Answer :
Answer:
The answer is "40"
Step-by-step explanation:
Given value:
[tex]\ vector \ field: \\ \\ \bar F =x^2i+y^2j+zk = pi+Qj+Rk[/tex]
Divergence Theorem:
[tex]\iint_{S}^{}\bar F ds= \iiint_{E}^{} \bigtriangledown f dv[/tex]
[tex]\bigtriangledown f = \frac{\partial p}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\\\\[/tex]
[tex]=2x+2y+1[/tex]
the coordinates x=2, y=2, and z=2
[tex]=\int\limits^2_0 \int\limits^2_0 \int\limits^2_0 {(2x+2y+1)} \, dz \ dy \ dx \\\\[/tex]
[tex]=\int\limits^2_0 \int\limits^2_0 {((2x+2y+1)z)}^{2}_0 \, \ dy \ dx \\\\=\int\limits^2_0 \int\limits^2_0 {2(2x+2y+1)} \, \ dy \ dx \\\\[/tex]
[tex]=2\int\limits^2_0 {(2xy+y^2+y)^2_0} \, dx \\\\=2\int\limits^2_0 {(2xy+4+2)^2_0} \, dx \\\\=2[\frac{4x^2}{2}+4x+2x]^2_0\\\\=2[2x^2+6x]^2_0\\\\=2[2(2)^2+6(2)]\\\\=2[8+12]\\\\=2[20]\\\\=40[/tex]