
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
163.5k+ views
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.
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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Instantaneous Velocity - Formula based Examples for JEE

JEE Advanced 2025 Notes

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Total MBBS Seats in India 2025: Government and Private Medical Colleges
