Answer:[tex]18\sqrt{3}[/tex]
Step-by-step explanation:
Given data
we haven given a parabola and a straight line
Parabola is [tex]{y^2}={-\left ( x-10\right )[/tex]
line is [tex]x=7[/tex]
Find the point of intersection of parabola and line
[tex]y=\pm \sqrt{3}[/tex] when[tex]x=7[/tex]
Area enclosed is the shaded area which is given by
[tex]Area=\int_{0}^{\sqrt{3}}\left ( 10-y^2 \right )dy[/tex]
[tex]Area=_{0}^{\sqrt{3}}10y-_{0}^{\sqrt{3}}\frac{y^3}{3}[/tex]
[tex]Area=10\sqrt{3}-\sqrt{3}[/tex]
[tex]Area=9\sqrt{3}units[/tex]
Required area will be double of calculated because it is symmetrical about x axis=[tex]18\sqrt{3}units[/tex]
