Q. The parabola $y^{2} = 4x\ and\ x^{2} = 4\ y$ divide the square region bounded by the lines $x = 4,\ y = 4\$and the coordinate axes. If $S_{1},S_{2},S_{3}$ are respectively the areas of these parts numbered from top to bottom, then $S_{1}:S_{2}:S_{3}$ is
A) 1:1:1
B) 2:1:2
C) 1:2:3
D) 1:2:1
Explanation
(a)
$S_{1} = S_{3} = \int_{0}^{4}{\frac{x^{2}}{4}dx}$
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$= \frac{1}{4}\left\lbrack \frac{x^{3}}{3} \right\rbrack_{0}^{4} = \frac{16}{3}\ sq\ units$
$Now,\ \ \ S_{2} + S_{3} = \int_{0}^{4}\sqrt{4x}\ dx = 2 \times \left\lbrack \frac{x^{3/2}}{3/2} \right\rbrack_{0}^{4} = \frac{32}{3}sq\ units$
$\Longrightarrow \ s_{2} = \frac{16}{3}sq\ units$
$\therefore\ \ S_{1}:S_{2}:S_{3} = \frac{16}{3}:\frac{16}{3}:\frac{16}{3} = 1:1:1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.