question_answer

Area of the region enclosed between the curves \[x={{y}^{2}}-1\] and \[x=\,|y|\,\sqrt{1-{{y}^{2}}}\] is

A)  2 sq units           

B)  2/3 sq unit        

C)  1 sq unit             

D)  4/3 sq units

Correct Answer: A

Solution :

 Required area \[A=2\,\int_{0}^{1}{[y\,\sqrt{1-{{y}^{2}}}-({{y}^{2}}-1)]\,dy}\] \[A=2\,\int_{0}^{1}{y\,\sqrt{1-{{y}^{2}}}\,dy-2\,\int_{0}^{1}{({{y}^{2}}-1)\,dy}}\] Let          \[1-{{y}^{2}}={{t}^{2}}\] \[-2y\,\,dy=2t\,\,dt\] \[-y\,dy=t\,dt\] \[A=-\,2\,\,\int_{1}^{0}{{{t}^{2}}\,dt-2\,\left[ \frac{{{y}^{3}}}{3}-y \right]_{0}^{1}}\] \[A=\,2\,\,\int_{0}^{1}{{{t}^{2}}\,dt-2\,\left( \frac{1}{3}-1 \right)}\] \[A=2\,\left[ \frac{{{t}^{3}}}{3} \right]_{0}^{1}\,-2\,\left( -\frac{2}{3} \right)\] \[=2\,\left( \frac{1}{3} \right)+\frac{4}{3}=\frac{2+4}{3}\] = 2 sq units