Question

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

Answer 1

Answer:
`(32)/(3)`

Explanation:

Given

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

Find the point of intersection of two parabolas

`\Rightarrow \left((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√(x)-(x^2)/(2)}) \, dx\\\\\Rightarrow I=\left ( 4* (2)/(3)x^{(3)/(2)}-(x^3)/(2* 3) \right) _0^4\\\\\Rightarrow I=\left ( (8)/(3)* 8-(4^3)/(6) \right )-0\\\\\Rightarrow I=(128-64)/(6)\\\\\Rightarrow I=(64)/(6)\\\\\Rightarrow I=(32)/(3)`

Thus, the area bounded by the two parabolas is
`(32)/(3)` sq. units.