Question

Check if the following integrals are independent of path and evaluate the path-independent integrals if they exists.
(i) int[(2x3^y)dx + (x^2 e^y + x -2y)dy] from (0,0) to (1,2).
(ii) int[( x sin xy + y cos xy) dx + (x^2 e^y) + x-2y) dy] from (0,0) to (2,2).

Answer 1

To understand what to do, let's recall some definitions and theorems (without proof).

Definition. If $F(x,t)=P(x,j)\cdot\vec{i}+Q(x,j)\cdot\vec{j}$ is a vector field in 2-space and a path $C$ is defined by the vector function $\vec{r}=f(t)\cdot\vec{i}+g(t)\cdot\vec{j},\,\,\,a\le t\le b$ , then a line integral can be written as

$$\int_CP(x,y)dx+Q(x,y)dy=\int_C\vec{F}\cdot d\vec{r}$$

Definition. If the value of a line integral is the same for every path in a region connecting the initial point $A$ and terminal point $B$ , then integral is said to be independent.

Definition. A vector function $\vec{F}(x,y)$ in 2- or 3-space is said to be conservative if $\vec{F}(x,y)$ can be written as the gradient of a scalar function $\varphi(x,y)$ . The function $\varphi(x,y)$ is called a potential function of $\vec{F}(x,y)$ .

Theorem. In an open connected region $R$ , $\int_C\vec{F}\cdot d\vec{r}$ is independent of the path $C$ if and only if the vector field $\vec{F}(x,y)$ is conservative in $R$ .

Theorem ( Test for a Conservative Field ). Suppose $\vec{F}(x,y)=P(x,y)\cdot\vec{i}+Q(x,y)\cdot\vec{j}$ is a conservative vector field in an open region $R$ , and that $P(x,y)$ and $Q(x,y)$ are continuous and have continuous first partial derivatives in $R$ . Then

$$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}\qquad\qquad (1)$$

for all $(x,y)$ in $R$ . Conversely, if the equality (1) hold for all $(x,y)$ in a simply connected region $R$ , then $\vec{F}(x,y)=P(x,y)\cdot\vec{i}+Q(x,y)\cdot\vec{j}$ is conservative in $R$ .

All definitions and theorems are taken from the book: Dennis Zill, Warren S. Wright, Michael R. Cullen. Advanced Engineering Mathematics

Now we can move on to solving the tasks.

QUESTION(i)

$$\int_C\left[\left(2x\cdot 3^y\right)dx+\left(x^2 e^y + x -2y\right)dy\right]\longrightarrow\\[0.3cm] \left\{\begin{array}{l} P(x,y)=2x\cdot 3^{y}\\[0.3cm] Q(x,y)=x^2\cdot e^y+x-2y \end{array}\right.\longrightarrow\\[0.3cm] \left\{\begin{array}{l} \displaystyle\frac{\partial P}{\partial y}=2x\cdot3^y\cdot\ln 3\\[0.3cm] \displaystyle\frac{\partial Q}{\partial x}=2x\cdot e^y+1\end{array}\right.\longrightarrow\\[0.3cm] \frac{\partial P}{\partial y}=2x\cdot3^y\neq 2x\cdot e^y+1=\frac{\partial Q}{\partial x}$$

Conclusion,

$\vec{F}(x,y)=\left(2x\cdot 3^y\right)\cdot\vec{i}+\left(x^2\cdot e^y+x-2y\right)\cdot\vec{j}$ non-conservative vector field.

$$\boxed{\int_C\left[\left(2x\cdot 3^y\right)dx+\left(x^2 e^y + x -2y\right)dy\right]-\text{path-dependent}}$$

QUESTION(ii)

$$\int_C\left[\left( x\sin xy + y\cos xy\right)dx+\left(x^2 e^y + x -2y\right)dy\right]\longrightarrow\\[0.3cm] \left\{\begin{array}{l} P(x,y)=x\sin xy + y\cos xy\\[0.3cm] Q(x,y)=x^2\cdot e^y+x-2y \end{array}\right.\longrightarrow\\[0.3cm] \left\{\begin{array}{l} \displaystyle\frac{\partial P}{\partial y}=x^2\cdot\cos xy+\cos xy-xy\sin xy\\[0.3cm] \displaystyle\frac{\partial Q}{\partial x}=2x\cdot e^y+1\end{array}\right.\longrightarrow\\[0.3cm] \frac{\partial P}{\partial y}=x^2\cdot\cos xy+\cos xy-xy\sin xy\neq 2x\cdot e^y+1=\frac{\partial Q}{\partial x}$$

Conclusion,

$\vec{F}(x,y)=\left(x\sin xy+y\cos xy\right)\cdot\vec{i}+\left(x^2\cdot e^y+x-2y\right)\cdot\vec{j}$ non-conservative vector field.

$$\boxed{\int_C\left[\left( x\sin xy + y\cos xy\right)dx+\left(x^2 e^y + x -2y\right)dy\right]-\text{path-dependent}}$$

General conclusion,

Since, by the condition of the problem, it is necessary to calculate only integrals that are independent of the path, then we WILL NOT calculate anything since it turned out that these integrals depend on the integration path.