Question

Let \(\Delta\) be the area of the region \(\{(x,y)∈R^2:x^2+y^2≤21,y^2≤4x,x≥1\}\). Then \(\frac{1}{2}(\Delta-21\text{ sin}^{-1} (\frac{2}{\sqrt7}))\) is equal to

• A. √3-4/3
• B. 2√3-1/3
• C. √3-2/3
• D. 2√3-2/3

Answer 1

The Correct Option is A

Step 1: Understanding the geometry of the region.
We are given the region bounded by the circle \( x^2 + y^2 \leq 21 \), and the line \( y^2 \leq 4x \), with \( x \geq 1 \). The region described is part of the circle and a sector of the ellipse formed by the line.
Step 2: Finding the area \( \Delta \).
The area of the region can be split into two parts:
1. The area of the sector bounded by the circle from \( x = 1 \) to \( x = 3 \),
2. The area of the circular region from \( x = 1 \) to \( x = 3 \), corresponding to the equation \( x^2 + y^2 \leq 21 \).
Thus, the area \( \Delta \) is given by the integral: \[ \Delta = \int_1^3 \left( 2\sqrt{x} \, dx \right) + \int_1^3 \sqrt{21 - x^2} \, dx. \] Evaluating the first part gives: \[ \Delta = \frac{8}{3} (3\sqrt{3} - 1) + 21 \sin^{-1} \left( \frac{2}{\sqrt{7}} \right) - 6\sqrt{3}. \] Step 3: Calculating the desired expression. The expression we need to evaluate is: \[ \frac{1}{2} \left( \Delta - 21 \sin^{-1} \left( \frac{2}{\sqrt{7}} \right) \right). \] Substitute the value of \( \Delta \) from the previous calculation: \[ \frac{1}{2} \left( \Delta - 21 \sin^{-1} \left( \frac{2}{\sqrt{7}} \right) \right) = \frac{1}{2} \left( 2\sqrt{3} - \frac{8}{3} \right). \] Simplifying: \[ \frac{1}{2} \left( 2\sqrt{3} - \frac{8}{3} \right) = \sqrt{3} - \frac{4}{3}. \] Thus, the correct value is \( \sqrt{3} - \frac{4}{3} \), and the correct answer is option (1).