Question
Find the area included between the parabolas $y^2 = 4ax$ and $x^2 = 4 by$.
✓
Answer
To find area enclosed by $y^2 = 4ax ..(i) x^2= 4by$ ...(ii) Equation (i) represents a parabola with vertex (0. 0) at origin and axis as x-axis and equation (ii) represents a line parallel to vertex (0, 0) y-axis. points of intersection of parbola are (0, 0) and $\Big(4\text{a}\frac{1}{3}\ \text{b}\frac{2}{3}, 4\text{a}\frac{2}{3}\ \text{b}\frac{1}{3}\Big)$ A rough sketch of the equations is as below : 
The shaded region is requrid slides from x = 0 to $\text{x}=4\text{a}\frac{1}{3}\ \text{b}\frac{2}{3}$ Requrided area = Region OQAPO$=\int\limits_{0}^{4\text{a}\frac{1}{3}\text{b}\frac{2}{3}}(\text{y}_{1}-\text{y}_{2})\text{dx} $
$=\int\limits_{0}^{4\text{a}\frac{1}{3}\text{b}\frac{2}{3}}\Big(2\sqrt{\text{a}}.\sqrt{\text{x}}-\frac{\text{x}^{2}}{4\text{b}}\Big)\text{dx}$
$=\bigg[2\sqrt{\text{a}}.\frac{2}{3}\sqrt{\text{x}}-\frac{\text{x}^{3}}{12\text{b}}\Big)\text{dx}\bigg]^{4\text{a}\frac{1}{3}\text{b}\frac{2}{3}}$
$=\frac{32\sqrt{\text{a}}}{3}\text{a}\frac{1}{3}\ \text{b}\frac{2}{3}\ \text{a}\frac{1}{6}\ \text{b}\frac{1}{3}-\frac{64\text{ab}^{2}}{12\text{b}}$
$=\frac{32}{3}\text{ab}-\frac{16}{3}\text{ab}$
$=\frac{16}{3}\ \text{sq.}\ \text{units}$
The shaded region is requrid slides from x = 0 to $\text{x}=4\text{a}\frac{1}{3}\ \text{b}\frac{2}{3}$ Requrided area = Region OQAPO$=\int\limits_{0}^{4\text{a}\frac{1}{3}\text{b}\frac{2}{3}}(\text{y}_{1}-\text{y}_{2})\text{dx} $
$=\int\limits_{0}^{4\text{a}\frac{1}{3}\text{b}\frac{2}{3}}\Big(2\sqrt{\text{a}}.\sqrt{\text{x}}-\frac{\text{x}^{2}}{4\text{b}}\Big)\text{dx}$
$=\bigg[2\sqrt{\text{a}}.\frac{2}{3}\sqrt{\text{x}}-\frac{\text{x}^{3}}{12\text{b}}\Big)\text{dx}\bigg]^{4\text{a}\frac{1}{3}\text{b}\frac{2}{3}}$
$=\frac{32\sqrt{\text{a}}}{3}\text{a}\frac{1}{3}\ \text{b}\frac{2}{3}\ \text{a}\frac{1}{6}\ \text{b}\frac{1}{3}-\frac{64\text{ab}^{2}}{12\text{b}}$
$=\frac{32}{3}\text{ab}-\frac{16}{3}\text{ab}$
$=\frac{16}{3}\ \text{sq.}\ \text{units}$
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