The area of the region bounded by the curves \( x=at^2 \) and \( y=2at \) between the lines parallel to y-axis corresponding to \( t=1 \) and \( t=2 \) is equal to :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{56a^2}{3} \) Sq units
Eliminating t from \( x=at^2 \), \( y=2at \): \( t=\frac{y}{2a} \), so \( x=a\cdot\frac{y^2}{4a^2} \), that is
\( \displaystyle y^2=4ax \)
This is a parabola symmetric about the x-axis. The lines parallel to the y-axis are x = a(1)2 = a (t = 1) and x = a(2)2 = 4a (t = 2).
No x-axis is mentioned, so the closed region between the parabola and these lines is bounded by both branches and lies above and below the x-axis. Its area is twice the area above the x-axis, where \( y=2\sqrt{ax} \) (taking a > 0):
\( \displaystyle \text{Area}=2\int_{a}^{4a}2\sqrt{a}\,\sqrt{x}\,dx \)
\( \displaystyle =4\sqrt{a}\cdot\frac{2}{3}\left[x^{3/2}\right]_{a}^{4a} \)
\( \displaystyle =\frac{8\sqrt{a}}{3}\left(8a^{3/2}-a^{3/2}\right) \)
\( \displaystyle =\frac{8\sqrt{a}}{3}\cdot 7a^{3/2} \)
\( \displaystyle =\frac{56a^2}{3}\ \text{sq units} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{14}{3}a^2 \) Sq units — Incorrect. \( \frac{14}{3}a^2 \) is half of the area above the x-axis; it drops the factor 2 in \( y=2\sqrt{ax} \).
Option (2) \( \displaystyle\frac{112a^2}{3} \) Sq units — Incorrect. \( \frac{112a^2}{3} \) is twice the correct area.
Option (4) \( \displaystyle\frac{28}{3}a^2 \) Sq units — Incorrect. \( \frac{28}{3}a^2 \) is only the part above the x-axis; the lower branch \( y=-2\sqrt{ax} \) also bounds the region.
NCERT Exemplar:
Application of Integrals
