Answer to Question #206144 in Calculus for KABIR

Question

• Show that the area bounded by the curve, (x/2)^2 /3 + (y/4)^2/3 = 1 is 3π.

Answer 1

$\frac{(\frac{x}{2})^2}{3}+\frac{(\frac{y}{4})^2}{3}=1\\ \frac{x^2}{12}+\frac{y^2}{48}=1\\ y=\pm\sqrt{48(1-\frac{x^2}{12})}\\ y=\pm2\sqrt{12-x^2}\\$

To find x limit,

put y=0 in the given equation,

$\frac{x^2}{12}+\frac{0^2}{48}=1\\ x=\pm\sqrt{12}=\pm2\sqrt{3}$

area of the given curve

$=4×\int_a^bydx\\ =4×\int_0^{2\sqrt{3}}2\sqrt{12-x^2}dx\\ =8\int_0^{2\sqrt{3}}\sqrt{(2\sqrt{3})^2-x^2}dx\space (\text{By using integration formula)}\\ =8[\frac{x}{2}\sqrt{12-x^2}+\frac{12}{2}sin^{-1}(\frac{x}{2\sqrt{3}})+c]_0^{2\sqrt{3}}\\ =8[0+6.\frac{\pi}{2}+c-0-0-c]\\ =24\pi$ Thus, the area of the given curve is 24π.