Q.Find the area of the part of the circle $x^2 + y^2 = 9$ that lies in the first quadrant.

$x^2+y^2=8x$ is a circle of radius $4$ centred at $(4,0)$; integrate $y=\sqrt{8x-x^2}$ across its diameter and double for symmetry, which reproduces $\pi r^2$.

Step 1 — identify the circle. $x^2-8x+y^2=0 \Rightarrow (x-4)^2+y^2=16$: centre $(4,0)$, radius $r=4$.

Step 2 — set up the calculus integral. Solving for $y$: $y=\pm\sqrt{8x-x^2}$, with $x$ ranging over $[0,8]$ (where $8x-x^2\ge0$). By symmetry about the $x$-axis,

$\text{Area}=2\int_0^8\sqrt{8x-x^2}\,dx = 2\int_0^8\sqrt{16-(x-4)^2}\,dx.$

Step 3 — substitute $u=x-4$ ($du=dx$, limits $u=-4$ to $4$):

$\text{Area}=2\int_{-4}^{4}\sqrt{16-u^2}\,du = 2\left[\frac u2\sqrt{16-u^2}+8\sin^{-1}\frac u4\right]_{-4}^{4}.$

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Use symmetry to compute the area of one quadrant of the circle by integrating $y=\sqrt{16-x^2}$, then multiply by 4.

The circle $x^2+y^2=16$ has radius $4$. By the symmetry of a circle about both axes, its total area equals $4\times$(area in the first quadrant).

Area in the first quadrant (bounded by the curve $y=\sqrt{16-x^2}$, from $x=0$ to $x=4$):

$A_1 = \displaystyle\int_0^4 \sqrt{16-x^2}\,dx$

Using the standard result $\displaystyle\int \sqrt{a^2-x^2}\,dx = \dfrac{x}{2}\sqrt{a^2-x^2} + \dfrac{a^2}{2}\sin^{-1}\left(\dfrac{x}{a}\right) + C$, with $a=4$:

$A_1 = \left[\dfrac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\left(\dfrac x4\right)\right]_0^4$

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