Area Under the Curve
Area bounded by parabola and lines
Grade 12
Question:
<p>The area of the region bounded by <equation>x^2 = 4y</equation>, <equation>y = 2</equation>, <equation>y = 4</equation> and the Y-axis in the first quadrant is</p>
<p>(a) <equation>8(\sqrt{4} - \sqrt{2})</equation> sq units</p>
<p>(b) <equation>8(\sqrt{4} + \sqrt{2})</equation> sq units</p>
<p>(c) <equation>\frac{8}{3}(\sqrt{4} - \sqrt{2})</equation> sq units</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: When integrating with respect to y, express x in terms of y and use the Y-axis bounds. The parabola is symmetric about Y-axis, so consider the first quadrant portion.
<p><strong>Solution:</strong> To determine the required area, integrate <equation>x</equation> w.r.t. <equation>y</equation> and take <equation>y = 2</equation> as lower limit and <equation>y = 4</equation> as upper limit.</p><p>The given curve <equation>x^2 = 4y</equation> is a parabola, which is symmetrical about the Y-axis.</p><p>From <equation>x^2 = 4y</equation>, we get <equation>x = 2\sqrt{y}</equation> (considering first quadrant)</p><p>Required area = <equation>\int_2^4 x \, dy = \int_2^4 2\sqrt{y} \, dy = 2\left[\frac{2y^{3/2}}{3}\right]_2^4</equation></p><p><equation>= \frac{4}{3}[4^{3/2} - 2^{3/2}] = \frac{4}{3}[8 - 2\sqrt{2}] = \frac{8}{3}(4 - \sqrt{2})</equation> sq units</p><p>∴ Answer is (c)</p>
Correct Answer: C