question_answer

The area (in square unit) bounded by the curves \[4y={{x}^{2}}\]and \[2y=6-{{x}^{2}}\]is

A)  \[8\]               

B)  \[6\]

C)  \[4\]               

D)  \[10\]

Correct Answer: A

Solution :

Given curves are \[{{x}^{2}}=4y\] and \[{{x}^{2}}=-2(y-3)\] The point of intersections are \[(2,1)\] and \[(-2,1)\] \[\therefore \] Required area \[=2\int_{0}^{1}{x\,dy+2\int_{1}^{3}{x\,\,dy}}\] \[=2\int_{0}^{1}{\sqrt{4y}}\,\,dy+2\int_{1}^{3}{\sqrt{6-2y}}\,dy\] \[=4\left[ \frac{{{y}^{3/2}}}{3/2} \right]_{0}^{1}+2\left[ \frac{{{(6-2y)}^{3/2}}}{3/2}\times \frac{1}{-2} \right]_{1}^{3}\] \[=\frac{8}{3}[1-0]-\frac{2}{3}[0-{{4}^{3/2}}]\] \[=\frac{8}{3}+\frac{16}{3}=8\,sq\,unit\]