Consider the contour \(C\) shown in the figure below. For the vector \(\vec{F} = (x+2y)\hat{e}_x + (2x+4y)\hat{e}_y\), the integral \(\oint_C \vec{F}\cdot d\vec{l} =\)________.
Here \(d\vec{l}\) represents an infinitesimal length along the contour \(C\).
Show Hint
Check the curl of F first. If it is zero everywhere, the closed loop integral is zero no matter what shape or orientation the contour has.
Step 1: Check if the field could have a scalar potential.
A 2D vector field $\vec{F} = P\hat{e}_x + Q\hat{e}_y$ has a scalar potential $\phi(x,y)$ with $\vec{F} = \nabla\phi$ exactly when $\partial Q/\partial x = \partial P/\partial y$ everywhere. Here $P = x+2y$ and $Q = 2x+4y$, so $\partial Q/\partial x = 2$ and $\partial P/\partial y = 2$. These match, so a potential function does exist.
Step 2: Build the potential function $\phi(x,y)$.
We need $\partial \phi/\partial x = P = x + 2y$. Integrating with respect to $x$ while treating $y$ as fixed:
\[ \phi(x,y) = \frac{x^2}{2} + 2xy + g(y) \]
for some function $g(y)$ still to be found.
Step 3: Fix $g(y)$ using the other component.
Differentiate the trial $\phi$ with respect to $y$: $\partial \phi/\partial y = 2x + g'(y)$. This must equal $Q = 2x+4y$, so $g'(y) = 4y$, which integrates to $g(y) = 2y^2$, dropping the constant since it will not affect the final answer. So
\[ \phi(x,y) = \frac{x^2}{2} + 2xy + 2y^2 \]
Step 4: Use the fundamental theorem of line integrals.
For a field that is the gradient of a potential, the line integral between any two points depends only on the potential value at the start and end points, not on the path taken:
\[ \int_A^B \vec{F}\cdot d\vec{l} = \phi(B) - \phi(A) \]
The contour $C$ is closed, so its start point and end point are the same point, say $O(0,0)$.
Step 5: Evaluate for the closed loop.
\[ \oint_C \vec{F}\cdot d\vec{l} = \phi(O) - \phi(O) = 0 \]
This works regardless of the shape of $C$, whether it passes through $P$, $Q$, $R$ or any other points, and regardless of direction of travel, because the start and end point coincide for any closed loop.
Final Answer:
Because $\vec{F}$ has a potential function, its integral around any closed path is zero.
\[ \boxed{0} \]
