What is the Correct Double Integral for the Given Area and Function?

[kliker]

Homework Statement

calculate the integral of f(x,y) = xy over the area that is defined from the equations below

x^2+y^2 <= 1, x>=0, y>=1/2

The Attempt at a Solution

first i tried to figure out the graph

[PLAIN]http://img571.imageshack.us/img571/9039/33928873.jpg

so i want to find the value from the black surface and z = xy right?

so i say that it is equals to

[URL]http://latex.codecogs.com/gif.latex?\int_{1/2}^{sqrt(1-x^2)}\int_{0}^{1}(xy)dxdy[/URL]

what am i doing wrong?

 

[LCKurtz]

Homework Statement

calculate the integral of f(x,y) = xy over the area that is defined from the equations below

x^2+y^2 <= 1, x>=0, y>=1/2

The Attempt at a Solution

first i tried to figure out the graph

[PLAIN]http://img571.imageshack.us/img571/9039/33928873.jpg

so i want to find the value from the black surface and z = xy right?

so i say that it is equals to

[URL]http://latex.codecogs.com/gif.latex?\int_{1/2}^{sqrt(1-x^2)}\int_{0}^{1}(xy)dxdy[/URL]

what am i doing wrong?

Several things. For one thing your outer integral must have constant limits. If you wish to integrate in the x direction first, x must go from x = 0 to x on the circle, which will be a function of y. Then the y limits will be constants and you can tell from your picture, y doesn't go from 0 to 1. What are the correct y limits?

 

[kliker]

from 0 to 1 is for x variable

i have dx first then dy

do you mean something else?

thanks in advance

Edit: yea you're right

but again if I put the second integral outside the first one inside and change dxdy to dydx i get the wrong answers

 

[kliker]

i find 1/16 which is 0.0625 but the result is 9/128 which is 0,07

how i find these results:

first i calculate the integral of xy from 1/2 to sqrt(1-x^2) dy

i find (3*x)/8-x^3/2

then i calculate the integral of (3*x)/8-x^3/2 from 0 to 1 dx and i find 1/16

 

[LCKurtz]

i find 1/16 which is 0.0625 but the result is 9/128 which is 0,07

how i find these results:

first i calculate the integral of xy from 1/2 to sqrt(1-x^2) dy

i find (3*x)/8-x^3/2

then i calculate the integral of (3*x)/8-x^3/2 from 0 to 1 dx and i find 1/16

It would be easier to follow what you are doing if you would first write out the double integral, with limits, that you are doing. If you integrate x second it does not go clear to 1 as you can see from your picture. It is probably easier to set it up integrating x first.

 

[kliker]

this is the integral

[PLAIN]http://img687.imageshack.us/img687/4043/qwe.gif
[STRIKE]
why doesn't x go from 0 to 1? I want the black part of the shape only, where x goes from 0 to 1[/STRIKE]

edit: oh...

you're right, so there is the problem... but how can I find x? hm..

Edit2:

ok found x,

y = 1/2
x^2+y^2 = 1

so x = +- sqrt(3)/2

let me check now

Edit3:

YEA! It works now :)

the final integral

[PLAIN]http://img202.imageshack.us/img202/9528/asdfs.gif

Thanks a lot for your help LCKurtz :)

 

[LCKurtz]

[PLAIN]http://img202.imageshack.us/img202/9528/asdfs.gif

Thanks a lot for your help LCKurtz :)

You're welcome. Don 't forget to change that upper limit on your first integral when you write it up; it isn't 1 either.