The area enclosed by the curve \(x=8\cos t, y=4\sin t\) between the lines \(x=0\) and \(x=4\sqrt{3}\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(8\sqrt{3}+\displaystyle\frac{32\pi}{3}\) units
Eliminating t with \(\cos^2t + \sin^2t = 1\):
\( \displaystyle \frac{x^2}{64} + \frac{y^2}{16} = 1 \)
This is an ellipse with a = 8, b = 4. Solving for y:
\( \displaystyle y = \pm\frac{1}{2}\sqrt{64-x^2} \)
The region enclosed by the curve between x = 0 and \(x = 4\sqrt{3}\) runs from the lower half of the ellipse to the upper half. The ellipse is symmetric about the x-axis, so the area is twice the part above the x-axis:
\( \displaystyle A = 2\int_{0}^{4\sqrt{3}}\frac{1}{2}\sqrt{64-x^2}\,dx = \int_{0}^{4\sqrt{3}}\sqrt{64-x^2}\,dx \)
Using the standard integral \(\int\sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C\) with a = 8:
\( \displaystyle A = \left[\frac{x}{2}\sqrt{64-x^2} + 32\sin^{-1}\frac{x}{8}\right]_{0}^{4\sqrt{3}} \)
At \(x = 4\sqrt{3}\): \(\sqrt{64-48} = 4\) and \(\sin^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{3}\). At x = 0 both terms are 0.
\( \displaystyle A = \frac{4\sqrt{3}}{2}\times4 + 32\times\frac{\pi}{3} \)
\( \displaystyle = 8\sqrt{3} + \frac{32\pi}{3} \)
Why the other options are incorrect
Option (1) \(8\sqrt{2}+\displaystyle\frac{32\pi}{3}\) units — Incorrect. The first term is \(8\sqrt{3}\), from \(\frac{4\sqrt{3}}{2}\times4\), not \(8\sqrt{2}\).
Option (2) \(4\sqrt{3}+\displaystyle\frac{16\pi}{3}\) units — Incorrect. This is only the part above the x-axis, half of the region the curve encloses between the two lines.
Option (4) \(32\pi\) units — Incorrect. \(\pi ab = 32\pi\) is the area of the whole ellipse, from x = −8 to x = 8.