question_answer

The volume of the solid formed by rotating the area enclosed between the curve \[y={{x}^{2}}\] and the line \[y=1\] about \[y=1\] is (in cubic units [UPSEAT 2003]

A)            \[9\pi /5\]                             

B)            \[4\pi /3\]

C)            \[8\pi /3\]                             

D)            \[7\pi /5\]

Correct Answer: B

Solution :

           Volume of the solid formed by rotating the area enclosed between the curve \[y={{x}^{2}}\] and line \[y=1\] will be \[\int_{0}^{1}{2\pi \,xdy}\]= \[2\int_{0}^{1}{\pi \sqrt{y}dy}\]= \[\frac{4\pi }{3}[{{y}^{3/2}}]_{0}^{1}=\frac{4\pi }{3}\].