The area of the region bounded by the curve y | vedclass

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

Question

(D) The given equation is $y = 2 \sqrt{1 - x^2}$.Squaring both sides,we get $y^2 = 4(1 - x^2)$,which implies $x^2 + \frac{y^2}{4} = 1$.This is the equ

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(D) The given equation is $y = 2 \sqrt{1 - x^2}$.Squaring both sides,we get $y^2 = 4(1 - x^2)$,which implies $x^2 + \frac{y^2}{4} = 1$.This is the equation of an ellipse with semi-major axis $a = 2$ (along the $Y$-axis) and semi-minor axis $b = 1$ (along the $X$-axis).Since the curve is $y = 2 \sqrt{1 - x^2}$,it represents the upper half of the ellipse.The area bounded by the curve and the $X$-axis is the area of the upper semi-ellipse.The area of the full ellipse is given by $A = \pi ab = \pi 	imes 1 	imes 2 = 2 \pi$.Therefore,the area of the upper semi-ellipse is $\frac{1}{2} 	imes 2 \pi = \pi$ sq. units.

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

Similar Questions

Find the area of the region bounded by $x^{2}=4y$,$y=2$,$y=4$ and the $y-$axis in the first quadrant.

Find the area of the region bounded by $y^{2}=9x$,$x=2$,$x=4$ and the $x$-axis in the first quadrant.

The area bounded by the $x$-axis and the curve $y = \sin x$ between $x = 0$ and $x = \pi$ is:

Find the area of the region bounded by the curve $y^{2}=x$,the lines $x=1$,$x=4$,and the $x$-axis in the first quadrant.

If the straight line $x=b$ divides the area enclosed by $y=(1-x)^2$,$y=0$,and $x=0$ into two parts $R_1 (0 \leq x \leq b)$ and $R_2 (b \leq x \leq 1)$ such that $R_1 - R_2 = \frac{1}{4}$,then $b$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module