Why Is the Calculated Charge Enclosed Within the Cube Incorrect?

[vrinda mukund]

The displacement vector in a cube is defined by the relation D =(4x*xy + z)i+(4xy)j-(z)k c/m2 . Then charge enclosed within the cube of side 1m is …………..
I know that the displacement vector holds the relation
∇ . D = ρ
Where ρ is the charge density. I found ∇ . D and calculated its volume integral with x,y and z varying between 0 and 1. What I got was 3 coloumb but the correct answer is 4 colomb. Can someone help me with this?

 

[vanhees71]

It's not exactly clear to me, how your $\vec{D}$ reads. Please use the LaTeX formulas. I read it as follows

$$\vec{D}=\begin{pmatrix}<br /> 4 x^2 y +z\\<br /> 4 x y \\<br /> z<br /> \end{pmatrix}$$

Then

$$\vec{\nabla} \cdot \vec{D}=4xy+4x+1.$$

The next question is, where your cube is located. I used Mathematica to calculate for your cube, which you give as $[0,1] \times [0,1] \times [0,1]$. Then I get 5 (whatever units you use, since these are also inconsistent).

 

[vrinda mukund]

sir, actually D is having -z as k component. ∇ . D = 8xy + 4x -1 then.(u have written it as 4xy+4x+1). nothing is mentioned in the question about the location of the cube. i didn't understand your [0 X 1] technique sir. can you please explain?