Question

The area enclosed between the parabolas $y^2=16x$ and $x^2=16y$ is

• A. $\frac{64}{3}sq.units$
• B. $\frac{256}{3}sq.units$
• C. $\frac{16}{3}sq.units$
• D. None of these

Answer 1

The correct option is B $\frac{256}{3}sq.units$

Both parabolas cut each other at $(0,0)$ and $(16,16)$

Area enclosed by these parabolas

$={\int }_0^{16}4\sqrt{x}dx-{\int }_0^{16}\frac{x^2}{16}dx$

$=[\frac{2\times 4\times x^{3/2}}{3}{]}_0^{16}-[\frac{x^3}{16\times 3}{]}_0^{16}$

$=\frac{2\times 4^4}{3}-\frac{4^4}{3}=\frac{256}{3}$ sq. units