First slide

Area of bounded Regions

Question

$T h e a r e a o f t h e p l a n e r e g i o n b o u n d e d b y t h e c u r v e s$ $x + 2 y^{2} = 0 and x + 3 y^{2} = 1 is$

Moderate

A

$\frac{1}{2}$
B

$\frac{2}{3}$
C

$\frac{4}{3}$
D

$\frac{5}{3}$

Solution

$\begin{matrix} Equations of Parabolas : y^{2} = \frac{- x}{2} and y^{2} = \frac{1}{3} (1 - x) \\ On solving, we get x = - 2, y = \pm 1 \\ required Area = 2 [\frac{1}{\sqrt{3}} \int_{- 2}^{1} \sqrt{(1 - x)} d x - \frac{1}{\sqrt{2}} \int_{- 2}^{0} \sqrt{- x} d x] \\ = 2 \{{[\frac{1}{\sqrt{3}} \times \frac{- 2}{3} {(1 - x)}^{3 / 2}]}_{- 2}^{1} - {[\frac{1}{\sqrt{2}} \times \frac{- 2}{3} {(- x)}^{3 / 2}]}_{- 2}^{0}\} \\ = 2 \{(\frac{2}{3 \sqrt{3}} \cdot 3 \sqrt{3}) - (\frac{2}{3 \sqrt{2}} \cdot 2 \sqrt{2})\} \\ = \frac{4}{3} . \end{matrix}$

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