Let \(f\) be a function of two real variables \(x\) and \(y\), defined as: \[ f(x,y) = x^2 + y^2 \] The value of the line integral \(\int \vec{V}f \cdot d\vec{l}\) from point P to point Q along the path \(L\) shown in the figure below is Here \(\vec{V} = \hat{x}\dfrac{\partial}{\partial x} + \hat{y}\dfrac{\partial}{\partial y}\), with \(\hat{x}\) and \(\hat{y}\) as the unit vectors along X and Y axes, respectively.
Show Hint
\(\vec{V}\) is just \(\nabla\), so the integral equals \(f(Q)-f(P)\); both P and Q lie on \(x^2+y^2=1\).
Step 1: Identify what the operator \(\vec{V}\) represents.
The vector \(\vec{V} = \hat{x}\dfrac{\partial}{\partial x} + \hat{y}\dfrac{\partial}{\partial y}\) is exactly the gradient operator \(\nabla\) in two dimensions. So \(\vec{V}f\) is the gradient of \(f\), namely \(\nabla f\).
Step 2: Recall the fundamental theorem for line integrals of a gradient.
For any scalar function \(f\), the line integral of its gradient along a path from a starting point P to an ending point Q depends only on the values of \(f\) at the two endpoints, not on the shape of the path in between. This is because \(\nabla f \cdot d\vec{l} = df\), an exact differential, so the integral telescopes:
\[ \int_{P}^{Q} \nabla f \cdot d\vec{l} = f(Q) - f(P) \]
This means we do not need to parametrize the quarter-circle path at all; we only need the value of \(f\) at P and at Q.
Step 3: Evaluate \(f\) at P and Q.
From the figure, P = (1, 0) and Q = (0, 1). Using \(f(x,y) = x^2 + y^2\):
\[ f(P) = f(1,0) = 1^2 + 0^2 = 1 \]
\[ f(Q) = f(0,1) = 0^2 + 1^2 = 1 \]
Step 4: Compute the line integral.
\[ \int_{P}^{Q} \nabla f \cdot d\vec{l} = f(Q) - f(P) = 1 - 1 = 0 \]
Both points lie on the unit circle \(x^2+y^2=1\), where \(f\) is constant and equal to 1, so the change in \(f\) between them is zero regardless of which path, the quarter circle or any other, connects them.
Final Answer:
The line integral evaluates to 0, so options (B), (C) and (D), which give nonzero values, are incorrect.
\[ \boxed{0} \]
