This is sort of like the cylinders that meet at right angles, except the intersections do not create a square, but instead a 'diamond'.
The lengths of the sides of the square in the former problem was
\(\displaystyle \sqrt{r^{2}-y^{2}}\), therefore the area of the square was
\(\displaystyle r^{2}-y^{2}\)
Now, we have to use a little trig to find the length of the side.
I wish I had a picture. Hey, I do have a picyure, albeit, hackneyed and comical. It's the best I can do with that infernal 'paint'. I hope it's enough to give you the idea.
Try:
\(\displaystyle \L\\8\int_{0}^{r}csc({\theta})(r^{2}-y^{2})dy\)
Also,
\(\displaystyle \L\\4\int_{-r}^{r}csc({\theta})(r^{2}-y^{2})dy\)
