Answer to Question #214761 in Calculus for Maryam

Question

Show that line integral is independent of path by finding a potential function for F

F(x,y) = (sin x − x sin y) j + (cos y + y cos x) i

Answer 1

Let $P(x, y)=\cos y+y\cos x$ and $Q(x, y)=\sin x-x\sin y.$ Then

$$\dfrac{\partial P}{\partial y}=-\sin y+\cos x$$

$$\dfrac{\partial Q}{\partial x}=\cos x-\sin y$$

$$\dfrac{\partial P}{\partial y}=\cos x-\sin y=\dfrac{\partial Q}{\partial x}$$

Then $F(x, y)$ is conservative.

$$\exist f , \nabla f=\vec F$$

$$\dfrac{\partial f}{\partial x}=P(x, y)=\cos y+y\cos x$$

$$\dfrac{\partial f}{\partial y}=Q(x, y)=\sin x-x\sin y$$

$$f=\int(\cos y+y\cos x)dx+\varphi(y)$$

$$=x\cos y+y\sin x+\varphi(y)$$

$$\dfrac{\partial f}{\partial y}=-x\sin y+\sin x+\varphi'(y)=\sin x-x\sin y$$

$$=>\varphi'(y)=0=>\varphi(y)=C$$

$$f(x, y)=x\cos y+y\sin x+C$$

Therefore the line integral is independent of path.