Question Number 139272 by bobhans last updated on 25/Apr/21
$$ \\ $$What is the area bounded by the parabola y^2=4ax and x^2 = 4ay?
Answered by EDWIN88 last updated on 25/Apr/21
![case(1) If a>0 ⇒Area = ∫_0 ^(4a) (2(√(ax)) −(x^2 /(4a)))dx ⇒Area = [ (4/3)(√a) x^(3/2) −(x^3 /(12a)) ]_0 ^(4a) ⇒Area = (4/3).4a(√(4a^2 )) −((64a^3 )/(12a)) ⇒Area = ((32a^2 )/3)−((16a^2 )/3) = ((16a^2 )/3).](https://www.tinkutara.com/question/Q139281.png)
$$\mathrm{case}\left(\mathrm{1}\right)\:\mathrm{If}\:\mathrm{a}>\mathrm{0} \\ $$$$\:\Rightarrow\mathrm{Area}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{4a}} {\int}}\left(\mathrm{2}\sqrt{\mathrm{ax}}\:−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4a}}\right)\mathrm{dx} \\ $$$$\Rightarrow\mathrm{Area}\:=\:\left[\:\frac{\mathrm{4}}{\mathrm{3}}\sqrt{\mathrm{a}}\:\mathrm{x}^{\mathrm{3}/\mathrm{2}} \:−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{12a}}\:\right]_{\mathrm{0}} ^{\mathrm{4a}} \: \\ $$$$\Rightarrow\mathrm{Area}\:=\:\frac{\mathrm{4}}{\mathrm{3}}.\mathrm{4a}\sqrt{\mathrm{4a}^{\mathrm{2}} }\:−\frac{\mathrm{64a}^{\mathrm{3}} }{\mathrm{12a}}\: \\ $$$$\Rightarrow\mathrm{Area}\:=\:\frac{\mathrm{32a}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{16a}^{\mathrm{2}} }{\mathrm{3}}\:=\:\frac{\mathrm{16a}^{\mathrm{2}} }{\mathrm{3}}.\: \\ $$