Question
Determine the area under the curve $\text{y}=\sqrt{\text{a}^2-\text{x}^2}$ included between the lines x = 0 and x = a.
✓
Answer
We have $\text{y}=\sqrt{\text{a}^2-\text{x}^2}$ $\Rightarrow\ \text{y}^2=\text{a}^2-\text{x}^2$ $\Rightarrow\ \text{x}^2+\text{y}^2=\text{a}^2$ Graph of above function is a semi-circle lying above x-axis. The graph is as shown is the following figure.
From the figure, are of shaded region, $\text{A}=\int\limits^\text{a}_0\sqrt{\text{a}^2-\text{x}^2}\text{ dx}$ $=\bigg[\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\bigg]^\text{a}_0$ $=\bigg[0+\frac{\text{a}^2}{2}\sin^{-1}1-0\frac{\text{a}^2}{2}\sin^{-1}0\bigg]$ $=\frac{\text{a}^2}{2}\cdot\frac{\pi}{2}=\frac{\pi\text{a}^2}{4}\text{ sq. units}$
From the figure, are of shaded region, $\text{A}=\int\limits^\text{a}_0\sqrt{\text{a}^2-\text{x}^2}\text{ dx}$ $=\bigg[\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\bigg]^\text{a}_0$ $=\bigg[0+\frac{\text{a}^2}{2}\sin^{-1}1-0\frac{\text{a}^2}{2}\sin^{-1}0\bigg]$ $=\frac{\text{a}^2}{2}\cdot\frac{\pi}{2}=\frac{\pi\text{a}^2}{4}\text{ sq. units}$
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