Question

If f = x^3 + y^3 + z^3 – 3xyz, find div grad f and curl grad f.

Answer 1

Answer: $\nabla . \bold F =6x+6y+6z$ and $\nabla \times \bold F=0$

Given: $f = x^3 + y^3 + z^3-3xyz$

To Find: div grad f and curl grad f.

Step-by-step explanation:

Step 1:

Given that $f = x^3 + y^3 + z^3-3xyz$

$gradient \ f = \bold F = \nabla f$

Gradient of a function is a vector field It is obtained by applying the vector operator $\nabla$ to the scalar function f. So we can calculate as given below,

$\bold F=\nabla (x^3 + y^3 + z^3-3xyz)$ and $\nabla = \frac{\partial }{\partial x} \hat i+\frac{\partial }{\partial y}\hat j +\frac{\partial }{\partial z}\hat k$

So,

$\bold F =(3x^2-3yz)\hat i+(3y^2-3xyz)\hat j+(3z^2-3xyz)\hat k$

The divergence of a vector field measures the density of change in the strength of the vector field. Now, div grad f is given below

$\nabla . \bold F = \frac{\partial}{\partial y} (3x^2-3yz)\hat i+ \frac{\partial}{\partial y}3y^2-3xyz)\hat j +\frac{\partial}{\partial y} (3z^2-3xyz)\hat k\\\\\; \; \;\; \;\nabla . \bold F=6x+6y+6z$

Step 2:

The curl of a vector field measures the rotation of the vector field about a point.

Now, curl grad f is given below

$\nabla \times \bold F= (\frac{\partial (3z^2+3xy)}{\partial y}-\frac{(3y^2+3xz)}{\partial z}) \hat i- (\frac{\partial (3z^2+3xy)}{\partial x}-\frac{(3x^2+3yz)}{\partial z}) \hat j+ (\frac{\partial (3y^2+3xz)}{\partial x}-\frac{(3x^2+3yz)}{\partial y}) \hat k\\\\\nabla \times \bold F=0$

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