Question

Find the areas included between the parabolas
(i) $y^2=4ax$ and $x^2=4ay$
(ii) $y^2=4ax$ and $x^2=4by$

Answer 1

R.E.F image

$\left(i\right)y^2=4ax,x^2=4ay$

$x^{4^3}=16a^2.4ax$

$x^3=64a^3$

$x=4a$

$y=4a$

Area =${\int }_0^{4a}2\sqrt{ax}-\frac{x^2}{4a}dx$

$={[\frac{2\sqrt{a}2x^{2/3}}{3}-\frac{1}{4a}\frac{x^3}{3}]}_0^{4a}$

$=\frac{4\sqrt{a}}{3}(4a{)}^{3/2}-\frac{1}{4a}\frac{(4a{)}^3}{3}$

$=\frac{4}{3}\times 8a^{\frac{3}{2}+\frac{1}{2}}-\frac{1}{3}\frac{16a^2}{3}$

$=\frac{32-16}{3}{a}^2=\frac{16a^2}{3}$

$y^2=4ax,x^2=4by$

$y^3=16a4by$

$y^3=64ab$
$y=4(ab{)}^{1/3}$

$4(ab{)}^{2/3}=4ax$

$x=4a^{2/3}{b}^{2/3}$
$x=4b^{2/3}$

$a=1/3$

$Area={\int }_0^{4b^{2/3/a^{1/3}}}2\sqrt{a}\sqrt{x}-\frac{x^2}{4b}dx$

$=[\frac{22\sqrt{a}}{3}{x}^{3/2}-\frac{x^3}{12b}{]}_0^{4b^{2/3}{a}^{1/3}}$

$=\frac{4\sqrt{a}}{3}(\frac{4b^{2/3}}{a^{1/3}}{)}^{3/2}-(\frac{4b^{2/3}}{a^{1/3}})\frac{1}{12b}$

$=\frac{4\sqrt{a}}{3}8\frac{b\frac{2}{3}\frac{3}{2}}{a\frac{1}{3}\frac{3}{2}}-\frac{64b\frac{2}{3}\times 3}{a\frac{1}{3}\times 3}\frac{1}{12b}$

$=\frac{32}{3}{a}^{12}\frac{b}{a^{12}}-\frac{64}{12}\frac{b}{a}$

$=\frac{32b}{3}-\frac{16}{3}\frac{b}{a}$

$=\frac{32ab-16b}{3a}$