# The area of the region bounded by the curve $y=\sqrt {16-x^2}$ and x-axis is \begin{array}{1 1}(A)\;8\pi \;sq.units & (B)\;20\pi\; sq.units\\(C)\;16\pi\; sq.units & (D)\;256\pi\; sq.units \end{array}

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• The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
• $\int\sqrt{a^2-x^2}dx=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\Large\frac{x}{a}\big)$$+c Step 1: Given curve is y=\sqrt{16-x^2}. Consider the curve y=\sqrt{16-x^2} Squaring on both sides we get y^2=16-x^2 or x^2+y^2=16 Clearly this is a curve whose centre is (0,0) and radius 4. Required area A=2\int_0^4y dx A=2\int_0^4\sqrt{16-x^2}dx \int\sqrt{a^2-x^2}dx=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\Large\frac{x}{a}\big) Step 2: On integrating we get, \;\;=2\begin{bmatrix}\large\frac{x}{2}\normalsize\sqrt{16-x^2}+\large\frac{16}{2}\normalsize\sin^{-1}\big(\large\frac{x}{4}\big)\end{bmatrix}_0^4 On applying the limits we get, \;\;=2\begin{bmatrix}\large\frac{4}{2}\normalsize\sqrt{16-16}+8\sin^{-1}(1)\end{bmatrix} But \sin^{-1}(1)=\large\frac{\pi}{2} Hence A=2\times 8 \times\large\frac{\pi}{2}$$=8\pi$ sq.units.
Hence A is the correct option.
edited May 6, 2013