describe the region of integration (double integrals)

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February 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
February 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 :

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

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

Now, the region $y = \sqrt{4-x^2}$ can be expressed $y^2+x^2 = 2^2$ , which is a circle of radius 2 with its center at the original, and the region $y = 0$ 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!
February 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 $\sqrt{4- x^2}$ and 0 so you want to draw y= 0 and $\sqrt{4- x^2}$ . The region is that bounded by those three lines and one curve. If you recognize that $x^2+ y^2= 4$ as a circle of radius
2, you should see that $y= \sqrt{4- x^2}$ 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.
February 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
February 15th 2009, 08:10 AM
Krizalid

$x+y$ 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.