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Question
Determine the area under the curve y = `sqrt(a^2-x^2)` included between the lines x = 0 and x = a.
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Solution
We have,
\[y = \sqrt{a^2 - x^2}\]
\[ \Rightarrow y^2 = a^2 - x^2 \]
\[ \Rightarrow x^2 + y^2 = a^2 \]
\[\text{ Since in the given equation }x^2 + y^2 = a^2, \text{ all the powers of both } x\text{ and }y\text{ are even, the curve is symmetrical about both the axis }. \]
\[ \therefore\text{ Required area = area enclosed by circle in first quadrant }\]
\[(a, 0 ), ( - a, 0)\text{ are the points of intersection of curve and }x -\text{ axis }\]
\[(0, a), (0, - a)\text{ are the points of intersection of curve and }y -\text{ axis }\]
\[\text{ Slicing the area in the first quadrant into vertical stripes of height }= \left| y \right|\text{ and width }= dx\]
\[ \therefore\text{ Area of approximating rectangle }= \left| y \right| dx\]
\[\text{ Approximating rectangle can move between }x = 0\text{ and }x = a\]
\[A =\text{ Area of enclosed curve in first quadrant }= \int_0^a \left| y \right| dx\]
\[ \Rightarrow A = \int_0^a \sqrt{a^2 - x^2} d x\]
\[ = \left[ \frac{1}{2}x\sqrt{a^2 - x^2} + \frac{1}{2} a^2 \sin^{- 1} \frac{x}{a} \right]_0^a \]
\[ = \frac{1}{2} a^2 \sin^{- 1} 1\]
\[ = \frac{1}{2} a^2 \frac{\pi}{2} \]
\[ = \frac{a^2 \pi}{4}\text{ sq units }\]
