## Calculus: Early Transcendentals 8th Edition

a) $\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}$ b) $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}$ is not independent of path.
Consider $F(x,y)=Pi+Qj$ In order to get the conservative field troughout the domain $D$, we must have the following condition $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$ Here, $P$ and $Q$ are the first-order partial derivatives on the domain $D$. Now, $\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}=\dfrac{y^2-x^2}{(x^2+y^2)^2}$ This shows that the vector field $F$ is conservative. Hence, $\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}$ b) $\int_{C_1} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_0^{\pi} (-\sin \theta i+\cos \theta j) \cdot (-\sin \theta i+\cos \theta j) d\theta= \int_0^{\pi} (1) d\theta =\pi$ and $\int_{C_2} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_0^{-\pi} [-\sin \theta i+\cos \theta j] \cdot [-\sin \theta i+\cos \theta j] d\theta= \int_0^{-\pi} (1) d\theta =-\pi$ Here, we can see that the two integrals are different. Therefore, the line integral $\int_C \overrightarrow{F} \cdot \overrightarrow{dr}$ is not independent of the path.