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(A) The given equation of the circle $x^{2}+y^{2}=8x$ can be expressed as $(x-4)^{2}+y^{2}=16$. Thus,the centre of the circle is $(4, 0)$ and the radi

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$\frac{4}{3}(8+3\pi)$

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(A) The given equation of the circle $x^{2}+y^{2}=8x$ can be expressed as $(x-4)^{2}+y^{2}=16$. Thus,the centre of the circle is $(4, 0)$ and the radius is $4$.Its intersection with the parabola $y^{2}=4x$ gives:$x^{2}+4x=8x$$x^{2}-4x=0$$x(x-4)=0$$x=0, x=4$Thus,the points of intersection of these two curves are $O(0, 0)$ and $P(4, 4)$ above the $x-$ axis.The required area of the region $OPQCO$ included between these two curves above the $x-$ axis is:Area $= \int_{0}^{4} \sqrt{4x} \, dx + \int_{4}^{8} \sqrt{16-(x-4)^{2}} \, dx$$= 2 \int_{0}^{4} x^{1/2} \, dx + \int_{0}^{4} \sqrt{4^{2}-t^{2}} \, dt$ (where $t = x-4$)$= 2 	imes \frac{2}{3} [x^{3/2}]_{0}^{4} + [\frac{t}{2} \sqrt{16-t^{2}} + \frac{16}{2} \sin^{-1}(\frac{t}{4})]_{0}^{4}$$= \frac{4}{3} (8) + [0 + 8 \sin^{-1}(1) - 0]$$= \frac{32}{3} + 8(\frac{\pi}{2}) = \frac{32}{3} + 4\pi = \frac{4}{3}(8+3\pi)$

Find the area lying above the $x-$ axis and included between the circle $x^{2}+y^{2}=8x$ and inside the parabola $y^{2}=4x$.

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