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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals

The area of the region in the first quadrant enclosed by the x-axis,the line y=x and the circle $x^2+y^2=32$ is \begin{array}{1 1}(A)\;16\pi\; sq.units & (B)\;4\pi \;sq.units\\(C)\;32\pi\; sq.units & (D)\;24\pi\;sq.units \end{array}

1 Answer

  • 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)$$+c$
Step 1:
Given curves are $x^2+y^2=32$ and the line $y=x$
The points of intersection can be obtained as follows:
$x^2+x^2=32\Rightarrow 2x^2=16$
$x^2=4\Rightarrow x=\pm 4$
The points of intersection are $(0,0)$ and $(4,4)$
Hence the required area is the shaded portion shown in the fig.
A=Area of the triangle OAC and the curve ABC.
$\;\;\;=\large\frac{1}{2}$$\times OC\times AC+\int_4^{4\sqrt 2}\sqrt{32-x^2}$
Area of the triangle $OAC=\large\frac{1}{2}$$\times 4\times 4=8$ sq. units=$A_1$
Step 2:
Area of the curve $ABC=\int_4^{4\sqrt 2}\sqrt{32-x^2}dx.$
$\;\;=\begin{bmatrix}\large\frac{x}{2}\normalsize\sqrt{32-x^2}+\large\frac{32}{2}\normalsize\sin^{-1}\big(\large\frac{x}{4\sqrt 2}\big)\end{bmatrix}_4^{4\sqrt 2}$
On applying limits we get,
$\;\;=\large\frac{4\sqrt 2}{2}$$\sqrt {32-32}+16\sin^{-1}\big(\large\frac{4\sqrt 2}{4\sqrt 2}\big)-\large\frac{4}{2}$$\sqrt{32-16}+$$16\sin^{-1}\big(\large\frac{4}{4\sqrt 2}\big)$
$\;\;=2\sqrt 2\times 0+16\sin^{-1}(1)-2\sqrt{16}-16\sin^{-1}\big(\large\frac{1}{\sqrt 2}\big)$
But $\sin^{-1}(1)=\large\frac{\pi}{2}$ and $\sin^{-1}\big(\large\frac{1}{\sqrt{2}}\big)=\large\frac{\pi}{4}$
$\;\;=16.\large\frac{\pi}{2}$$-2\sqrt {16}+16\large\frac{\pi}{4}$
$\;\;=8\pi-2\times 4-4\pi=4\pi-8$=$A_2$
Step 3:
Hence the total area $A=A_1+A_2$
$\qquad\qquad\qquad\qquad=8+4\pi-8=4\pi$ sq.units.
Hence B is the correct option.
answered May 6, 2013 by sreemathi.v
edited May 6, 2013 by sreemathi.v

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