The area (in square units) lying in the first | vedclass

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

Question

(A) The given circle is $x^2+y^2=4$,which has a radius $r=2$. The area is bounded by $x=0$ (the $y$-axis) and $x=2$ in the first quadrant. The area $A

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(A) The given circle is $x^2+y^2=4$,which has a radius $r=2$. The area is bounded by $x=0$ (the $y$-axis) and $x=2$ in the first quadrant. The area $A$ is given by the integral of $y$ with respect to $x$ from $0$ to $2$. Since $y^2 = 4-x^2$,we have $y = \sqrt{4-x^2}$ in the first quadrant. $A = \int_0^2 \sqrt{4-x^2} dx$ Using the standard integral formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a}) + C$: $A = \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}(\frac{x}{2}) ight]_0^2$ $A = \left( \frac{2}{2}\sqrt{4-4} + 2\sin^{-1}(1) ight) - \left( 0 + 2\sin^{-1}(0) ight)$ $A = (0 + 2 	imes \frac{\pi}{2}) - 0 = \pi 	ext{ square units.}$

The area (in square units) lying in the first quadrant and bounded by the circle $x^2+y^2=4$ and the lines $x=0$ and $x=2$ is

Similar Questions

The area bounded by the curve $y = x |\sin(\frac{x}{2})|$ and the $X$-axis between the lines $x = 0$ and $x = 4\pi$ (in square units) is: (in $\pi$)

Area of the region bounded by curve $y^2=x$,$X$-axis and lines $x=1$ and $x=4$ in the first quadrant is . . . . . . sq. units.

The area bounded by the circle $x^{2}+y^{2}=16$ and the lines $x=0$ and $x=2$ is

Area enclosed by the ellipse $2x^2 + 3y^2 = 1$ is . . . . . . sq. units.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module