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 $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$. Here,$a^2 = 16 \implies a = 4$ and $b^2 = 9 \implies b = 3$. The ellips

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) The given equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$. Here,$a^2 = 16 \implies a = 4$ and $b^2 = 9 \implies b = 3$. The ellipse is symmetrical about both the $x-$axis and $y-$axis. Therefore,the total area bounded by the ellipse is $4 	imes$ (Area of the region in the first quadrant). Area in the first quadrant $= \int_{0}^{4} y \, dx$. From the equation,$y = 3 \sqrt{1 - \frac{x^2}{16}} = \frac{3}{4} \sqrt{16 - x^2}$. Area $= \int_{0}^{4} \frac{3}{4} \sqrt{16 - x^2} \, dx = \frac{3}{4} \left[ \frac{x}{2} \sqrt{16 - x^2} + \frac{16}{2} \sin^{-1} \left( \frac{x}{4} ight) ight]_{0}^{4}$. $= \frac{3}{4} \left[ (0 + 8 \sin^{-1}(1)) - (0 + 8 \sin^{-1}(0)) ight] = \frac{3}{4} \left[ 8 	imes \frac{\pi}{2} ight] = \frac{3}{4} [4 \pi] = 3 \pi$. Total area $= 4 	imes 3 \pi = 12 \pi$ square units.

Find the area of the region bounded by the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$. (in $\pi$)

Similar Questions

The area bounded by the curve $x(x^2 + p) = y - 1$ and the line $y = 1$ is:

The area (in sq. units) of the region bounded by the $X$-axis and the curve $y=1-x-6x^2$ is

The area of the region bounded by the curve $y=|x-5|$,the $X$-axis,and the lines $x=5$ and $x=6$ is . . . . . . sq. units.

The area of the region (in sq. units) enclosed by the curve $y=x^3-19x+30$ and the $x$-axis is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module