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Q.
The ratio in which the area bounded by the curves $y^{2} = 12x$ and $x^{2} = 12y$ is divided by the line $x = 3$ is
Application of Integrals
Solution:
We have $y^{2} = 12x$ and $x^{2} = 12y$, parabolas with vertex $\left(0,0\right)$.
The curves intersect at $x = 0,12$
Required ratio $=\frac{\int_{0}^{3}\left(\sqrt{12x}-\frac{x^{2}}{12}\right) dx}{\int_{3}^{12}\left(\sqrt{12 x}-\frac{x^{2}}{12}\right)dx}=\frac{\left|\sqrt{12}\cdot\frac{2}{3}x^{\frac{3}{2}}-\frac{x^{3}}{36}\right|_{0}^{3}}{\left|\sqrt{12}\cdot\frac{2}{3}x^{\frac{3}{2}}-\frac{x^{3}}{36}\right|_{3}^{12}}$
$=\frac{15}{49}$ sq. units
