009, 08:10 AM
Krizalid
$\displaystyle x+y$ is just what you have to integrate that's all.

You just need to graph where the double integral is taken and that's all, it's quite easy, not hard.