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  • Find the area of the region bounded by the curve y^2 = 2x and x^2 + y^2 = 4x.

Q: Find the area of the region bounded by the curve y^2 = 2x and x^2 + y^2 = 4x.

Answers (1)

y\textsuperscript{2} = 2x depicts a parabola having no negative values for x and lying to the right of the Y axis with passing through origin.

The other equation of x\textsuperscript{2} + y\textsuperscript{2} = 4x depicts a circle. 

The general equation of circle is given by x^{2}+y^{2}+2 g x+2 f y+c=0

Centre of circle is (-\mathrm{g},-\mathrm{f})$ \text{and radius is }$\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}

In x^{2}+y^{2}-4 x=0,2 g=-4 \Rightarrow g=-2$ 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. 
\\ \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 \\

Therefore, the point of interaction have been found to be (0, 0), (2, 2), and (2, -2) 

Below is the diagram of the area to be calculated. 

On integrating we will get area for 1st quadrant only, but since it is symmetrical we can multiple it by 2. 
Area of shaded region = area under the circle – area under the parabola  \ldots (1) \\

For finding the area under circle 
\\ 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}} \\

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 parabola, 
\\ \Rightarrow y^{2}=2 x \\ \Rightarrow y=\sqrt{2} \sqrt{x} \\

Integrate the above equation from 0 to 2 
\\ \Rightarrow \int_{0}^{2} y d x=\sqrt{2} \int_{0}^{2} x^{\frac{1}{2}} d x \\ \Rightarrow \quad \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} \\

\begin{aligned} &\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{y} \mathrm{d} \mathrm{x}=\frac{2^{\frac{3}{2}+\frac{3}{2}}}{3}\\ &\Rightarrow \int_{0}^{2} \mathrm{y} \mathrm{dx}=\frac{2^{3}}{3}\\ &\Rightarrow \int_{0}^{2} \mathrm{ydx}=\frac{8}{3}\\ &\text { Using (i) } \end{aligned} \\
\Rightarrow$ area of shaded in $1^{\text {st }}$ quadrant $=\pi-\frac{8}{3}$ unit $^{2}

After multiplying it by 2 

\text { The area required of shaded region }=2\left(\pi-\frac{8}{3}\right) \text { unit }^{2}

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