Respuesta :
Answer:
Required total charge is [tex]256\pi[/tex] coulombs per square meter.
Step-by-step explanation:
Given electric charge is dristributed over the disk,
[tex]x^2=y^2\leq 16[/tex] so that the charge density at (x,y) is,
[tex]\rho (x,y)=2x+2y+2x^2+2y^2[/tex]
To find total charge on the disk let Q be the total charge and [tex]x=r\cos\theta,y=r\sin\theta[/tex] so that,
[tex]Q={\int\int}_Q\rho(x,y) dA[/tex] where A is the surface of disk.
[tex]=\int_{0}^{2\pi}\int_{0}^{4}(2x+2y+2x^2+2y^2)dA[/tex]
[tex]=\int_{0}^{2\pi}\int_{0}^{4}(2r\cos\theta+2r\sin\theta+2r^2 \cos^{2}\theta+2r^2\sin^2\theta)rdrd\theta[/tex]
[tex]=2\int_{0}^{2\pi}\int_{0}^{4}r^2(\cos\theta+\sin\theta)drd\theta+2\int_{0}^{2\pi}\int_{0}^{4}r^3drd\theta[/tex]
[tex]=\frac{2}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)\Big[r^3\Big]_{0}^{4}d\theta+2\int_{0}^{2\pi}\Big[\frac{r^4}{4}\Big]d\theta[/tex]
[tex]=\frac{128}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)d\theta+128\int_{0}^{2\pi}d\theta[/tex]
[tex]=\frac{128}{3}\Big[\sin\theta-\cos\theta\Big]_{0}^{2\pi}+128\times 2\pi[/tex]
[tex]=\frac{128}{3}\Big[\sin 2\pi-\cos 2\pi-\sin 0+\cos 0\Big]+256\pi[/tex]
[tex]=256\pi[/tex]
Hence total charge is [tex]256\pi[/tex] coulombs per square meter.
