Consider the ellipse (x^2)/16+(y^2)/9=1 . Which of the following stat…

The correct answer is Option (4) → (A) and (C) only

In the first quadrant the ellipse is \(y = \frac{3}{4}\sqrt{16-x^{2}}\), and the ellipse is symmetric about both axes, so the whole area is four times the first-quadrant part:

    \( \displaystyle \text{First quadrant} = \frac{3}{4}\int_{0}^{4}\sqrt{16-x^{2}}\,dx \)

    \( \displaystyle \text{Whole} = 4\times\frac{3}{4}\int_{0}^{4}\sqrt{16-x^{2}}\,dx = 3\int_{0}^{4}\sqrt{16-x^{2}}\,dx \)

The integral is a quarter of the disc of radius 4:

    \( \displaystyle \int_{0}^{4}\sqrt{16-x^{2}}\,dx = \left[\frac{x}{2}\sqrt{16-x^{2}}+8\sin^{-1}\frac{x}{4}\right]_{0}^{4} = 8\times\frac{\pi}{2} = 4\pi \)

    Whole area \(= 3\times4\pi = 12\pi\), first-quadrant part \(= 3\pi\)

Integrating in y instead, \(x = \frac{4}{3}\sqrt{9-y^{2}}\) and \(\int_{0}^{3}\sqrt{9-y^{2}}\,dy = \frac{9\pi}{4}\):

    \( \displaystyle \text{Whole} = 4\times\frac{4}{3}\int_{0}^{3}\sqrt{9-y^{2}}\,dy = \frac{16}{3}\times\frac{9\pi}{4} = 12\pi \)

So the ellipse encloses \(12\pi = 3\int_{0}^{4}\sqrt{16-x^{2}}\,dx\), its first-quadrant part is \(3\pi\), and the y-integral needs the factor \(\frac{16}{3}\).

Checking the statements

(A) The area of the region enclosed by the ellipse is \(\displaystyle 3\int_{0}^{4}\sqrt{16-x^{2}}\,dx\) — Correct. \(4\times\frac{3}{4} = 3\), and the value is \(3\times4\pi = 12\pi\).

(B) The area of the region enclosed by the ellipse is \(\displaystyle \frac{4}{3}\int_{0}^{3}\sqrt{9-y^{2}}\,dy\) — Incorrect. This equals \(\frac{4}{3}\times\frac{9\pi}{4} = 3\pi\), only the first-quadrant part.

(C) The area of the region enclosed by the ellipse is \(12\pi\) sq. units — Correct. \(\pi ab = \pi\times4\times3 = 12\pi\).

(D) The area of the part of the region in the first quadrant is \(\displaystyle 3\int_{0}^{4}\sqrt{16-x^{2}}\,dx\) — Incorrect. That expression equals \(12\pi\), the whole area; the first-quadrant part is \(3\pi\).