Question

Area bounded by the curves $x^2+y^2\le 8x$ and $y^2\ge 4x$ lies in the first quadrant is equal to

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

Answer 1

Answer: (B)

Given $x^2+y^2\le 8x$
Let $x^2+y^2=8x$, a circle with centre $(4,0)$ and radius $4\dots \left(i\right)$
$y^2\ge 4x$, $y^2=4x$ a parabola with vertex $(0,0)$ $\dots \left(ii\right)$
$\therefore$ From $(i)$ and $(ii)$, we have $x^2=4x$
$\Rightarrow x=0,x=4$

Required area = area of shaded region
$=\underset{0}{\overset{4}{\int }}\sqrt{4^2-{\left(x-4\right)}^2}-\underset{0}{\overset{4}{\int }}2\sqrt{x}dx$
$={\left[\frac{\left(x-4\right)}{2}\sqrt{4^2-{\left(x-4\right)}^2}+\frac{16}{2}si{n}^{-1}\left(\frac{x-4}{4}\right)\right]}_0^4-2\times \frac{2}{3}{\left[x^{3/2}\right]}_0^4$
$=8\times \frac{\pi }{2}-\frac{4}{3}\times {\left(4\right)}^{3/2}$
$=\left(4\pi -\frac{32}{3}\right)=\frac{32}{3}\left(\frac{3\pi }{8}-1\right)$ sq. units