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  • Choose the correct answer. The area of the circle x^2 + y^2 = 16 exterior to the parabola y^2 = 6x is

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Q : 18        The area of the circle  \small x^2+y^2=16  exterior to the parabola  \small y^2=6x  is

                   (A) \small \frac{4}{3}(4\pi -\sqrt{3} )        (B)  \small \frac{4}{3}(4\pi +\sqrt{3} )         (C)  \small \frac{4}{3}(8\pi -\sqrt{3} )        (D)  \small \frac{4}{3}(8\pi +\sqrt{3} )

Answers (1)

best_answer

The area of the shaded region is to be found.

Required area =ar(DOC)+ar(DOA)

The region to the left of the y-axis is half of the circle with radius 4 units and centre origin.

Area of the shaded region to the left of y axis is ar(1) = \frac{\pi 4^{2}}{2}=8\pi\ units

For the region to the right of y-axis and above x axis

\\x^{2}+y^{2}=16\\ y=\sqrt{16-x^{2}}     

\\y^{2}=6x\\ y=\sqrt{6x}

The parabola and the circle in the first quadrant intersect at point 

\left ( 2,2\sqrt{3} \right )

Remaining area is 2ar(2) is

\\ar(2)=\int_{0}^{2}\left ( \sqrt{16-x^{2}}-\sqrt{6x} \right )dx\\ =[ \frac{x}{2}\sqrt{16-x^{2}}+\frac{16}{2}sin^{-1}(\frac{x}{4})-\frac{2\sqrt{6}}{3}x^{\frac{3}{2}} ]{_{0}}^{2}\\ =[\sqrt{12}+\frac{16}{2}\times \frac{\pi }{6}-\frac{4\sqrt{12}}{3}]\\ =\frac{4\pi }{3}-\frac{2\sqrt{3}}{3}

Total area of shaded region is

\\ar(1)+2ar(2)\\ =8\pi +\frac{8\pi}{3}-\frac{4\sqrt{3}}{3}\\ =\frac{4}{3}(8\pi -\sqrt{3})\ units

Posted by

Sayak

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