The area (in sq. units) of the region \(\left\{(x, y): \displaystyle\frac{x^{2}}{4}+\frac{y^{2}}{9}\le 1,\ \frac{x^{2}}{4}+y^{2}\ge 1\right\}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(4\pi\)
The region is inside the ellipse \(\displaystyle\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\) (semi-axes 2 and 3) and outside the ellipse \(\displaystyle\frac{x^{2}}{4}+\frac{y^{2}}{1} = 1\) (semi-axes 2 and 1).
Since \(\displaystyle\frac{y^{2}}{9} \le y^{2}\), every point with \(\displaystyle\frac{x^{2}}{4}+y^{2} \le 1\) also has \(\displaystyle\frac{x^{2}}{4}+\frac{y^{2}}{9} \le 1\). So the smaller ellipse lies inside the larger one, and the area is the difference of the two areas.
Using area \(= \pi ab\) for an ellipse:
Larger ellipse: \(\pi\times2\times3 = 6\pi\)
Smaller ellipse: \(\pi\times2\times1 = 2\pi\)
\(A = 6\pi-2\pi = 4\pi\)