The area of the region bounded by a²y²= x² (a²–x²) is __ sq.units

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$\frac{4 a}{3}$

$\frac{4 a^{2}}{3}$

$\frac{5 a^{2}}{8}$

$\frac{4 a^{2}}{5}$

answer is C.

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Detailed Solution

$A = 4 \int_{0}^{a} \frac{x}{a} \sqrt{a^{2} - x^{2}} dx; p u t a^{2} - x^{2} = t \Rightarrow - 2 x dx = dt; A = \frac{4}{2 a} \int_{a^{2}}^{0} - \sqrt{t} dt = \frac{2}{a} (\frac{2}{3} a^{3}) = \frac{4}{3} a^{2}$

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