Area Ques 60
Question
- Find the area bounded by the curves $x^{2}+y^{2}=4$, $x^{2}=-\sqrt{2} y$ and $x=y$.
$(1986,5$ M)
Show Answer
Answer:
Correct Answer: 60.$(\frac{1}{3}-\pi) \text { sq units }$
Solution:
Formula:
- 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-\left|(\frac{x^{2}}{2})^{0} _{-\sqrt{2}}\right| -\left|\frac{x^{3}}{3 \sqrt{2}}\right|^{-\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} $
BETA
