The area (in sq. units) of the part of the circle x2+y2=36,which is outside the parabola y2=9x,is:

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24π+33

12π+33

12π−33

24π−33

answer is D.

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Detailed Solution

The circle and curve intersect at x=3∴Area =πr2−2∫039xdx+∫3636−x2dx =36π−123−2x236−x2+18sin−1x636 =36π−123−29π−932+3π =24π−33

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