Question

The area of the region 
\(\begin{array}{l} \left\{\left(x,y\right);\left|x-1\right|\leq y \leq \sqrt{5-x^2} \right\}\end{array}\)
is equal to

• A. \(\frac{5}{2}sin^{−1}⁡(\frac{3}{5})−\frac{1}{2}\)
• B. \(\frac{5π}{4}−\frac{3}{2}\)
• C. \(\frac{3π}{4}+\frac{3}{2}\)
• D. \(\frac{5π}{4}−\frac{1}{2}\)

Answer 1

The Correct Option is D

The given problem requires finding the area of a region defined by the set of inequalities: 

\(\left\{\left(x,y\right); \left|x-1\right|\leq y \leq \sqrt{5-x^2} \right\}\)

Step 1: Understand the inequalities

1. The inequality \(\left|x-1\right| \leq y\) implies two conditions:
   • \(x-1 \leq y\)
   • \(y \leq \sqrt{5-x^2}\) represents the upper half of a circle centered at the origin with a radius of \(\sqrt{5}\).

Step 2: Intersecting the inequalities

1. The region of integration is the intersection of these two conditions, hence lies within the quarter circle and between the lines \(y = x - 1\) and \(y = 1 - x\).

Step 3: Calculate the area

The area bounded by these inequalities is calculated by integrating over the quarter-circle. This involves converting the Cartesian coordinates to polar coordinates for simplification.

Step 4: Use symmetry and geometry

1. The area enclosed by the arc of the circle from \(x = -\sqrt{5}\) to \(x = \sqrt{5}\) is a sector of \(90^\circ\) or \(\pi/4\) radians.
2. Calculate the area of the sector: 

\[\frac{\pi (\sqrt{5})^2}{4} = \frac{5\pi}{4}\]

1. The area below the lines is given by the triangular area under the line \(y = x - 1\) over the range \([-1,1]\): 

\[\int_{-1}^{1} (x-1) \, dx = \left[ \frac{x^2}{2} - x \right]_{-1}^{1} = \frac{-1}{2}\]

Step 5: Combine areas

The area of the region is:

\[\text{Total Area} = \text{Area of sector} - \text{Area below lines} = \frac{5\pi}{4} - \left(-\frac{1}{2}\right) = \frac{5\pi}{4} - \frac{1}{2}\]

 

Conclusion

Thus the area of the given region is \(\frac{5\pi}{4} - \frac{1}{2}\).

Answer 2

\(\begin{array}{l} A=\displaystyle\int\limits_{-1}^1\left(\sqrt{5-x^2}-\left(1-x\right)\right)dx+\displaystyle\int\limits_{1}^2\left(\sqrt{5-x^2}-\left(x-1\right)\right)dx\end{array}\)
\(\begin{array}{l} \left.A=2\left(\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}\sin^{-1}\frac{x}{\sqrt{5}}\right)-2x\right|_0^1 \left.+\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}\sin^{-1}\frac{x}{\sqrt{5}}-\frac{x^2}{2}+x\right|_1^2\end{array}\)
\(\begin{array}{l} =\left(\frac{5\pi}{4}-\frac{1}{2}\right)\text{ sq. units} \end{array}\)