line vector in conservative vector field. inconsistent results.

Question

Studying a problem related to line integral in vector field in conservative field.

$\vec{F} = (2xy + z^3)\hat{\text{i}} + (x^2)\hat{\text{j}} + (3xz^2)\hat{\text{k}}$

$\phi = x^2y +xz^3$

$\vec{F} = \nabla \phi$

find line integral from point (1,-2,1) to (3,1,4)

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here's the solution the textbook gives:

$d\vec{r} = dx\hat{\text{i}} + dy\hat{\text{j}} + dz\hat{\text{k}}$

$\text{work done} = \int \limits^{P_2}_{P_1} \vec{F} \cdot d\vec{r}$

$\text{work done} = \int \limits^{P_2}_{P_1} (2xy+z^3)dx + (x^2)dy + (3xz^2)dz$

now for the part i don't understand:

$\text{work done} = \int \limits^{P_2}_{P_1} d(x^2y+xz^3) = ... = 202$

how did they go from dx and dy to d(...)? and what does d(...) syntax mean?

when i try to do this integral i get:

which is off by a factor of 2... and not in agreement with the result found using just the scalar potential...

$\phi(3,1,4) - \phi(1,-2,1) = 202$

Answer 1

The syntax means,

$$df(x,y,z)=\frac{df}{dx}dx + \frac{df}{dy}dy +\frac{df}{dz}dz$$

So, if $f(x,y,z) =x^2y+xz^3$,

$$d(x^2y+xz^3) = (2xy+z^3)dx + x^2dy + 3xz^2dz$$

Edit:

You were integrating over $x$, $y$ and $z$ independently as if they are unrelated. Specifically, you integrate over $x$ holding $y$ and $z$ constant, which is incorrect. All three variables change simultaneously along the path $\phi$. The factor of 2 is merely an coincidence.

Answer 2

By an easy inspection we get that $\phi(x,y,z)=x^{2}y+z^{3}x$.

It is clear that $\bigtriangledown\,\phi=f$ . Since the field is conservative, we have:

$I=\phi(3,1,4)-\phi(1,-2,1)=202$, if my calculations are correct!!