Correct Answer - Option 2 :
\(\dfrac{32}{3}\)
Calculation:
⇒ x2 = 2y , y2 = 16x
⇒ x2/2 = √16x
⇒ x2/2 = 4√x
⇒ x = 4
The value of x lies between 0 and 4
The area enclosed by the curves = || \(\mathop \smallint \limits_{0\;}^{4\;} [\frac{{{x^2}}}{2} - \;\sqrt {16x} ]dx\)||
⇒ || \(\left[ {\frac{{{x^3}}}{6}} \right]_0^4 - 4\left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}\;} \right]_0^4\)||
⇒ || \(\left[ {\frac{{64}}{6} - 0} \right] - \frac{8}{3}\left[ {{4^{\frac{3}{2}}} - 0} \right]\)||
⇒ ||32/3 - 64/3 || = ||-32/3|| = 32/3
∴ The area enclosed by the curves x2 = 2y and y2 = 16x is 32/3.
