Answer: [tex]\dfrac{32}{3}[/tex]
Step-by-step explanation:
Given
The parabolas are [tex]x^2=2y[/tex] and [tex]y^2=16x[/tex]
Find the point of intersection of two parabolas
[tex]\Rightarrow \left(\dfrac{x^2}{2}\right)^2=16x\\\\\Rightarrow x^4=64x\\\Rightarrow x(x^3-64)=0\\\Rightarrow x=0,4[/tex]
Obtain y using x
[tex](x,y)\rightarrow (0,0)\ \text{and }(4,8)[/tex]
Area enclosed between the two is
[tex]\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}[/tex]
Thus, the area bounded by the two parabolas is [tex]\dfrac{32}{3}[/tex] sq. units.
