Question

Calculate the divergence of the vector field f
A)f=(xy,yz,y^(2)-x^(3))
B)f=(x,y,z)
C)f=(x-2z(x^(2)),z-xy,(z^(2)x^(2))

Answer 1

Answer on Question #52627 – Math – Vector Calculus

Calculate the divergence of the vector field $f$:

a) $f = (xy, yz, y^2 - x^3)$;

b) $f = (x, y, z)$;

c) $f = (x - 2zx^2, z - xy, z^2x^2)$.

Solution

First of all, the definition of divergence is the following:

$$div \, f = \frac{\partial f_x}{\partial x} + \frac{\partial f_y}{\partial y} + \frac{\partial f_z}{\partial z}.$$

Now, we can calculate divergence of the vector field $f$ in three cases:

a) $div \, f = \frac{\partial(xy)}{\partial x} + \frac{\partial(yz)}{\partial y} + \frac{\partial(y^2 - x^3)}{\partial z} = y \frac{\partial(x)}{\partial x} + z \frac{\partial(y)}{\partial y} + 0 = y + z$;

b) $div \, f = \frac{\partial(x)}{\partial x} + \frac{\partial(y)}{\partial y} + \frac{\partial(z)}{\partial z} = 1 + 1 + 1 = 3$;

c) $div \, f = \frac{\partial(x - 2zx^2)}{\partial x} + \frac{\partial(z - xy)}{\partial y} + \frac{\partial(z^2x^2)}{\partial z} = \left(\frac{\partial x}{\partial x} - 2z \frac{\partial(x^2)}{\partial x}\right) + \left(\frac{\partial z}{\partial y} - x \frac{\partial y}{\partial y}\right) + x^2 \frac{\partial(z^2)}{\partial z} = 1 - 4zx - x + 2zx^2$.

Answer:

a) $div \, f = y + z$;

b) $div \, f = 3$;

c) $div \, f = 1 - 4zx - x + 2zx^2$.

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