Question

Determine whether there is a function phi such that F = grad(phi), where:
a) F = (xz-y)i + (x^2y+z^3)j + (3xz2-xy)k.
b) F = 2xe^(-y)i + (cosz - x^2e^(-y))j - (ysinz)k. If so, find it.

Answer 1

Determine whether there is a function $\varphi(x, y, z)$ such that $F = \text{grad}\big(\varphi(x)\big)$, where

1) $F = (xz - y)i + (x^2 + z^3)j + (3xz^2 - xy)k$

**Solution:** Set initial point $(0,0,0)$.

So

$$\varphi(x, y, z) = \int_{0}^{x} P(t, 0, 0) dt + \int_{0}^{y} Q(x, t, 0) dt + \int_{0}^{z} R(x, y, t) dt,$$

where $P(x,y,z) = (xz - y), Q(x,y,z) = (x^2 + z^3), R(x,y,z) = (3xz^2 - xy)$.

$$\begin{aligned} \varphi(x, y, z) &= \int_{0}^{x} (0 - 0) dt \\ &+ \int_{0}^{y} (x^2) dt \\ &+ \int_{0}^{z} (3xt^2 - xy) dt = x^2 t \left| \begin{array}{c} y \\ 0 \end{array} \right| + (xt^3 - xyt) \left| \begin{array}{c} z \\ 0 \end{array} \right| = x^2 y + xz^3 - xyz. \end{aligned}$$

**Answer:** $\varphi(x, y, z) = x^2 y + xz^3 - xyz$

2) $F = 2xe^{-y}i + (\cos z - x^2 e^{-y})j - y\sin zk$

**Solution:** Set initial point $(0,0,0)$.

So

$$\varphi(x, y, z) = \int_{0}^{x} P(t, 0, 0) dt + \int_{0}^{y} Q(x, t, 0) dt + \int_{0}^{z} R(x, y, t) dt,$$

where $P(x,y,z) = 2xe^{-y}, Q(x,y,z) = (\cos z - x^2 e^{-y})j, R(x,y,z) = y\sin z$.

$$\begin{array}{l} \varphi(x, y, z) \\ = \int_{0}^{x} 2 \, t \, dt \\ + \int_{0}^{y} (1 - x^{2} e^{-t}) \, dt \\ + \int_{0}^{z} y \, s \, i n t \, dt \\ = t^{2} \left|_{0}^{x} + (t - x^{2}) \right|_{0}^{y} \\ = x^{2} e^{-y} + 2y - y \cos z. \end{array}$$

Answer: $\varphi(x, y, z) = x^{2} e^{-y} + 2y - y \cos z$