1. ## 2-dimensional integral

Region A is the area between the unit circle and the circle of radius 2, in the top right quadrant only. Find

$\displaystyle \int_A{xy}~dA$

Ok so that's the question. We're doing the theory behind changing variables from x,y to u,v or r,θ.

In standard x,y coordinates:

$\displaystyle \int_0^2\int_{\sqrt(1-y^2)}^{\sqrt(4-y^2)}{xy}~dx~dy$

which comes out as 3. I'd write it all out but I'm not fluent in Latex and it would take hours.

In r,θ:

$\displaystyle \int_1^2\int_0^{\pi/2}{sin\theta cos\theta r^3}~dx~dy$ which comes out as 15/8. I'm pretty sure the second one is the one intended to be correct, but not sure what I'm missing. Perhaps because the x,y integral doesn't limit to the top right quadrant? I just can't work it out.

2. Originally Posted by Thomas154321
Region A is the area between the unit circle and the circle of radius 2, in the top right quadrant only. Find

$\displaystyle \int_A{xy}~dA$

Ok so that's the question. We're doing the theory behind changing variables from x,y to u,v or r,θ.

In standard x,y coordinates:

$\displaystyle \int_0^2\int_{\sqrt(1-y^2)}^{\sqrt(4-y^2)}{xy}~dx~dy$

which comes out as 3. I'd write it all out but I'm not fluent in Latex and it would take hours.

In r,θ:

$\displaystyle \int_1^2\int_0^{\pi/2}{sin\theta cos\theta r^3}~dx~dy$ which comes out as 15/8. I'm pretty sure the second one is the one intended to be correct, but not sure what I'm missing. Perhaps because the x,y integral doesn't limit to the top right quadrant? I just can't work it out.
Perhaps it would be best to work out the area enclosed by the circle of radius 2 and the coordinate axes, then subtract the area enclosed by the 1 radius circle and the axes. I believe that will yield the correct answer.

So:

$\displaystyle \int_0^2 \int_0^{\sqrt{4-y^2}} xy dx dy$

$\displaystyle \int_0^1 \int_0^{\sqrt{1-y^2}} xy dx dy$

$\displaystyle \text{Total Area} = \int_0^2 \int_0^{\sqrt{4-y^2}} xy dx dy-\int_0^1 \int_0^{\sqrt{1-y^2}} xy dx dy$

3. Originally Posted by Thomas154321
Region A is the area between the unit circle and the circle of radius 2, in the top right quadrant only. Find

$\displaystyle \int_A{xy}~dA$

Ok so that's the question. We're doing the theory behind changing variables from x,y to u,v or r,θ.

In standard x,y coordinates:

$\displaystyle \int_0^2\int_{\sqrt(1-y^2)}^{\sqrt(4-y^2)}{xy}~dx~dy$

which comes out as 3. I'd write it all out but I'm not fluent in Latex and it would take hours.

In r,θ:

$\displaystyle \int_1^2\int_0^{\pi/2}{sin\theta cos\theta r^3}~dx~dy$ which comes out as 15/8. I'm pretty sure the second one is the one intended to be correct, but not sure what I'm missing. Perhaps because the x,y integral doesn't limit to the top right quadrant? I just can't work it out.
You need to switch $\displaystyle dx\;dy\; = \; r \; dr\; d\theta$ (you have the r)

so

$\displaystyle \int_0^{\pi/2} \int_1^2 \; r sin\theta \;r cos\theta \;r \; dr\; d\theta = \frac{15}{8}$

4. I did that, hence my $\displaystyle r^3$. I'm more confident that my polar integral is correct than the x,y one.

5. Originally Posted by danny arrigo
You need to switch $\displaystyle dx\;dy\; = \; r \; dr\; d\theta$ (you have the r)

so

$\displaystyle \int_0^{\pi/2} \int_1^2 \; r sin\theta \;r cos\theta \;r \; dr\; d\theta = \frac{15}{8}$
The OP already did this. The 2nd integral was not the problem. It yielded the correct answer. The 1st integral was the anomaly.

6. Originally Posted by Thomas154321
I did that, hence my $\displaystyle r^3$. I'm more confident that my polar integral is correct than the x,y one.
Sorry. To do your problem in $\displaystyle x\; \text{and}\;y$ you'll need two integrals. The problem is that your left boundary changes from $\displaystyle y = [0,1]\;\text{and}\; [1,2]$

so

$\displaystyle \int_0^1\int_{\sqrt{1-y^2}}^{\sqrt{4-y^2}}{xy}~dx~dy\;\;+\;\;\int_1^2\int_{0}^{\sqrt{4-y^2}}{xy}~dx~dy$

7. Originally Posted by Thomas154321
I did that, hence my $\displaystyle r^3$. I'm more confident that my polar integral is correct than the x,y one.
It is. See my post.

The reason your first integral didn't work out is because the x limits are not ALWAYS true for the area we are considering.

Consider the small slice of the region ABOVE the point where the smaller circle meets the y axis, and BELOW the point there the big circle meets the y axis. Is it true for that region, to say that the you enter the region through the boundary of the smaller circle and exit through the boundary of the larger circle? No, it is not. In fact, for that small portion, you enter through the y axis, and exit through the boundary of the larger circle. Your limits do not account for that small slice of the region.

Employ my method.

8. Both your methods work, thanks very much.