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Q4:
Area of the region bounded by the curve $y^2 = 4x$, $y$-axis and the line $y = 3$ is
Solution :
Given Variables & Initial Setup
The region is bounded by:
- The parabola: $y^2 = 4x$
- The $y$-axis: $x = 0$
- The horizontal line: $y = 3$
To find the area with respect to the $y$-axis, we express $x$ in terms of $y$ from the given equation of the parabola:
$x = \frac{y^2}{4}$ [By algebraic rearrangement]
Step 1: Establishing the Limits of Integration
The region starts from the origin where $y = 0$ and extends up to the line $y = 3$. Therefore, the definite integral limits are $[0, 3]$.
Step 2: Defining the Integral for Area
The area $A$ under the curve bounded by the $y$-axis is given by the integral of $x$ with respect to $y$:
$A = \int_{c}^{d} x \, dy = \int_{0}^{3} \frac{y^2}{4} \, dy$
Step 3: Calculating the Integral
Applying the Power Rule for integration, $\int y^n \, dy = \frac{y^{n+1}}{n+1}$:
$A = \frac{1}{4} \int_{0}^{3} y^2 \, dy$
$A = \frac{1}{4} \left[ \frac{y^3}{3} \right]_{0}^{3}$ [By the Fundamental Theorem of Calculus]
$A = \frac{1}{4} \left( \frac{3^3}{3} - \frac{0^3}{3} \right)$
$A = \frac{1}{4} \left( \frac{27}{3} \right)$
$A = \frac{1}{4} (9)$
$A = \frac{9}{4} \text{ sq. units}$
Visual Representation
Final Solution: The area of the region is $\frac{9}{4}$ or $2.25 \text{ square units}$.
More Questions from Class 12 Mathematics Application of Integrals EXERCISE 8.1
- Q1: Find the area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
- Q2: Find the area of the region bounded by the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$.
- Q3: Area lying in the first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ is
CBSE Solutions for Class 12 Mathematics Application of Integrals
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