Trig Substitutions — Question 1

Question 1 – Solution

Geometrically, $y=\sqrt{a^2-x^2}$ is the upper semicircle $x^2+y^2=a^2$. On $0\le x\le a$, the region is one quarter of a disk of radius $a$. Hence its area is $$I=\frac14\pi a^2.$$ For an analytic verification, let $$x=a\sin\theta,\qquad dx=a\cos\theta\,d\theta.$$ Since $0\le x\le a$, the new bounds are $0\le\theta\le\pi/2$. Also, $$\begin{align*} \sqrt{a^2-x^2} &=\sqrt{a^2-a^2\sin^2\theta}\\ &=a\cos\theta, \end{align*}$$ because $\cos\theta\ge0$ on this interval. Therefore, $$\begin{align*} I&=a^2\int_0^{\pi/2}\cos^2\theta\,d\theta\\ &=\frac{a^2}{2}\int_0^{\pi/2}(1+\cos2\theta)\,d\theta\\ &=\frac{a^2}{2}\left[\theta+\frac12\sin2\theta\right]_0^{\pi/2} =\frac{\pi a^2}{4}. \end{align*}$$ Thus both methods give $$\boxed{\displaystyle \int_0^a\sqrt{a^2-x^2}\,dx=\frac{\pi a^2}{4}}.$$