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.

Question

The torque due to the force $ \left( 2\hat{i} + \hat{j} + 2\hat{k} \right) $ about the origin, acting on a particle whose position vector is $ \hat{i} + \hat{j} + \hat{k} $, would be:

Answer 1

Instead of using the shortcut that a gradient field's line integral only depends on endpoints, let's check the answer by directly parametrizing the quarter-circle path and integrating along it.

Parametrize the path: the quarter unit circle from P $(1,0)$ to Q $(0,1)$ can be written as $x = \cos\theta$, $y = \sin\theta$, with $\theta$ running from $0$ (at P) to $\pi/2$ (at Q).
Write $\vec{V}f = \nabla f$: since $f = x^2+y^2$, $\partial f/\partial x = 2x$ and $\partial f/\partial y = 2y$, so $\nabla f = 2x\hat{x} + 2y\hat{y}$.
Write $d\vec{l}$ along the path: $d\vec{l} = (dx)\hat{x} + (dy)\hat{y}$, with $dx = -\sin\theta\,d\theta$ and $dy = \cos\theta\,d\theta$.
Form the dot product: $\nabla f \cdot d\vec{l} = 2x\,dx + 2y\,dy = 2\cos\theta(-\sin\theta\,d\theta) + 2\sin\theta(\cos\theta\,d\theta) = -2\sin\theta\cos\theta\,d\theta + 2\sin\theta\cos\theta\,d\theta = 0$.
Integrate: since the integrand is zero at every point along the path, $\int_0^{\pi/2} 0\,d\theta = 0$.

The two terms cancel at every single point along the quarter circle, not only at the endpoints, because the unit circle is a level curve of $f = x^2+y^2 = 1$: $f$ never changes as we move along it, so its gradient has no component along the path direction anywhere on the circle.

Let's summarize:

Direct parametrization confirms the integrand is zero everywhere along the path, not only that the endpoint values match.
This happens because P and Q both sit on the same level curve $x^2+y^2=1$ of $f$.

So the line integral evaluates to $0$, the same result as the endpoint-difference shortcut.

Answer 2

Instead of using the shortcut that a gradient field's line integral only depends on endpoints, let's check the answer by directly parametrizing the quarter-circle path and integrating along it.

Parametrize the path: the quarter unit circle from P $(1,0)$ to Q $(0,1)$ can be written as $x = \cos\theta$, $y = \sin\theta$, with $\theta$ running from $0$ (at P) to $\pi/2$ (at Q).
Write $\vec{V}f = \nabla f$: since $f = x^2+y^2$, $\partial f/\partial x = 2x$ and $\partial f/\partial y = 2y$, so $\nabla f = 2x\hat{x} + 2y\hat{y}$.
Write $d\vec{l}$ along the path: $d\vec{l} = (dx)\hat{x} + (dy)\hat{y}$, with $dx = -\sin\theta\,d\theta$ and $dy = \cos\theta\,d\theta$.
Form the dot product: $\nabla f \cdot d\vec{l} = 2x\,dx + 2y\,dy = 2\cos\theta(-\sin\theta\,d\theta) + 2\sin\theta(\cos\theta\,d\theta) = -2\sin\theta\cos\theta\,d\theta + 2\sin\theta\cos\theta\,d\theta = 0$.
Integrate: since the integrand is zero at every point along the path, $\int_0^{\pi/2} 0\,d\theta = 0$.

The two terms cancel at every single point along the quarter circle, not only at the endpoints, because the unit circle is a level curve of $f = x^2+y^2 = 1$: $f$ never changes as we move along it, so its gradient has no component along the path direction anywhere on the circle.

Let's summarize:

Direct parametrization confirms the integrand is zero everywhere along the path, not only that the endpoint values match.
This happens because P and Q both sit on the same level curve $x^2+y^2=1$ of $f$.

So the line integral evaluates to $0$, the same result as the endpoint-difference shortcut.

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The Correct Option is A

Solution and Explanation

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