Find the area of the region bounded by the el | vedclass

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

Question

(A) The given equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$.This can be written as $\frac{y^{2}}{9}=1-\frac{x^{2}}{4} \Rightarrow y^{

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) The given equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$.This can be written as $\frac{y^{2}}{9}=1-\frac{x^{2}}{4} \Rightarrow y^{2}=9\left(1-\frac{x^{2}}{4}ight) \Rightarrow y=3 \sqrt{1-\frac{x^{2}}{4}}$ (taking the positive value for the first quadrant).Since the ellipse is symmetrical about both the $x$-axis and $y$-axis,the total area is $4$ times the area in the first quadrant.Area $= 4 \int_{0}^{2} y \, dx = 4 \int_{0}^{2} 3 \sqrt{1-\frac{x^{2}}{4}} \, dx = 4 	imes \frac{3}{2} \int_{0}^{2} \sqrt{4-x^{2}} \, dx = 6 \int_{0}^{2} \sqrt{2^{2}-x^{2}} \, dx$.Using the formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1} \left(\frac{x}{a}ight)$:Area $= 6 \left[ \frac{x}{2} \sqrt{4-x^{2}} + \frac{4}{2} \sin^{-1} \left(\frac{x}{2}ight) ight]_{0}^{2}$$= 6 \left[ (0 + 2 \sin^{-1}(1)) - (0 + 2 \sin^{-1}(0)) ight] = 6 \left[ 2 	imes \frac{\pi}{2} ight] = 6 \pi$ square units.

Find the area of the region bounded by the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$. (in $\pi$)

Similar Questions

The area of the region bounded by the curve $y = \cos x$,$x = \frac{\pi}{2}$,$x = \frac{3\pi}{2}$,and $y = 0$ is . . . . . . sq. units.

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 . . . . . .

The area bounded by the parabola $y^2 = 2x$ and the ordinates $x = 1$ and $x = 4$ is:

The area bounded by the curve $y = \log x$,the $x$-axis,and the ordinates $x = 1$ and $x = 2$ is:

The area of the region bounded by the curve $y = 2^{kx}$ and the lines $x = 0$ and $x = 2$ in the first quadrant is $\frac{2}{\ln 2}$. Find the value of $k$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module