Question

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

Answer 1

$y^2=4x$ $⟹x=\frac{y^2}{4}$ --- (i)

$x^2=4y$ --- (ii)

Substitute (i) in (ii),

${\left(\frac{y^2}{4}\right)}^2=4y$

$\frac{y^4}{16}=4y$ $⟹y^3=64$ $⟹y=4$

$⟹x=\frac{y^2}{4}=\frac{4^2}{4}=4$

Thus the parabolas intersect at $(4,4)$ and $(0,0)$

$Area={\int }_0^{4}ydx={\int }_0^4(y_{p1}-y_{p2})dx={\int }_0^4(\sqrt{4x}-\frac{x^2}{4})dx$

$={\int }_0^4(2\sqrt{x}-\frac{x^2}{4})dx$

$={|\frac{2x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^3}{4\left(3\right)}|}_0^4$

$=\left(\frac{4}{3}\right(4{)}^{\frac{3}{2}}-\frac{4^3}{12})-\left(0\right)$

$Area=5.33sq.unit$