Question

The area bounded by the curve y² = 8x and x² = 8y is ___________ .

Options

• \[\frac{16}{3}\]sq. units

• \[\frac{3}{16}\]sq. units

• \[\frac{14}{3}\]sq. units

• \[\frac{3}{14}\]sq. units

• None of these

Answer 1

None of these

 

Point of intersection of both the parabolas y² = 8x and x² = 8y is obtained by solving the two equations
\[y^2 = 8x\text{ and }x^2 = 8y \]
\[ \therefore \frac{y^4}{64} - 8y = 0\]
\[ \Rightarrow y\left( y^3 - 8^3 \right) - 0\]
\[ \Rightarrow y = 0\text{ or }y = 8\]
\[ \Rightarrow x = 0\text{ or }x = 8 \]
\[ \therefore O\left( 0, 0 \right)\text{ and }A\left( 8, 8 \right)\text{ are the points of intersection .} \]
\[\text{ Area of the shaded region }= \int_0^8 \left| y_2 - y_1 \right| dx\]
\[ = \int_0^8 \left( y_2 - y_1 \right)dx\]
\[ = \int_0^8 \left( \sqrt{8x} - \frac{x^2}{8} \right)dx\]
\[ = \left[ \frac{\sqrt{8}}{\frac{3}{2}} x^\frac{3}{2} - \frac{1}{8} \times \frac{x^3}{3} \right]_0^8 \]
\[ = \frac{2}{3} \times \sqrt{8} \times 8^\frac{3}{2} - \frac{1}{8} \times \frac{8^3}{3} - 0\]
\[ = \frac{2}{3} \times \sqrt{8} \times 8\sqrt{8} - \frac{8^2}{3}\]
\[ = \frac{2}{3} \times 8^2 - \frac{8^2}{3}\]
\[ = \frac{8^2}{3}\left( 2 - 1 \right)\]
\[ = \frac{64}{3}\text{ sq units }\]