Question

Find the area of the region bounded by the parabola $y^2=16x$ and the line $x=4$.

Answer 1

Given that:

Parabola $y^2=16x$

at $x=4$, we have

$y=\pm 8$

Hence,

Parabola and line intersect at points $(4,8)$ and $(4,-8)$.

Parabola has a vertex $(0,0)$.

$y^2=16x$

$\Rightarrow x=\frac{y^2}{16}$

$\Rightarrow 4-\frac{y^2}{16}=0$

Area $={\int }_{-8}^8(4-\frac{y^2}{16})dy$

$\Rightarrow {[4y-\frac{y^3}{16\times 3}]}_{-8}^8$

$\Rightarrow [32-\frac{8\times 8\times 8}{8\times 6}]-[-32-\frac{(-8{)}^3}{8\times 6}]$

$\Rightarrow \frac{128}{3}.$

The same answer will come while integrate with respect to $x$.