Question

Prove that the area common to the two parabolas y = 2x² and y = x² + 4 is \[\frac{32}{3}\] sq. units.

Answer 1

We have, two parabolas y = 2x² and y = x² + 4
To find point of intersection , solve the two equations 
\[2 x^2 = x^2 + 4\]
\[ \Rightarrow x^2 = 4\]
\[ \Rightarrow x = \pm 2\]
\[ \therefore y = 4\]
\[\text{ Thus A }(2, 4)\text{ and A'}( - 2 , 4 )\text{ are points of intersection of the two parabolas }\]
\[\text{ Shaded area }= 2 \times\text{ area }\left(\text{ OCAO }\right)\]
\[\text{ Consider a vertical stip of length }\left| y_2 - y_1 \right| \text{ and width }dx \]
\[\text{ Area of approximating rectangle }= \left| y_2 - y_1 \right| dx \]
\[\text{ The approximating rectangle moves from }x = 0\text{ to } x = 2\]
\[\text{ Area }\left( OCAO \right) = \int_0^2 \left| y_2 - y_1 \right|dx = \int_0^2 \left( y_2 - y_1 \right)dx .............\left\{ \because \left| y_2 - y_1 \right| = y_2 - y_1\text{ as }y_2 > y_1 \right\}\]
\[ = \int_0^2 \left( x^2 + 4 - 2 x^2 \right)dx\]
\[ = \int_0^2 \left( 4 - x^2 \right)dx\]
\[ = \left[ \left( 4x - \frac{x^3}{3} \right) \right]_0^2 \]
\[ = 8 - \frac{8}{2}\]
\[ = \frac{16}{3}\text{ sq units }\]
\[\text{ Shaded area }= 2 \times\text{ area }\left( OCAO \right) = 2 \times \frac{16}{3} = \frac{32}{3}\text{ sq units }\]