Question

The area bounded by the curves $x^2+y^2\le 8$ and $y^2\ge 4x$ lying in the first quadrant is not equal to

• A. $32(\frac{\pi }{8}-\frac{1}{3})$
• B. $\frac{32}{3}(\frac{3\pi }{8}-1)$
• C. $4\pi -\frac{32}{3}$
• D. $\frac{1}{3}(12\pi -32)$

Answer 1

The correct option is B $\frac{32}{3}(\frac{3\pi }{8}-1)$

Given $x^2+y^2\le 8x$

Let $x^2+y^2=8x$ ...(1)

$y^2\ge 4x,y^2=4x$ ...(2)

From (1) and (2), we have

$x^2=4x\Rightarrow x=0,x=4$

$\therefore$ Set of points $($ or points of intersection $)$ are $(0,0),(4,4)$.

Required area $={\int }_0^4\sqrt{4^2-{(x-4)}^2}-{\int }_0^{4}2\sqrt{x}dx(\because x^2+y^2=8x)$

$\Rightarrow {(x-4)}^2+y^2=4^2\Rightarrow y^2=4^2-{(x-4)}^2$

$={[\frac{(x-4)}{2}\sqrt{4^2-{(x-4)}^2}+\frac{16}{2}{\sin }^{-1}\left(\frac{x-4}{4}\right)]}_0^4-2\times \frac{2}{3}{\left[x^{3/2}\right]}_0^4$

$=8\times \frac{\pi }{2}-\frac{4}{3}\times {\left(4\right)}^{3/2}=(4\pi -\frac{32}{3})=\frac{32}{3}(\frac{3\pi }{8}-1)$ sq. units.