91Ó°ÊÓ

${âˆ\neg }_{\mathcal{R}}\left(3xy-4x^{2}y\right)dA$

Here, the region R is the unit disk. In polar coordinates, the region of integration is described as follows,

$R=\left\{\right(r,\mathrm{θ})âˆ\pounds 0≤r≤1,0≤\mathrm{θ}≤2\mathrm{Ï€}\}.$

In this case,

$$\begin{gathered} x=r\cos \mathrm{θ},y=r\sin \mathrm{θ},x^2+y^2=r^2,\text{and} \\ dA=rdrd\mathrm{θ}. \end{gathered}$$

Then, evaluate the required integral as follows:

$$\begin{gathered} {âˆ\neg }_R\left(3xy-4x^{2}y\right)dA \\ ={∫}_0^{2\mathrm{Ï€}}{∫}_0^1\left(3\left(r\cos \mathrm{θ}\right)\left(r\sin \mathrm{θ}\right)-4\left(r\cos \mathrm{θ}{)}^2\right(r\sin \mathrm{θ})\right)rdrd\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}{∫}_0^1\left(3r^2\sin \mathrm{θ}\cos \mathrm{θ}-4r^3\sin \mathrm{θ}{\cos }^2\mathrm{θ}\right)rdrd\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}{∫}_0^1\left(3r^3\sin \mathrm{θ}\cos \mathrm{θ}-4r^4\sin \mathrm{θ}{\cos }^2\mathrm{θ}\right)drd\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}\left[{∫}_0^1\left(3r^3\sin \mathrm{θ}\cos \mathrm{θ}-4r^4\sin \mathrm{θ}{\cos }^2\mathrm{θ}\right)dr\right]d\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}{\left[\frac{3r^4}{4}\sin \mathrm{θ}\cos \mathrm{θ}-\frac{4r^5}{5}\sin \mathrm{θ}{\cos }^2\mathrm{θ}\right]}_0^{1}d\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}\left[\left(\frac{3·1^4}{4}\sin \mathrm{θ}\cos \mathrm{θ}-\frac{4·1^5}{5}\sin \mathrm{θ}{\cos }^2\mathrm{θ}\right)-\left(\frac{3·0^4}{4}\sin \mathrm{θ}\cos \mathrm{θ}-\frac{4·0^5}{5}\sin \mathrm{θ}{\cos }^2\mathrm{θ}\right)\right]d\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}\left[\left(\frac{3}{4}\sin \mathrm{θ}\cos \mathrm{θ}-\frac{4}{5}\sin \mathrm{θ}{\cos }^2\mathrm{θ}\right)-0\right]d\mathrm{θ} \\ ={∫}_0^{2\mathrm{Ï€}}\left(\frac{3}{4}\cos \mathrm{θ}-\frac{4}{5}{\cos }^2\mathrm{θ}\right)\sin \mathrm{θ}d\mathrm{θ} \\ =-{\left[\frac{3}{4}·\frac{{\cos }^2\mathrm{θ}}{2}-\frac{4}{5}·\frac{{\cos }^3\mathrm{θ}}{3}\right]}_0^{2\mathrm{Ï€}} \\ =-\left[\left(\frac{3}{4}·\frac{{\cos }^{2}2\mathrm{Ï€}}{2}-\frac{4}{5}·\frac{{\cos }^{3}2\mathrm{Ï€}}{3}\right)-\left(\frac{3}{4}·\frac{{\cos }^{2}0}{2}-\frac{4}{5}·\frac{{\cos }^{3}0}{3}\right)\right] \\ =-\left[\left(\frac{3}{4}·\frac{1^2}{2}-\frac{4}{5}·\frac{1^3}{3}\right)-\left(\frac{3}{4}·\frac{1^2}{2}-\frac{4}{5}·\frac{1^3}{3}\right)\right] \\ =0 \end{gathered}$$

Therefore, the required integral is ${âˆ\neg }_R\left(3xy-4x^{2}y\right)dA=0$.