# integrate $\frac{\partial f}{\partial x}=\cos \left[ (x+a)(y+b) \right]$

If a function f(x,y) satisfies: $$\frac{\partial f}{\partial x}=\cos \left[ \left((\tan ^{-1} (y/x)+a\right)(\sqrt{x^2+y^2}+b) \right]\\ \frac{\partial f}{\partial y}=\sin \left[ \left((\tan ^{-1} (y/x)+a\right)(\sqrt{x^2+y^2}+b) \right]$$ how to write f(x,y) as an explicit function of x and y by integration? I'm struggling with multivariable integral.

A simpler form: $$\frac{\partial f}{\partial x}=\cos \left[ \left(\theta+a\right)(r+b) \right]\\ \frac{\partial f}{\partial y}=\sin \left[ \left(\theta+a\right)(r+b) \right]\\ \theta=\tan ^{-1} (y/x)\\ r=\sqrt{x^2+y^2}$$ but how to put things together? Help!

• Are you sure such a function exists? Try computing $\frac{\partial^2f}{\partial x\partial y}$ and $\frac{\partial^2f}{\partial y\partial x}$ and see what you get. – Joey Zou Aug 28 '16 at 4:46
• Thanks! I corrected the question. – whitegreen Aug 28 '16 at 5:16
• Even with the correction the same issue remains. Can you compute $\frac{\partial^2f}{\partial x\partial y}$ and $\frac{\partial^2f}{\partial y\partial x}$ and see if they are the same? – Joey Zou Aug 28 '16 at 6:47

The following holds for any $C^1$ function $f(x,y)$, including yours. Consider the vector field $\mathbf F = \nabla f$. Since $\mathbf F$ is conservative, you can retrieve the value of $f$ at any point $(x,y) \in \mathbb R^2$ with the following vector line integral $$f(x,y) = f(0,0)+\int_{(0,0)}^{(x,y)}\nabla f\cdot d\mathbf s = f(0,0)+\int_{(0,0)}^{(x,y)}\frac{\partial f}{\partial x}\,dx + \frac{\partial f}{\partial y}\,dy.$$ The notation "$\int_{(0,0)}^{(x,y)}$" is used to denote that the integral may be taken over any piecewise-smooth curve in $\Bbb R^2$ beginning at the origin and ending at $(x,y)$. Since $\mathbf F$ is conservative, and the vector line integral of a conservative field is path-independent, this notation makes sense.