Question
Find the area between the parabolas $y^2=7 x$ and $x^2=7 y$.
✓
Answer
For finding the points of intersection of the two parabolas,
we equate the values of $y^2$ from their equations.
From the equation $x ^2=7 y , y ^2=\frac{x^4}{49}$
$
\begin{aligned}
& \therefore \frac{x^4}{49}=7 x \\
& \therefore x ^4=343 x \\
& \therefore x ^4-343 x =0 \\
& \therefore x \left( x ^3-343\right)=0
\end{aligned}
$
$\therefore x =0$ or $x ^3=343$, i.e. $x =7$
When $x =0, y =0$
When $x=7,7 y=49$
$
\therefore y =7
$
$\therefore$ the points of intersection are $O (0,0)$ and $A (7,7)$
Required area $=$ area of the region $OBACO$
$=($ area of the region $O D A C O)-($ area of the region $O D A B O$ )
Now, area of the region ODACO $=$ area under the parabola $y^2=7 x$
i.e. $y=\sqrt{ } 7 \sqrt{ } x$
$
\begin{aligned}
& =\int_0^7 \sqrt{7} \sqrt{x} d x=\sqrt{7}\left[\frac{x^{\frac{3}{2}}}{3 / 2}\right]_0^7 \\
& =\sqrt{7} \times \frac{2}{3}\left[7^{\frac{3}{2}}-0\right]=\frac{2 \sqrt{7}}{3}[7 \sqrt{7}-0] \\
& =\frac{98}{3}
\end{aligned}
$
Area of the region $ODABO =$ Area under the parabola
$
x^2=7 y
$
i.e. $y =\frac{x^2}{7}$
$
=\int_0^7 \frac{x^2}{7} d x=\frac{1}{7}\left[\frac{x^3}{3}\right]_0^7=\frac{1}{7}\left[\frac{7^3}{3}-0\right]
$
$=\frac{7^2}{3}=\frac{49}{3}$
$\therefore$ required area $=\frac{98}{3}-\frac{49}{3}=\frac{49}{3}$ sq units.
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