Answer
The total area of the region is $2a^3/3$
Work Step by Step
$y=x\sqrt{a^2-x^2}$ for $a>0$ and $y=0$
1) Draft the graph:
The graph is drafted below, for $a=3$. We would use integration with respect to $x$ here.
2) Find the limits of integration:
Here the region can be divided into 2 parts, from $x=-a$ to $x=0$ and from $x=0$ to $x=a$, with the curves being in reverse position with each other. However, since the two parts are symmetric with each other, we only need to calculate the area of one of them then double the result for the total area of the region.
We would calculate the area of the part from $x=0$ to $x=a$, so the upper limit is $a$ and the lower one is $0$.
3) Find the area:
The region from $x=0$ to $a$ is bounded above by $y=x\sqrt{a^2-x^2}$ and below by $y=0$. Therefore, the area of the region is $$A=\int^a_0x\sqrt{a^2-x^2}dx$$
- We set $a^2-x^2=u$, which means $$du=-2xdx$$ $$xdx=-\frac{du}{2}$$
- For $x=0$, $u=a^2$ and for $x=a$, $u=0$
So, $$A=-\frac{1}{2}\int^0_{a^2}\sqrt udu=-\frac{1}{2}\int^0_{a^2} u^{1/2}du$$ $$A=-\frac{1}{2}\Big(\frac{2u^{3/2}}{3}\Big)\Big]^0_{a^2}=-\frac{1}{3}u^{3/2}\Big]^0_{a^2}$$ $$A=-\frac{1}{3}(0-a^{3})$$ $$A=\frac{a^{3}}{3}$$
The total area of the region is $$T=2A=\frac{2a^3}{3}$$
