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  • Find the area of the region included between y^2 = 9x and y = x

Find the area of the region included between y^2 = 9x and y = x

Answers (1)

The equations y\textsuperscript{2} = 9x has no negative values of x. Therefore it lies on right of Y axis passing through (0, 0).

 

And y = x depicts a straight line through origin. 

For finding the area, picture shown below. 

 

 

 

 

For finding the point of interaction solve the two equations simultaneously. 
\\Put \: \: y=x \text{ in } y^{2}=9 x \Rightarrow x^{2}=9 x \Rightarrow x=9\\Put\: \: x=9 in y=x \text{ we get } y=9 \\

 

Therefore, point of interaction is (9, 9) 

 

The area between the parabola and line = area under parabola – area under line \ldots (1) \\

Let us find area under parabola

\\\Rightarrow y^{2}=9 x \\ \Rightarrow y=3 \sqrt{x} \\ \text{Integrate from 0 to 9}\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=\int_{0}^{9} 3 \mathrm{x}^{\frac{1}{2}} \mathrm{dx}\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=3\left[\frac{\mathrm{x} \frac{1}{2}+1}{\frac{1}{2}+1}\right]_{0}^{9}\\ \Rightarrow \int_{0}^{9} y \mathrm{dx}=3\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{9}\\ \\
\\\Rightarrow \int_{0}^{9} \mathrm{ydx}=3 \frac{2}{3}\left[9 \frac{3}{2}-0\right]\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=2\left(3^{2}\right)^{\frac{3}{2}}\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=2\left(3^{3}\right)\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=54\\

Now let us find area under straight line \mathrm{y}=\mathrm{x}

 y=x

Integrate from 0 to 9

\\\Rightarrow \int_{0}^{9} \mathrm{ydx}=\int_{0}^{9} \mathrm{xdx} \\\Rightarrow \int_{0}^{9} \mathrm{ydx}=\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{9} \\\Rightarrow \int_{0}^{9} \mathrm{ydx}=\left(\frac{9^{2}}{2}-0\right)\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=\frac{81}{2}\\ \Rightarrow \int_{0}^{9} y \mathrm{dx}=40.5

Using (i)

\Rightarrow  area between parabola and line  =54-40.5=13.5 unit ^{2}

Therefore, area was found to be 13.5 unit\textsuperscript{2}. \\

Posted by

infoexpert22

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