Question

if the force field defined by vector F=(3x^2yz-3y)i^ + (x^3z - 3x)j^ + (x^3y+2z)k^ conservative? if so, find the scalar potenial associated with the vector F.

Answer 1

Answer on Question #40755, Math, Vector Calculus

If the force field defined by vector F = (3x^2yz - 3y)i^ + (x^3z - 3x)j^ + (x^3y + 2z)k^ conservative? if so, find the scalar potential associated with the vector $F$.

Solution

$$\vec{F} = (3x^2yz - 3y)\vec{i} + (x^3z - 3x)\vec{j} + (x^3y + 2z)\vec{k} = M(x,y,z)\vec{i} + N(x,y,z)\vec{j} + P(x,y,z)\vec{k}$$

Then, $\vec{F}$ is conservative if and only if

$$\frac{\partial P}{\partial x} = \frac{\partial M}{\partial z}, \frac{\partial N}{\partial x} = \frac{\partial M}{\partial y}, \frac{\partial P}{\partial y} = \frac{\partial N}{\partial z}.$$ $$\frac{\partial P}{\partial x} = \frac{\partial}{\partial x}(x^3y + 2z) = 3x^2y.$$ $$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x^3y + 2z) = x^3.$$ $$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(3x^2yz - 3y) = 3x^2z - 3.$$ $$\frac{\partial M}{\partial z} = \frac{\partial}{\partial z}(3x^2yz - 3y) = 3x^2y.$$ $$\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x^3z - 3x) = 3x^2z - 3.$$ $$\frac{\partial N}{\partial z} = \frac{\partial}{\partial z}(x^3z - 3x) = x^3.$$

So $\frac{\partial P}{\partial x} = \frac{\partial M}{\partial z} = 3x^2y$, $\frac{\partial N}{\partial x} = \frac{\partial M}{\partial y} = 3x^2z - 3$, $\frac{\partial P}{\partial y} = \frac{\partial N}{\partial z} = x^3$ and the force field $\vec{F}$ is conservative.

Let's find the scalar potential $f$ associated with the vector $\vec{F}$

$$\frac{\partial f}{\partial x} = M = (3x^2yz - 3y),$$ $$\frac{\partial f}{\partial y} = N = (x^3z - 3x),$$ $$\frac{\partial f}{\partial z} = P = (x^3y + 2z).$$

If we integrate the first of the three equations with respect to $x$, we find that

$$f(x,y,z) = \int (3x^2yz - 3y)dx = x^3yz - 3yx + g(y,z).$$

where $g(y,z)$ is a constant dependent on $y$ and $z$ variables. We then calculate the partial derivative with respect to $y$ from this equation and match it with the equation of above.

$$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^3yz - 3yx + g(y,z)) = x^3z - 3x + \frac{\partial g}{\partial y} = (x^3z - 3x).$$

This means that the partial derivative of $g$ with respect to $y$ is 0, thus eliminating $y$ from $g$ entirely and leaving at as a function of $z$ alone.

$$f(x, y, z) = x^3 y z - 3 y x + h(z).$$

We then repeat the process with the partial derivative with respect to $z$

$$\frac{\partial f}{\partial z} = \frac{\partial}{\partial z} \left(x^3 y z - 3 y x + h(z)\right) = x^3 y + \frac{d h}{d z} = (x^3 y + 2 z)$$

which means that

$$\frac{d h}{d z} = (2 z)$$

so we can find $h(z)$ by integrating:

$$h(z) = z^2 + c.$$

Therefore,

$$f(x, y, z) = x^3 y z - 3 y x + z^2 + c.$$