Area under the curveMedium

Question

The area of the region bounded by the curves $x + 3y^{2} = 0$ and $x + 4y^{2} = 1$ is equal to :

Options

A.$\frac{1}{3}$
B.$\frac{2}{3}$
C.$\frac{4}{3}$
D.$\frac{5}{3}$

Solution

Sol. [Key Concept: Integrate $(x_{right} - x_{left})$ in $y$ between the curves; the area reduces to $\int(1 - y^{2})dy$.]

$x = - 3y^{2}$ and $x = 1 - 4y^{2}$

$${A = \int_{- 1}^{1}\mspace{2mu}\left( \left( 1 - 4y^{2} \right) + 3y^{2} \right)dy }{A = \int_{- 1}^{1}\mspace{2mu}\left( 1 - y^{2} \right)dy = 2\int_{0}^{1}\mspace{2mu}\left( 1 - y^{2} \right)dy }{A = \frac{4}{3} }$$

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