Area enclosed by the parabola ay = 3(a^2 - x^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given equation of the parabola is $ay = 3(a^2 - x^2)$,which can be written as $y = \frac{3}{a}(a^2 - x^2)$.The parabola meets the $x$-axis whe

Vedclass · Accepted Answer

$4a^2 	ext{ sq. units}$

Vedclass · Answer

(A) The given equation of the parabola is $ay = 3(a^2 - x^2)$,which can be written as $y = \frac{3}{a}(a^2 - x^2)$.The parabola meets the $x$-axis where $y = 0$,so $\frac{3}{a}(a^2 - x^2) = 0$,which gives $x^2 = a^2$,or $x = \pm a$.The required area $A$ is given by the integral of $y$ with respect to $x$ from $x = -a$ to $x = a$:$A = \int_{-a}^{a} \frac{3}{a}(a^2 - x^2) dx$.Since the function is even,we can write:$A = 2 \int_{0}^{a} \frac{3}{a}(a^2 - x^2) dx = \frac{6}{a} \int_{0}^{a} (a^2 - x^2) dx$.Evaluating the integral:$A = \frac{6}{a} [a^2x - \frac{x^3}{3}]_{0}^{a} = \frac{6}{a} (a^3 - \frac{a^3}{3}) = \frac{6}{a} (\frac{2a^3}{3}) = 4a^2 	ext{ sq. units}$.

Area enclosed by the parabola $ay = 3(a^2 - x^2)$ and $x$-axis is

Similar Questions

Let the area of the bounded region $\{(x, y): 0 \leq 9x \leq y^2, y \geq 3x-6\}$ be $A$. Then $6A$ is equal to . . . . . .

Area of the region bounded by the curve $y=x|x|$,$X$-axis and lines $x=0$ and $x=1$ is . . . . . . sq. units.

The area of the region bounded by $y=||x-3|-4|-5$ and the $X$-axis is

The area of the region bounded by the curve $y = -2\sqrt{x}$ and the lines $x = 0$,$x = 1$,and $y = 0$ is . . . . . . sq. units.

The area of the region bounded by the curve $y = \sin(\pi x)$ and the $X$-axis for $x \in [0, 2]$ is . . . . . . sq. units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module