Question

Question No. 1
The divergence of F = (xyz)i + (3x2y)j + (xz2 - y2z)k at (2,-1,1) is
Option
15
16
14
-16​

Answer 1

Answer:

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Answer 2

Answer:

The value of Divergence for the given equation is 14.

Step-by-step explanation:

If F(x,y) is a vector field, then its divergence is written as :

$divF(x,y)=del.F(r)\\del.F(r)=\frac{dabaF_{1} }{dabax} +\frac{dabaF_{2} }{dabay} +\frac{dabaF_{3} }{dabaz}$

Step 1 of 2

$del.F=\frac{deba(xyz)}{deba(x)}+ \frac{deba(3x2y)}{deba(y)}+\frac{deba(xz^{2}-y^{2}z )}{deba(z)}$

$=yz+6x+2zx-y^{2}$

Step 2 of 2

Now we will substitute the values of x, y, and z to get the divergence at given points for a given equation.

So as given

$x=2, y=-1,z=1$

We get the final divergence for the given equation as :

$Divergence =yz+6x+2zx-y^{2}\\=(-1)(1)+6(2)+2(1)(2)-(-1)^{2} \\=-1+12+4-1\\=14$

So we the value of divergence for the given equation as 14.