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

We are given a circular table top with a radius of 8 cm. The equation of the circle is: \[ x^2 + y^2 = 64 \] The scratch passes through the origin and makes an angle of \( \frac{\pi}{4} \) with the positive x-axis. The equation of the scratch (a straight line) is: \[ y = x \] We need to find the area enclosed by the x-axis, the scratch, and the circular table top in the first quadrant. The limits of integration are determined by the points where the line intersects the circle. Substitute \( y = x \) into the equation of the circle: \[ x^2 + x^2 = 64 \] \[ 2x^2 = 64 \quad \Rightarrow \quad x^2 = 32 \quad \Rightarrow \quad x = 4\sqrt{2} \] Thus, the point of intersection is at \( x = 4\sqrt{2} \), and the corresponding value of \( y \) is also \( y = 4\sqrt{2} \). Now, the area in the first quadrant is given by the integral of the function \( y = x \) from \( x = 0 \) to \( x = 4\sqrt{2} \), minus the area under the curve of the circle in the same range. The area enclosed by the x-axis, the scratch, and the circle in the first quadrant is: \[ A = \int_0^{4\sqrt{2}} x \, dx \] This integral evaluates to: \[ A = \left[ \frac{x^2}{2} \right]_0^{4\sqrt{2}} = \frac{(4\sqrt{2})^2}{2} = \frac{32}{2} = 16 \, \text{cm}^2 \] Thus, the area of the region is \( \boxed{16 \, \text{cm}^2} \).