Answer
$$\int3y\sqrt{7-3y^2}dy=-\frac{(7-3y^2)^{3/2}}{3}+C$$
Work Step by Step
$$A=\int3y\sqrt{7-3y^2}dy$$
We set $u=7-3y^2$.
Then $$du=-6ydy=-2\times3ydy$$
That means, $$3ydy=-\frac{1}{2}du$$
Therefore, $$A=-\frac{1}{2}\int\sqrt udu=-\frac{1}{2}\int u^{1/2}du$$ $$A=-\frac{1}{2}\times\frac{u^{3/2}}{\frac{3}{2}}+C=-\frac{u^{3/2}}{3}+C$$ $$A=-\frac{(7-3y^2)^{3/2}}{3}+C$$