Question

The area included between the parabolas y² = 4x and x² = 4y is (in square units)

Options

• 4/3

• 1/3

• 16/3

• 8/3

Answer 1

16/3

 

We have, 
\[x = \frac{y^2}{4} . . . . . \left( 1 \right)\]
\[ x^2 = 4y . . . . . \left( 2 \right)\]
Points of intersection of two parabola is given by,
\[\left( \frac{y^2}{4} \right)^2 = 4y\]
\[ \Rightarrow y^4 - 64y = 0\]
\[ \Rightarrow y\left( y^3 - 64 \right) = 0\]
\[ \Rightarrow y = 0, 4\]
\[\Rightarrow x = 0, 4\]
Therefore, the points of intersection are A(0, 0) and C(4, 4).
Therefore, the area of the required region ABCD,
\[= \int_0^4 \left( y_1 - y_2 \right) d x ..............\left(\text{where, }y_1 = 2\sqrt{x}\text{ and }y_2 = \frac{x^2}{4} \right)\]
\[ = \int_0^4 \left( 2\sqrt{x} - \frac{x^2}{4} \right) d x\]
\[ = \left[ 2 \times \frac{2 x^\frac{3}{2}}{3} - \frac{x^3}{12} \right]_0^4 \]
\[ = \left( 2 \times \frac{2 \left( 4 \right)^\frac{3}{2}}{3} - \frac{\left( 4 \right)^3}{12} \right) - \left( 2 \times \frac{2 \left( 0 \right)^\frac{3}{2}}{3} - \frac{\left( 0 \right)^3}{12} \right)\]
\[ = \left( \frac{32}{3} - \frac{16}{3} \right) - 0\]
\[ = \frac{16}{3}\text{ square units }\]