I'm trying to find the potential function of this coservative vector field $$F\left(x,y,z\right)=(2xyz^2\ ,x^2z^2+z\cos yz, 2x^2yz+y\cos yz\ ) $$ so I can use the fundamental theorem of calculus to compute de line integral where C is the line segment from $(0,0,0)$ to $(5,0,5)$
I got this one
$$f\left(x,y,z\right)=x^2yz^2 +\ \sin{\left(zy\right)+K}$$
but when I evaluate the starting and ending points I got
$$=5^2\bullet0\bullet5^2\mathrm{\ }+\ \sin{\left(0\right)+K-0^2\bullet0\bullet0^2\mathrm{\ } +\ \sin{\left(0\right)-K}}=0$$
Can anybody tell me what's going on here. I saw someboy getting $$f=x^2 y z^2+cos(yz)+k$$ so that confuses me. Did I make a mistake?
This is what I did
$$\frac{\partial f}{\partial y} \left(x,y,z\right)= x^2z^2+\frac{\partial C}{\partial y}\left(y,z\right)$$
$$x^2z^2+\frac{\partial C}{\partial y}\left(y,z\right)=x^2z^2+zcosyz$$
$$\frac{\partial C}{\partial y}\left(y,z\right)=zcosyz$$
$$C\left(y,z\right)=\int{zcosyz\ dy=\sin{\left(zy\right)}+G\left(z\right)}$$
$$f\left(x,y,z\right)=x^2yz^2 +\ \sin{\left(zy\right)+G(z)}$$
$$\frac{\partial f}{\partial z}\mathrm{\ }\left(x,y,z\right)=2x^2yz+\cos{\left(zy\right)}y+\frac{\partial G}{\partial z}\left(z\right)$$
$$2x^2yz+\cos{\left(zy\right)}y+\frac{\partial G}{\partial z}\left(z\right)=2x^2yz+ycosyz
\frac{\partial G}{\partial z}\left(z\right)=0$$
$$G=K$$
$$f\left(x,y,z\right)=x^2yz^2 +\ \sin{\left(zy\right)+K}$$
edit I fixed F(5,0,5)-F(0,0,0)
Best Answer
Your potential function is correct. Your evaluation looks odd, I was expecting to see $$f(5,0,5) - f(0,0,0)$$ which gives $$(0+0+K) - (0 + 0 + K) = 0$$