<s>[INST] Path Integral Problem Could anyone help understand why my answer disagrees with the given answer?
let the exact differential $dz = 3x^2(x^2+y^2)dx + 2y(x^3+y^4)dy$
Find $\int\limits_c dz$ from (1,2) to (2,1).
Here's my answer:
$\frac{\partial z}{\partial x}=3x^2(x^2+y^2)$
$\therefore$
$z = \int(3x^4+3y^2x^2) dx$
$\therefore$
$z = \frac{3}{5}x^5+y^2x^3+f(y)$
and
$\frac{\partial z}{\partial y} = 2y(x^3+y^4)$
$\therefore$
$z=\int(2x^3y+2y^5)dy$
$\therefore$
$z = x^3y^2+\frac{1}{3}y^6+g(x)$
so, in order to have $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial y}$ we need
$f(y)=\frac{1}{3}y^6$ and $g(x)=\frac{3}{5}x^5$
$\therefore$
$z = \frac{3}{5}x^5+y^2x^3+\frac{1}{3}y^6$
$\therefore$
$\int\limits_c dz$ from (1,2) to (2,1) = 
$\left[\frac{3}{5}x^5+y^2x^3+\frac{1}{3}y^6\right]_{(1,2)}^{(2,1)}$
= $(\frac{96}{5}+8+\frac{1}{3})-(\frac{3}{5}+4+\frac{64}{3})$
= $\frac{8}{5}$
But the answer given in the book I'm using is $\frac{9}{2}$
Thanks,
Mitch.
 [/INST] I write this only to confirm that I get the same $z$ as you (one could add an arbitrary constant, but that does not really change anything), and the same final answer. It looks like the book has a misprint.
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