The area of the shaded region in the given fi | vedclass

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

Question

(A) The equation of the ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$,which can be written as $\frac{x^2}{4^2}+\frac{y^2}{3^2}=1$. Here,$a=4$ and $b=3$.

Vedclass · Accepted Answer

$3(\pi-2)$

Vedclass · Answer

(A) The equation of the ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$,which can be written as $\frac{x^2}{4^2}+\frac{y^2}{3^2}=1$. Here,$a=4$ and $b=3$.From the figure,the shaded region is bounded by the ellipse in the first quadrant,specifically between $x=0$ and $x=4$,and the line connecting $(0,3)$ and $(4,0)$.The equation of the line passing through $(0,3)$ and $(4,0)$ is $\frac{x}{4}+\frac{y}{3}=1$,which gives $y = \frac{3}{4}(4-x)$.The area of the shaded region is the area under the ellipse minus the area under the line in the first quadrant.Area $= \int_{0}^{4} \frac{3}{4}\sqrt{16-x^2} \, dx - \int_{0}^{4} \frac{3}{4}(4-x) \, dx$.$= \frac{3}{4} \left[ \frac{x}{2}\sqrt{16-x^2} + \frac{16}{2}\sin^{-1}(\frac{x}{4}) \right]_{0}^{4} - \frac{3}{4} \left[ 4x - \frac{x^2}{2} \right]_{0}^{4}$.$= \frac{3}{4} [0 + 8(\frac{\pi}{2})] - \frac{3}{4} [16 - 8] = 3\pi - 6 = 3(\pi-2)$ sq. units.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

The area of the shaded region in the given figure of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is . . . . . . sq. units.

Similar Questions

The values of a function $f(x)$ at different values of $x$ are as follows:
$x$ $0$ $1$ $2$ $3$ $4$ $5$
$f(x)$ $2$ $3$ $6$ $11$ $18$ $27$

Then, the approximate area (in square units) bounded by the curve $y=f(x)$ and the $x$-axis between $x=0$ and $x=5$, using the Trapezoidal rule, is:

Area of the region bounded by $y=|x|$ and $y=-|x|+2$ is

The area of the region bounded by the curve $y = \tan x$,the $X$-axis,and the line $x = \frac{\pi}{3}$ is

The area lying between the curves $y^{2}=4x$ and $y=2x$ is

Area of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module