Question

A woman discovered a scratch along a straight line on a circular table top of radius 8 cm. She divided the table top into 4 equal quadrants and discovered the scratch passing through the origin inclined at an angle \( \frac{\pi}{4} \) anticlockwise along the positive direction of x-axis. Find the area of the region enclosed by the x-axis, the scratch and the circular table top in the first quadrant, using integration.

Hint:

Quick Tip: When calculating the area of a region enclosed by a circle and a line, set up the appropriate integral for the area under the line, and subtract the area under the circle in the given range.

Answer 1

A circular table top has a radius of 8 cm, with its equation as \[ x^2 + y^2 = 64 \]. A scratch, represented by a straight line passing through the origin at an angle of \( \frac{\pi}{4} \) with the positive x-axis, has the equation \( y = x \). The objective is to determine the area in the first quadrant bounded by the x-axis, the scratch, and the circular table top. The intersection points of the line and the circle define the integration limits. Substituting \( y = x \) into the circle's equation yields \( x^2 + x^2 = 64 \), which simplifies to \( 2x^2 = 64 \), resulting in \( x^2 = 32 \) and \( x = 4\sqrt{2} \). Consequently, the intersection point is at \( x = 4\sqrt{2} \), with a corresponding \( y = 4\sqrt{2} \). The area in the first quadrant is calculated by integrating the function \( y = x \) from \( x = 0 \) to \( x = 4\sqrt{2} \). The area enclosed by the x-axis, the scratch, and the circle in the first quadrant is given by the integral: \[ A = \int_0^{4\sqrt{2}} x \, dx \] Evaluating this integral gives: \[ A = \left[ \frac{x^2}{2} \right]_0^{4\sqrt{2}} = \frac{(4\sqrt{2})^2}{2} = \frac{32}{2} = 16 \, \text{cm}^2 \] The area of the specified region is \( \boxed{16 \, \text{cm}^2} \).