For what value(s) of $a$ is $$ \int_0 ^a \sqrt {9-x^2} \ dx = \pi \ \text{ ?} $$
I solved this question by using integral substitution, but I was told that it is better to obtain this solution by using the graphichal method.
I tried it, but I wasn't able to have any useful insight.
When I did it using trignomteric substituton, I ended up with the following equation for $a$, which I wasn't sure how to simplify (hence I need for a better method):
$$ \Rightarrow \arcsin \left(\frac{a}{3}\right) + \frac {1}{2} \sin \left[ 2 \arcsin \left(\frac{a}{3} \right) \right] = \frac {2}{9} \pi $$
Graphically, the integral $$ \int_0 ^a \sqrt {9-x^2} \ dx = \pi $$
represents the shaded area in the diagram, which is the area sum of the circle sector of angle $x$ and the right triangle of base length $a=r\sin x$, i.e.
$$\pi = \frac12 x r^2 + \frac 12 r^2\sin x\cos x$$
where $r=3$ is the radius of the circle. Rewrite the equation as,
$$x + \frac12\sin 2x = \frac{2\pi}9$$
which does not have a close form solution and can only be solved numerically. But, a decent approximation can be obtained with $\sin 2x \approx 2x$ to obtain $x = \frac\pi9$. Thus, $a = 3\sin \frac\pi9 = 1.03$, compared with exact numerical result $1.07$