Question

. Find the area included between
x^2+y^2= 4x
and
y^2=x
above x-axis

Answer 1

Find point of intersection:

$x^2=3x\implies x=3,y=\sqrt{3}$

$S=S_1+S_2$

where $S_1$ is area under the graph of $x=y^2$ ,

$S_2$ is area under the graph of $x^2+y^2=4x$ at $x\isin [3,4]$

$S_1=\int^3_0 \sqrt{x}dx=\frac{2x\sqrt{x}}{3}|^3_0=2\sqrt{3}$

S₂ is half of area of sector with central angle $\pi/3$ minus area of triangle with vertices: (2,0),(3,0),(3,$\sqrt{3}$ ).

$S_2=R^2\alpha/4-\sqrt{3}/2$

where $R=2$ ,

$\alpha=tan^{-1}(\sqrt{3})=\pi/3$

$S_2=4\pi/12-\sqrt{3}/2=\pi/3-\sqrt{3}/2$

$S=2\sqrt{3}+\pi/3-\sqrt{3}/2=\pi/3+1.5\sqrt{3}=3.645$