Q. 15.0( 4 Votes )

# Find the area of the region bounded by the curves y^{2} = 9x, y = 3x.

Answer :

y^{2} = 9x is the equation of a parabola and y = 3x is a straight line passing through origin

let us first roughly draw those equations

In y^{2} = 9x negative values of x are not allowed hence the graph is on right of X-axis that is the parabola opening to the right

To find point of intersection solve both the equations simultaneously

Put y = 3x in y^{2} = 9x

⇒ (3x)^{2} = 9x

⇒ 9x^{2} = 9x

⇒ x^{2} – x = 0

⇒ x(x – 1) = 0

⇒ x = 0 and x = 1

Put x = 1 and x = 0 in y = 3x we will get y = 3 and y = 0 respectively hence the parabola and straight line intersect at (1, 3) and (0, 0)

We have to find the area between parabola and straight line

That area will be the area under parabola minus the area under the straight line from x = 0 to x = 1 as shown in figure below

⇒ area between parabola and straight line = area under parabola – area under straight line …(i)

Let us calculate the area under parabola

y^{2} = 9x

⇒ y = 3√x

Integrate from 0 to 1

Now let us calculate area under the line y = 3x that is area of triangle OAB

⇒ y = 3x

Integrate from 0 to 1

Using (i)

⇒ area between parabola and straight line = 2 – = 1/2 unit^{2}

Hence area of the region bounded by the curves y^{2} = 9x and y = 3x is 1/2 unit^{2}

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