Answer
Result : See proof
Work Step by Step
Step 1 Recall that the Conservative Field Test states that if $\vec{F}(x, y) = (e^y + ye^x)i + (xe^y + e^x)j$ is a conservative vector field in some open region $D$ where $f(x,y)$ and $g(x,y)$ are continuous functions with continuous first partial derivatives, then: \[ \frac{\partial \vec{F}}{\partial x} = \frac{\partial \vec{F}}{\partial y} \] holds true at each point in $D$. Step 2 Suppose that $f(x,y)$ and $g(x,y)$ are differentiable functions on the $xy$-plane, and the equation: \[ \frac{\partial f}{\partial y}(x,y) = \frac{\partial g}{\partial x}(x,y) \] holds true at each point on some open region $D$. Step 3 By the Conservative Field Test, we find that there exists a function $\Phi(x,y) = (e^y + ye^x)i + (xe^y + e^x)j$ that is conservative. So, the given statement is true. Result True Additional Calculations We need to find the line integral of the force field $\vec{F}(x, y) = (e^y + ye^x)i + (xe^y + e^x)j$ along the curve $\vec{r}(t) = \sin(\pi t/2)i + \ln t j$, where $1 \leq t \leq 2$. We note that: \[ \vec{r}(t) = (\frac{t^2}{2}\cos(\pi t/2)i + \ln t j) \] and \[ \vec{F}(\vec{r}(t)) = (t^2e^{\ln t} + t\ln te^{\frac{t^2}{2}\cos(\pi t/2)})i + (t\ln te^{\ln t} + e^{\frac{t^2}{2}\cos(\pi t/2)})j \] Then, we can calculate the solution by the integral: \[ \int_C \vec{F} \cdot d\vec{r} = \Phi(\vec{r}(2)) - \Phi(\vec{r}(1)) = \Phi(0, \ln 2) - \Phi(1, 0) = \ln 2 - 1 \]
