Question

Find the area enclosed by x2=2y and y2=16x​

Answer 1

Answer: $\dfrac{32}{3}$

Step-by-step explanation:

Given

The parabolas are $x^2=2y$ and $y^2=16x$

Find the point of intersection of two parabolas

$\Rightarrow \left(\dfrac{x^2}{2}\right)^2=16x\\\\\Rightarrow x^4=64x\\\Rightarrow x(x^3-64)=0\\\Rightarrow x=0,4$

Obtain y using x

$(x,y)\rightarrow (0,0)\ \text{and }(4,8)$

Area enclosed between the two is

$\Rightarrow I=\int\limits^4_0 ({4\sqrt{x}-\dfrac{x^2}{2}}) \, dx\\\\\Rightarrow I=\left ( 4\times \dfrac{2}{3}x^{\frac{3}{2}}-\dfrac{x^3}{2\times 3} \right) _0^4\\\\\Rightarrow I=\left ( \dfrac{8}{3}\times 8-\dfrac{4^3}{6} \right )-0\\\\\Rightarrow I=\dfrac{128-64}{6}\\\\\Rightarrow I=\dfrac{64}{6}\\\\\Rightarrow I=\dfrac{32}{3}$

Thus, the area bounded by the two parabolas is  $\dfrac{32}{3}$ sq. units.