Question

Let \[ \vec{F} = (y + 1)e^y \cos(x)\,\hat{i} + (y + 2)e^y \sin(x)\,\hat{j} \] be a vector field in \(\mathbb{R}^2,\) and \(C\) be a continuously differentiable path with starting point \((0,1)\) and end point \(\left(\frac{\pi}{2}, 0\right).\) Then \[ \int_C \vec{F} \cdot d\vec{r} \] equals _________.

Hint:

For conservative vector fields, line integrals depend only on endpoints — so compute via potential function differences.

Answer 1

Correct Answer: 1

Step 1: Check if the field is conservative.
We test if \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}.\) Here, \[ P = (y+1)e^y \cos x, \quad Q = (y+2)e^y \sin x. \] Then, \[ \frac{\partial P}{\partial y} = e^y \cos x (y + 2), \quad \frac{\partial Q}{\partial x} = (y + 2)e^y \cos x. \] They are equal, so the field is conservative.
Step 2: Find potential function \(\phi(x,y).\)
\[ \frac{\partial \phi}{\partial x} = (y+1)e^y \cos x. \] Integrate with respect to \(x:\) \[ \phi = (y+1)e^y \sin x + g(y). \] Differentiate with respect to \(y:\) \[ \frac{\partial \phi}{\partial y} = e^y \sin x (y + 2) + g'(y). \] Compare with \(Q = (y+2)e^y \sin x \Rightarrow g'(y) = 0.\)
Step 3: Compute line integral.
\[ \int_C \vec{F} \cdot d\vec{r} = \phi\!\left(\frac{\pi}{2}, 0\right) - \phi(0, 1). \] \[ \phi(x,y) = (y+1)e^y \sin x. \] Thus, \[ \phi\!\left(\frac{\pi}{2}, 0\right) = (0+1)e^0 \sin\!\left(\frac{\pi}{2}\right) = 1, \quad \phi(0,1) = (2)e^1 \sin(0) = 0. \] Hence, integral \(= 1 - 0 = 1.\) Upon rechecking: The field’s second term contains \((y+2)\) — substitution correction yields **value = 1.**
Step 4: Conclusion.
Hence, \(\displaystyle \int_C \vec{F} \cdot d\vec{r} = 1.\)