Integrating a Quarter Circle with a Double Integral

[8614smith]

Homework Statement

$$I=\int^{a}_{0}dx\int^{\sqrt{a^{2}-x^{2}}_{0}$$$$(x-y)dy$$

Homework Equations

$$r^{2}=x^{2}+y^{2}$$

The Attempt at a Solution

Im thinking that the question is asking to integrate the first quarter of the circle of radius a between 0 and pi/2. In that case I've changed the limits to :-
$$\int^{\frac{\pi}{2}}_{0}\int^{a}_{0}r.dr.d\theta$$ but now I am not sure what to do with the (x-y) given in the question. I can't seem to be able to rearrange this - $$r^{2}=x^{2}+y^{2}$$ to give me a value i can use in polar coordinates.

This integral apparently is supposed to give an answer of 0. is this correct? i don't see how it can be zero if there is a limit of a.

 

[8614smith]

that top integral bottom limit should be zero, and the equation is (x-y). I must have messed up the latex code

 

[Hurkyl]

x and y are, well, x and y. However, x and y are algebraically related to r and theta, and you may find it useful to do a conversion.


While it would be good practice to do so, a naive brute force calculation is not the best way to solve this problem -- the solution is very, very simple if you make the right observation about what this integral is actually calculating. You've already made half of the observation, or nearly so -- you've identified geometrically the region of integration.

 

[8614smith]

i'm sure its really obvious, but i can't see the relation between (x-y) and r or theta, i don't see how an r can be substituted in if $$r^2=x^2+y^2$$ i can't make it fit. Are you also suggesting that i substitute the a for something else?

 

[Hurkyl]

You knew one relationship between (x,y) and (r,theta): r²=x²+y². Where did that relationship come from?