Question

The area of the upper half of the circle whose equation is \( (x - 1)^2 + y^2 = 1 \) is given by

• A. \( \frac{\pi}{4} \) sq units
• B. \( \int_0^2 \sqrt{2 - x^2} \, dx \) sq units
• C. \( \int_0^2 \sqrt{2x - x^2} \, dx \) sq units
• D. \( \int_0^1 \sqrt{2x - x^2} \, dx \) sq units

Hint:

When finding areas under curves or geometric shapes, using integrals with appropriate limits and expressions for the curve simplifies the process.

Answer 1

The Correct Option is C

The given equation of the circle is: \[ (x - 1)^2 + y^2 = 1 \] This represents a circle with center \( (1, 0) \) and radius 1. To find the area of the upper half of the circle, we need to solve for \( y \) in terms of \( x \): \[ y^2 = 1 - (x - 1)^2 \] Taking the square root, we get: \[ y = \sqrt{1 - (x - 1)^2} \] The area of the upper half of the circle is the integral of this expression with respect to \( x \) over the interval from \( x = 0 \) to \( x = 2 \): \[ \text{Area} = \int_0^2 \sqrt{1 - (x - 1)^2} \, dx \] Expanding the expression \( (x - 1)^2 \) and simplifying: \[ \text{Area} = \int_0^2 \sqrt{2x - x^2} \, dx \] Thus, the area of the upper half of the circle is given by the integral: \[ \int_0^2 \sqrt{2x - x^2} \, dx \]