Question

Determine the area under the curve y = `sqrt(a^2-x^2)` included between the lines x = 0 and x = a.

Answer 1

We have,

\[y = \sqrt{a^2 - x^2}\]

\[ \Rightarrow y^2 = a^2 - x^2 \]

\[ \Rightarrow x^2 + y^2 = a^2 \]

\[\text{ Since in the given equation }x^2 + y^2 = a^2, \text{ all the powers of both } x\text{ and }y\text{ are even, the curve is symmetrical about both the axis }. \]

\[ \therefore\text{ Required area = area enclosed by circle in first quadrant }\]

\[(a, 0 ), ( - a, 0)\text{ are the points of intersection of curve and }x -\text{ axis }\]

\[(0, a), (0, - a)\text{ are the points of intersection of curve and }y -\text{ axis }\]

\[\text{ Slicing the area in the first quadrant into vertical stripes of height }= \left| y \right|\text{ and width }= dx\]

\[ \therefore\text{ Area of approximating rectangle }= \left| y \right| dx\]

\[\text{ Approximating rectangle can move between }x = 0\text{ and }x = a\]

\[A =\text{ Area of enclosed curve in first quadrant }= \int_0^a \left| y \right| dx\]

\[ \Rightarrow A = \int_0^a \sqrt{a^2 - x^2} d x\]

\[ = \left[ \frac{1}{2}x\sqrt{a^2 - x^2} + \frac{1}{2} a^2 \sin^{- 1} \frac{x}{a} \right]_0^a \]

\[ = \frac{1}{2} a^2 \sin^{- 1} 1\]

\[ = \frac{1}{2} a^2 \frac{\pi}{2} \]

\[ = \frac{a^2 \pi}{4}\text{ sq units }\]