Area Under the Curve
Area enclosed by circle and parabola
Grade 12
Question:
<p>The area of the region, enclosed by the circle \(x^2 + y^2 = 2\) which is not common to the region bounded by the parabola \(y^2 = x\) and the straight line \(y = x\), is</p>
<p>(a) \(\frac{1}{3}(12\pi - 1)\)</p>
<p>(b) \(\frac{1}{6}(12\pi - 1)\)</p>
<p>(c) \(\frac{1}{6}(24\pi - 1)\)</p>
<p>(d) \(\frac{1}{3}(6\pi - 1)\)</p>
Step-by-Step Solution
Key Concept: Find the area between the parabola and line, then subtract from the circle's area to get the region not common to both.
<p><strong>Solution:</strong> Let the area of the region, enclosed by the parabola $y^2 = x$ and straight line $y = x$ be:</p><p>$$A = \int_0^1 (\sqrt{x} - x)\,dx$$</p><p>$$= \left[\frac{x^{3/2}}{3/2} - \frac{x^2}{2}\right]_0^1 = \left(\frac{2}{3} - \frac{1}{2}\right) = \frac{1}{6}$$</p><p>The area of the circle with radius $r = \sqrt{2}$ is $2\pi$.</p><p>The area not common to the region is $\frac{1}{6}(12\pi - 1)$.</p>
Correct Answer: B