Area of an irregular region

The area of an ellipse is given by $ab\pi$ for semi axes a and b, each covering half the max vertical and horizontal expanses of the ellipse respectively. As you can see in the diagram above, the kite portion of the ellipse can be divided into 4 congruent triangles each with base b and height a, thus each of them has an area of $\frac{ab}{2}$ and the total area of the kite, or just the 4 $\frac{ab}{2}$ triangles, is 2 $\text ab$ . The remaining area of the ellipse(The 4 congruent irregular outlying regions) is ( $\pi -2$ ) $\text ab$ . Divide this by 4 to get the area of each of the individual outlying irregular regions, which is $\frac{\pi -2}{4}$ $\text ab$ . The red region is just one of these 4 outlying irregular regions, so this is the answer! Now just simplify!