Question

Draw a rough sketch and find the area of the region bounded by the two parabolas y² = 4x and x² = 4y by using methods of integration.

Answer 1

To find the points of intersection between two parabola let us substitute \[x = \frac{y^2}{4}\] in \[x^2 = 4y\]

\[\left( \frac{y^2}{4} \right)^2 = 4y\]

\[ \Rightarrow y^4 - 64y = 0\]

\[ \Rightarrow y\left( y^3 - 64 \right) = 0\]

\[ \Rightarrow y = 0, 4\]

\[\Rightarrow x = 0, 4\]

Therefore, the points of intersection are \[A(4, 4)\] and \[C\left( 0, 0 \right)\] 

Therefore, the area of the required region ABCD = \[\int_0^4 y_1 d x - \int_0^4 y_2 d x\] where \[y_1 = 2\sqrt{x}\] and \[y_2 = \frac{x^2}{4}\]

Required Area

\[= \int_0^4 \left( 2\sqrt{x} \right) d x - \int_0^4 \left( \frac{x^2}{4} \right) d x\]

\[ = \left[ 2 \times \frac{2 x^\frac{3}{2}}{3} - \frac{x^3}{12} \right]_0^4 \]

\[ = \left[ \left( 2 \times \frac{2 \left( 4 \right)^\frac{3}{2}}{3} - \frac{\left( 4 \right)^3}{12} \right) - \left( 2 \times \frac{2 \left( 0 \right)^\frac{3}{2}}{3} - \frac{\left( 0 \right)^3}{12} \right) \right]\]
After simplifying we get, 

\[= \frac{32}{3} - \frac{16}{3} = \frac{16}{3}\] square units