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: Identify the components of the vector field.
Write \(\vec{F} = P\,\hat{e}_x + Q\,\hat{e}_y\), so here \(P(x,y) = x+2y\) and \(Q(x,y) = 2x+4y\). The contour \(C\) is the closed diamond shaped path through \(O(0,0)\), \(P(1,1)\), \(Q(0,2)\), \(R(-1,1)\), and back to \(O\).
Step 2: Recall Green's theorem for a closed contour in the plane.
For a closed curve \(C\) enclosing a region \(A\) in the xy-plane,
\[ \oint_C \vec{F}\cdot d\vec{l} = \oint_C (P\,dx + Q\,dy) = \pm\iint_A \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA \]
where the sign depends on whether the contour is traversed counterclockwise or clockwise. This turns a line integral around the boundary into an area integral of the field's curl over the enclosed region.
Step 3: Compute the two partial derivatives.
\[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(2x+4y) = 2 \]
\[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x+2y) = 2 \]
So the curl term is
\[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 - 2 = 0 \]
Step 4: Apply Green's theorem.
Since the curl term is 0 at every single point in the plane, not just on average, the area integral over any region \(A\) is \(\iint_A 0\,dA = 0\). This holds regardless of the size or shape of the diamond region enclosed by \(C\), and regardless of whether \(C\) is traced clockwise or counterclockwise, since a zero integrand gives zero area integral either way.
Step 5: Conclude the value of the line integral.
By Green's theorem, the closed line integral equals this area integral.
\[ \oint_C \vec{F}\cdot d\vec{l} = 0 \]
Why the other options are wrong:
2, 4, and 6 would only appear if the curl of \(\vec{F}\) were nonzero, so the enclosed area of the diamond would matter in the answer. Since the curl here is exactly zero everywhere, the enclosed area never enters the answer, and those nonzero values do not arise.
