Area enclosed by the parabola ay = 3(a^2 - x^ | vedclass

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(A) The given equation of the parabola is $ay = 3(a^2 - x^2)$,which can be written as $y = \frac{3}{a}(a^2 - x^2)$.The parabola meets the $x$-axis whe

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$4a^2 	ext{ sq. units}$

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(A) The given equation of the parabola is $ay = 3(a^2 - x^2)$,which can be written as $y = \frac{3}{a}(a^2 - x^2)$.The parabola meets the $x$-axis where $y = 0$,so $\frac{3}{a}(a^2 - x^2) = 0$,which gives $x^2 = a^2$,or $x = \pm a$.The required area $A$ is given by the integral of $y$ with respect to $x$ from $x = -a$ to $x = a$:$A = \int_{-a}^{a} \frac{3}{a}(a^2 - x^2) dx$.Since the function is even,we can write:$A = 2 \int_{0}^{a} \frac{3}{a}(a^2 - x^2) dx = \frac{6}{a} \int_{0}^{a} (a^2 - x^2) dx$.Evaluating the integral:$A = \frac{6}{a} [a^2x - \frac{x^3}{3}]_{0}^{a} = \frac{6}{a} (a^3 - \frac{a^3}{3}) = \frac{6}{a} (\frac{2a^3}{3}) = 4a^2 	ext{ sq. units}$.

Area enclosed by the parabola $ay = 3(a^2 - x^2)$ and $x$-axis is

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