Question

What is the area enclosed between the curves ${y^2} = 12x$ and the lines $x = 0$ and $y = 6$?
A. $2$ sq. unit
B. $4$ sq. unit
C. $6$ sq. unit
D. $8$ sq. unit

Answer 1

Hint: In this question, we are given the equation of curve and the lines. Firstly, calculate the coordinate of the point of intersection of the curve and the line using any of the method (substitution, elimination, or graphical method). Then, plot the graph of required point and the equation. Now, for the area integrate the difference of the equation of line and curve in terms of $x$ with respect to $dx$ from $x = 0$ to $x = 3$ and solve it further using integration formulas. Also, resolve the limits properly.



Formula Used:Integration formula –
$\int {cdx = cx} $ where $c$ is the constant
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $



Complete step by step solution:Given that,
Equation of the curve ${y^2} = 12x$ and the line $x = 0$, $y = 6$
Now, the intersection point of the curve and the line $y = 6$
At $y = 6$,
Put in ${y^2} = 12x$ we get $x = 3$
The coordinate of the point of intersection is $\left( {3,6} \right)$
Graph of the given equations and the required coordinates is attached below (figure 1);


Figure 1: A graph contains the equation of curve and the line
Now, ${y^2} = 12x$ is also written as $y = \sqrt {12x} $
To calculate the area, integrate the difference of line and the curve with respect to $dx$ from $x = 0$ to $x = 3$
Therefore, the area of the bounded by them will be
$A = \int\limits_0^3 {\left( {6 - \sqrt {12x} } \right)dx} $
As we know that, $\int {cdx = cx} $ where $c$ is the constant and $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
It implies that,
$A = \left[ {6x - \sqrt {12} \left( {\dfrac{{2{x^{\dfrac{3}{2}}}}}{3}} \right)} \right]_0^3$
On resolving the limits, we get
$A = 6\left( 3 \right) - \sqrt {12} \left( {\dfrac{{2{{\left( 3 \right)}^{\dfrac{3}{2}}}}}{3}} \right) - 6\left( 0 \right) - \sqrt {12} \left( {\dfrac{{2{{\left( 0 \right)}^{\dfrac{3}{2}}}}}{3}} \right)$
Here, ${3^{\dfrac{3}{2}}} = {\left( {{3^3}} \right)^{\dfrac{1}{2}}} = \sqrt {{3^3}} = \sqrt {27} $
So, $A = 18 - \sqrt {12} \left( {\dfrac{{2\sqrt {27} }}{3}} \right)$
$A = 18 - \sqrt {12} \left( {\dfrac{{2\sqrt {3 \times 3 \times 3} }}{3}} \right)$
$A = 18 - \left( {\dfrac{{2 \times 3\sqrt {12 \times 3} }}{3}} \right)$
$A = 18 - 2\left( {\sqrt {36} } \right)$
$A = 18 - 2\left( 6 \right)$
On solving, we get $A = 4$ sq. units



Option ‘B’ is correct

Note: Different methods are used to determine the area under the curve, with the antiderivative method being the most prevalent. The area under the curve can be calculated by knowing the curve's equation, borders, and the axis surrounding the curve. There exist formulas for obtaining the areas of regular shapes such as squares, rectangles, quadrilaterals, polygons, and circles, but no formula for finding the area under a curve. The integration procedure aids in solving the equation and determining the required area. Antiderivative methods are highly useful for determining the areas of irregular planar surfaces.