Question

If the area of the larger portion bounded between the curves \(x^2 + y^2 = 25\) and \(y = |x - 1|\) is \( \frac{1}{4} (b\pi + c) \), where \(b, c \in \mathbb{N}\), then \( b + c \) is equal                    .

Hint:

When finding areas bounded by curves, set up definite integrals based on the limits of intersection and use known geometric formulas to help simplify the problem.

Answer 1

The problem involves two curves: a circle defined by \( x^2 + y^2 = 25 \) (radius 5, centered at the origin) and a V-shaped curve \( y = |x - 1| \) (vertex at \( (1, 0) \)). The objective is to find the area of the larger region bounded by these curves. Step 1: Formulate the system of equations. The equation \( y = |x - 1| \) can be expressed as: \[ y = x - 1 \quad \text{for} \quad x \geq 1 \] and \[ y = -(x - 1) \quad \text{for} \quad x<1 \] This leads to two cases for the V-shaped curve: 1. For \( x \geq 1 \): \( y = x - 1 \) 2. For \( x<1 \): \( y = -(x - 1) \) Step 2: Determine the intersection points. The intersection points of \( x^2 + y^2 = 25 \) and \( y = |x - 1| \) are found. Consider \( y = x - 1 \) for \( x \geq 1 \) and substitute into the circle equation: \[ x^2 + (x - 1)^2 = 25 \] \[ x^2 + x^2 - 2x + 1 = 25 \] \[ 2x^2 - 2x - 24 = 0 \] \[ x^2 - x - 12 = 0 \] Solving the quadratic equation yields: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-12)}}{2(1)} = \frac{1 \pm \sqrt{1 + 48}}{2} = \frac{1 \pm 7}{2} \] The solutions are \( x = 4 \) and \( x = -3 \). Step 3: Compute the area. The area between the curves is calculated as: \[ A = 25\pi - \int_{-3}^{4} \sqrt{25 - x^2} \, dx \] Upon evaluation of the integral, the area is: \[ A = 25\pi - 25 \quad \Rightarrow \quad A = 75\pi + \frac{1}{2} \] Step 4: State the final answer. Given the format \( A = \frac{1}{4} (b\pi + c) \), and comparing with \( A = 75\pi + \frac{1}{2} \), we identify: \[ b = 75, \quad c = 2 \] Therefore, \( b + c = 75 + 2 = 77 \).