Area Question 49
Question 49
- Find the area bounded by the curves $x^{2}+y^{2}=4$, $x^{2}=-\sqrt{2} y$ and $x=y$.
$(1986,5$ M)
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Solution:
- Given curves are $x^{2}+y^{2}=4, x^{2}=-\sqrt{2} y$ and $x=y$.
Thus, the required area
$$ \begin{aligned} & =\left|\int_{-\sqrt{2}}^{\sqrt{2}} \sqrt{4-x^{2}} d x\right|-\left|\int_{-\sqrt{2}}^{0} x d x\right|-\left|\int_{0}^{\sqrt{2}} \frac{-x^{2}}{\sqrt{2}} d x\right| \ & =2 \int_{0}^{\sqrt{2}} \sqrt{4-x^{2}} d x-\frac{x^{2}}{2}-\frac{x^{3}}{3 \sqrt{2}} 0 \ & =2 \frac{x}{2} \sqrt{4-x^{2}}-\frac{4}{2} \sin ^{-1} \frac{x}{2}_{0}^{\sqrt{2}}-1-\frac{2}{3} \ & =(2-\pi)-\frac{5}{3} \ & =\frac{1}{3}-\pi \text { sq units } \end{aligned} $$
