Question 16
Find the area of the region bounded by the curve $y^2 = 2x$ and $x^2 + y^2 = 4x$.
Answer:
$y^2=2 x$ depicts a parabola having no negative values for x and lying to the right of the Y axis while passing through the origin. The other equation of $x^2+y^2=4 x$ depicts a circle.
The general equation of a circle is given by $x^2+y^2+2 g x+2 f y+c=0$
Centre of circle is $(-g,-f)$ and radius is $\sqrt{g^2+f^2-c}$
$
\text { In } x^2+y^2-4 x=0,2 g=-4 \Rightarrow g=-2 \text { and } f=c=0
$
Hence center is $(-(-2), 0)$ that is $(2,0)$ and radius is $\sqrt{(-2)^2+0^2-0}$ which is 2
For the point of interaction, solve the two equations simultaneously.
$
\begin{aligned}
& \text { Put } y^2=2 x \text { in } x^2+y^2=4 x \\
& \Rightarrow x^2+2 x=4 x \\
& \Rightarrow x^2-2 x=0 \\
& \Rightarrow x(x-2)=0 \\
& \Rightarrow x=0 \text { and } x=2 \\
& \text { Put } x=2 \text { in } y^2=2 x \\
& \Rightarrow y= \pm 2
\end{aligned}
$
Therefore, the point of interaction are $(0,0),(2,2)$, and $(2,-2)$
Below is the diagram of the area to be calculated.
On integrating, we will get an area for the 1st quadrant only, but since it is symmetrical, we can multiply it by.
Area of shaded region $=$ area under the circle - area under the parabola $\ldots$ (1)To findg the area under circle
$
\begin{aligned}
& x^2+y^2=4 x \\
& \Rightarrow y^2=4 x-x^2 \\
& \Rightarrow y=\sqrt{4 x-x^2} \\
& \Rightarrow y=\sqrt{-\left(-4 x+x^2+4-4\right)} \\
& \Rightarrow y=\sqrt{-\left(\left(x^2-4 x+4\right)-2^2\right)} \\
& \Rightarrow y=\sqrt{-\left((x-2)^2-2^2\right)} \\
& \Rightarrow y=\sqrt{2^2-(x-2)^2}
\end{aligned}
$
Integrate the above equation from 0 to 2
$
\Rightarrow \int_0^2 y d x=\int_0^2 \sqrt{2^2-(x-2)^2} d x
$
$
\text { Using } \int \sqrt{a^2-x^2}=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}
$
$
\Rightarrow \int_0^2 y d x=\left[\frac{x-2}{2} \sqrt{2^2-(x-2)^2}+\frac{2^2}{2} \sin ^{-1} \frac{x-2}{2}\right]_0^2
$
$
\Rightarrow \int_0^2 \mathrm{ydx}=\left[0-\left(\frac{0-2}{2} \sqrt{2^2-(0-2)^2}+\frac{2^2}{2} \sin ^{-1} \frac{0-2}{2}\right)\right]
$
$
\Rightarrow \int_0^2 y d x=\left[-\left(0+2 \sin ^{-1}(-1)\right)\right]
$
$
\Rightarrow \int_0^2 y d x=\pi
$
For the area under the parabola,
$
\begin{aligned}
& \Rightarrow y^2=2 x \\
& \Rightarrow y=\sqrt{2} \sqrt{x}
\end{aligned}
$
Integrate the above equation from 0 to 2
$
\begin{aligned}
& \Rightarrow \int_0^2 y d x=\sqrt{2} \int_0^2 x^{\frac{1}{2}} d x \\
& \Rightarrow \int_0^2 y d x=\sqrt{2}\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_0^2 \\
& \Rightarrow \int_0^2 y d x=2^{\frac{1}{2}}\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^2 \\
& \Rightarrow \int_0^2 \mathrm{ydx}=2^{\frac{1}{2}} \times \frac{2}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_0^2 \\
& \Rightarrow \int_0^2 \mathrm{ydx}=\frac{2^{\frac{1}{2}+1}}{3}\left(2^{\frac{3}{2}}-0\right) \\
& \Rightarrow \int_0^2 \mathrm{ydx}=\frac{2^{\frac{3}{2}+\frac{3}{2}}}{3} \\
& \Rightarrow \int_0^2 \mathrm{ydx}=\frac{2^3}{3} \\
& \Rightarrow \int_0^2 \mathrm{ydx}=\frac{8}{3}
\end{aligned}
$
Using (i)
$\Rightarrow$ area of shaded in $1^{\text {st }}$ quadrant $=\pi-\frac{8}{3}$ unit $^2$
After multiplying it by 2
The area required of shaded region $=2\left(\pi-\frac{8}{3}\right)$ unit $^2$
