A) \[\frac{1}{6}(24\pi -1)\]
B) \[\frac{1}{6}(12\pi -1)\]
C) \[\frac{1}{3}(12\pi -1)\]
D) \[\frac{1}{3}(6\pi -1)\]
Correct Answer: B
Solution :
[b] Area \[=2\pi -\int\limits_{0}^{1}{\left( \sqrt{x}-x \right)dx}\] \[=2\pi -\left[ \frac{2{{x}^{\frac{3}{2}}}}{3}-\frac{{{x}^{2}}}{2} \right]_{0}^{1}\] \[=2\pi -\left( \frac{1}{6} \right)\] \[=\frac{12\pi -1}{6}\] square unitsYou need to login to perform this action.
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