I've been working on this homework problem for a while now and I'm just not getting it right. I'd like for some extra eyes to look at this and hint to me where I'm going wrong. The cartesian for is:
$$ \int_{\sqrt{2}}^{2} \int_{\sqrt{4-y^2}}^{y} dxdy $$
Now, the lower limit of the first integral seems to imply that $x$ is moving on the circle $x^2+y^2=4$, i.e. a circle of radius 2. Then, the movements of $x$ have a projection onto the $y$ axis of $\sqrt{2} \leq y \leq 2$. However, as I try to put this on paper. I come up with all sorts of problems.
The limits placed on $x$ seem to indicate that $x$ follows a path more like the simple function $y=x$ since it's upper limit is simply the value of $y$. This also seems to make sense because, if I take the function $f(y)=\sqrt{4-y^2}$ then $f(\sqrt{2}) = \sqrt{2}$. Thus, it would seem that this is a valid relation:
$$ \int_{\sqrt{2}}^{2} \int_{\sqrt{4-y^2}}^{y} dxdy = \int_{\sqrt{2}}^{2} \int_{\sqrt{2}}^{y} dxdy $$
Is this true?
I have tried working this problem from the standpoint that $x$ moves along the path $x=y$, but this makes no sense when trying to determine my limits for $\theta$. The reason is, the line $y=x$ is at $\frac{\pi}{4}$ and $r$ moving until it reaches a point of $y=2$ in cartesian basically leaves me with the ordered pair $(\sqrt{2},2)$. This seems to leave me with a integral of value 0. This cannot be.
So, what am I getting wrong? I come out with similar nonsense trying to use $x$ moving along the circle $x^2 + y^2 = 4$ too.
Andy
This is your region of integration: the important thing is that $$ \sqrt{1-y^2}<x<y, $$ which is not true for your second integral, which is over a whole rectangle. (Using inequalities like this is one of the easier ways of understanding multiple integral regions: rearranging them tells you how to reparametrise much more easily.)
(Really, you don't want to use polar coordinates for this one: the top line $y=2$ is harder to parametrise)