The area of the region { (x, y) : x^2 + y^2 l | vedclass

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

Question

(D) The region is defined by the interior of the circle $x^2 + y^2 = 1$ and the region above the line $x + y = 1$.First,we find the intersection point

Vedclass · Accepted Answer

$\frac{\pi}{4} - \frac{1}{2}$

Vedclass · Answer

(D) The region is defined by the interior of the circle $x^2 + y^2 = 1$ and the region above the line $x + y = 1$.First,we find the intersection points of the circle $x^2 + y^2 = 1$ and the line $x + y = 1$:Substituting $y = 1 - x$ into the circle equation:$x^2 + (1 - x)^2 = 1$$x^2 + 1 + x^2 - 2x = 1$$2x^2 - 2x = 0$$2x(x - 1) = 0$This gives $x = 0$ and $x = 1$.For $x = 0$,$y = 1$. For $x = 1$,$y = 0$.So the intersection points are $A(1, 0)$ and $B(0, 1)$.The required area is the area under the circle arc minus the area under the line segment from $x = 0$ to $x = 1$:Area $= \int_0^1 (\sqrt{1 - x^2} - (1 - x)) \, dx$$= \left[ \frac{x\sqrt{1 - x^2}}{2} + \frac{1}{2} \sin^{-1}(x) - x + \frac{x^2}{2} \right]_0^1$$= \left( \frac{1 \cdot 0}{2} + \frac{1}{2} \sin^{-1}(1) - 1 + \frac{1}{2} \right) - (0 + 0 - 0 + 0)$$= \frac{1}{2} \cdot \frac{\pi}{2} - \frac{1}{2} = \frac{\pi}{4} - \frac{1}{2}$.

The area of the region $\{ (x, y) : x^2 + y^2 \le 1 \le x + y \}$ is

Similar Questions

The area of the region bounded by the curve $y=\sin x$ between $x=-\pi$ and $x=\frac{3\pi}{2}$ is

The area of the smaller part between the circle $x^2+y^2=4$ and the line $x=1$ is . . . . . . sq. units.

The area (in sq. units) of the region $A = \{(x, y) : (x-1)[x] \leq y \leq 2\sqrt{x}, 0 \leq x \leq 2\}$,where $[t]$ denotes the greatest integer function,is

The area bounded by the curve $y=x^3-3x^2+2x$ and the $X$-axis is (in square units)

The area bounded by the parabola $y=x^2$ and the line $y=x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module