Area Ques 8
- The area (in sq units) of the region $(x, y): y^2 \geq 2 x$ and $x^2+y^2 \leq 4 x, x \geq 0, y \geq 0$, is
(2016 Main)
(a) $\pi-\frac{4}{3}$
(b) $\pi-\frac{8}{3}$
(c) $\pi-\frac{4 \sqrt{2}}{3}$
(d) $\frac{\pi}{2}-\frac{2 \sqrt{2}}{3}$
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Answer:
Correct Answer: 8.(b)
Solution: (b) Given equations of curves are $y^2=2 x$
which is a parabola with vertex $(0,0)$ and axis parallel to $X$-axis. $\quad$ ……..(i)
And $x^2+2 x=4 x$
which is a circle with centre $(2,0)$ and radius $=2$ $\quad$ ……..(ii)
On substituting $y^2=2 x$ in Eq. (ii), we get
$ \begin{aligned} x^2+2 x=4 x \Rightarrow x^2 & =2 x \Rightarrow x=0 \text { or } x=2 \\ y & =0 \text { or } y= \pm 2 \quad \text{[using Eq. (i)]} \end{aligned} $
Now, the required area is the area of shaded region, i.e.
$ \begin{aligned} \text { Required area } & =\frac{\text { Area of a circle }}{4}-\int_0^2 \sqrt{2 x} d x \\ & =\frac{\pi(2)^2}{4}-\sqrt{2} \int_0^2 x^{1 / 2} d x=\pi-\sqrt{2}\left[\frac{x^{3 / 2}}{3 / 2}\right]_0^2 \\ & =\pi-\frac{2 \sqrt{2}}{3}[2 \sqrt{2}-0]=\left(\pi-\frac{8}{3}\right) \text { sq unit } \end{aligned} $
BETA
