Question

Find the area of the region bounded by $ {y^2} = 9x $ , $ x = 2 $ , $ x = 4 $ and the x-axis in the first quadrant.

Answer 1

Hint: We will first plot the graph of the given equation. Since it is given in the question that we need to find the area only in the first quadrant, hence we find the limits of $ y $ in the first quadrant from the equation of the curve. Then we will integrate to find the area of the region required within the limit derived.

Complete step-by-step answer:
We have a given curve $ {y^2} = 9x $ . Now we will plot $ {y^2} = 9x $ , $ x = 2 $ , $ x = 4 $ on the graph as shown:

 The required area that we need to find is $ BCEF $ . This can be expressed as:
 $ {\rm{Area }}\,{\rm{of}}\;BCFE = \int\limits_2^4 {y \cdot dx} $
We have given a curve $ {y^2} = 9x $ .
We will root of the above equation as
 $ \begin{array}{l}
y = \pm \sqrt {9x} \\
y = \pm 3\sqrt x
\end{array} $
We can see that $ BCEF $ is in the first quadrant, we will only consider $ 3\sqrt x $ for $ y $ . Now, we will substitute $ 3\sqrt x $ for $ y $ in the expression of an area of $ BCFE $ .
 $ \begin{array}{l}
{\rm{Area }}\,{\rm{of}}\;BCFE = \int\limits_2^4 {y \cdot dx} \\
{\rm{Area }}\,{\rm{of}}\;BCFE = 3\int\limits_2^4 {\sqrt x \cdot dx} \\
{\rm{Area }}\,{\rm{of}}\;BCFE = 3\int\limits_2^4 {{x^{\dfrac{1}{2}}} \cdot dx}
\end{array} $
 On integrating the above expression we get,
 $ \begin{array}{l}
{\rm{Area }}\,{\rm{of}}\;BCFE = 3\left[ {\dfrac{{{x^{\dfrac{1}{2} + 1}}}}{{\dfrac{1}{2} + 1}}} \right]_2^4\\
{\rm{Area }}\,{\rm{of}}\;BCFE = 3\left[ {\dfrac{{{x^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}}} \right]_2^4\\
{\rm{Area }}\,{\rm{of}}\;BCFE = 3 \times \dfrac{2}{3}\left[ {{x^{\dfrac{3}{2}}}} \right]_2^4
\end{array} $
 We will substitute the limits in the above expression we will get,
 $ \begin{array}{l}
{\rm{Area }}\,{\rm{of}}\;BCFE = 2\left[ {{{\left( 4 \right)}^{\dfrac{3}{2}}} - {{\left( 2 \right)}^{\dfrac{3}{2}}}} \right]_2^4\\
{\rm{Area }}\,{\rm{of}}\;BCFE = 2\left[ {{{\left( 4 \right)}^{\dfrac{3}{2}}} - {{\left( 2 \right)}^{\dfrac{3}{2}}}} \right]_2^4\\
{\rm{Area }}\,{\rm{of}}\;BCFE = 2\left[ {{{\left( 2 \right)}^3} - {{\left( {{2^{\dfrac{1}{2}}}} \right)}^3}} \right]_2^4\\
{\rm{Area }}\,{\rm{of}}\;BCFE = 2\left[ {{{\left( 2 \right)}^3} - {{\left( {\sqrt 2 } \right)}^3}} \right]_2^4
\end{array} $
 We will simplify the above expression as ‘
 $ \begin{array}{l}
{\rm{Area }}\,{\rm{of}}\;BCFE = 2\left[ {8 - 2\sqrt 2 } \right]\\
{\rm{Area }}\,{\rm{of}}\;BCFE = 16 - 4\sqrt 2
\end{array} $
Hence, the area bound is $ 16 - 4\sqrt 2 $ square units.

Note: The first thing that is to be kept in mind is that we only need to find the rea in the first quadrant. Hence the limits of integration will be found according to this. Secondly, We should have prior knowledge about the plotting of curves and lines on the graph.