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Calculate the divergence of the following vector functions:

(a)va=x2x^+3xz2y^-2xzz^(b)vb=xyx^+2yzy^+3zxz^(c)vc=y2x^+(2xy+z2)+2yzz^

Short Answer

Expert verified

(a) The divergence is 0.

(b) The divergence is 3x+y+2z.

(c) The divergence is2(x+y).

Step by step solution

01

Write the expression for the divergence of the function.

Determine the divergence of the function.

.v=x^x+y^y+z^z(vxx^+vyy^+vzz^)=vxx+y^vyy+z^vzz

02

Solve for the divergence of part (a).

Consider the given vector is,

va=x2x^+3xz2y-2xzz^

Solve for the divergence.

.va=xx2+y3xz2-z2xz=2x+0-2x=0


Therefore, the divergence is 0.

03

Solve for the divergence of part (b).

Consider the given vector is,

vb=xyx^+2yz2y+3zxz^

Solve for the divergence.

.va=xxy+y2yz-zxz=3x+y+2z

Therefore, the divergence is 3x+y+2z

04

Solve for the divergence of part (c).

Consider the given vector is,

va=y2x^+(2xy+z2)+2yzz^

Solve for the divergence.

.va=xy2+y(2xy+z2)+z2yz=0+(2x+0)+2y=2(x+y)

Therefore, the divergence is2(x+y).

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