Question

Consider the function \(f(x,y) = x^3 + y^2\).

If \(\vec{F} = \nabla f\), then \(\displaystyle\int \vec{F} \cdot d\vec{r}\) evaluated from \((0,0)\) to \((1,2)\) is ______ (answer in integer).

Hint:

A gradient field is path independent, so the integral is just the potential difference between the endpoints.

Answer 1

Correct Answer: 5

Step 1: Find the gradient explicitly.
$\nabla f = \left(\dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}\right) = (3x^2, 2y)$, so $\vec{F} = 3x^2\,\hat{i} + 2y\,\hat{j}$.

Step 2: Pick a straight line path and parametrize it.
Let $x = t$, $y = 2t$ for $t$ from 0 to 1, so $dx = dt$ and $dy = 2\,dt$.
Then $\vec{F}\cdot d\vec{r} = 3t^2\,dt + 2(2t)(2\,dt) = 3t^2\,dt + 8t\,dt$.

Step 3: Integrate over $t$.
$\int_0^1 (3t^2 + 8t)\,dt = \left[t^3 + 4t^2\right]_0^1 = 1 + 4 = 5$.
This matches the potential difference method exactly, since the field does not care which path is used.

Final Answer:
The value of the line integral is 5.
\[ \boxed{5} \]