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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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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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Toolbox:
  • 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.
answered May 6, 2013 by sreemathi.v
edited May 6, 2013 by sreemathi.v
 

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