Question

Evaluate the following integral.
int_0^a square root a^2-x^2 dx

Answer 1

that is the area under the circle with radius a, a quarter circle

Answer 2

Using geometry you can find the integral (the area ) without doing any calculus, what is the area of a circle with radius `a` ?

Answer 3

Thank you

Answer 4

Or:
\[
\begin{align*}
&\quad\int_0^a\sqrt{a^2-x^2}\,dx\\
&\qquad x^2=a^2\sin^2(\theta)\\
&\qquad x=a\sin(\theta)\\
&\qquad dx=a\cos(\theta)\,d\theta\\
&=\int_{x=0}^{x=a}\sqrt{a^2-a^2\sin^2(\theta)}\left(a\cos(\theta)\right)\,d\theta\\
&=\int_{x=0}^{x=a}a^2\cos^2(\theta)\,d\theta\\
&=a^2\int_{x=0}^{x=a}\frac{\cos(2\theta)+1}{2}\,d\theta\\
&=a^2\left[\frac{1}{4}\sin(2\theta)+\frac{1}{2}\theta\right]_{x=0}^{x=a}\\
&\qquad x=a\sin(\theta)\\
&\qquad\theta=\arcsin\left(\frac{x}{a}\right)\\
&\qquad x=a\implies \theta=\frac{\pi}{2}\\
&\qquad x=0\implies \theta=0\\
&=a^2\left[\frac{1}{4}\sin(2\theta)+\frac{1}{2}\theta\right]_{0}^{\frac{\pi}{2}}\\
&=a^2\left[\frac{1}{4}\sin(\pi)+\frac{1}{4}\pi-0\right]\\
&=\frac{1}{4}\pi a^2 \text{ as required}
\end{align*}
\]