GreenDivergence Handout Problem 1

Comments

• SighHillDoveN

  February 2018 edited February 2018

  For the integral $\int\limits_{C} F\cdot dr$,

  $\int\limits_{C} F\cdot dr=\int\limits_{0}^{2\pi}\int\limits_{0}^{1}r \bigg(\frac{\delta Q}{\delta x} - \frac{\delta P}{\delta y}\bigg)dx dy$

  $=\int\limits_{0}^{2\pi}\int\limits_{0}^{1}\bigg((-1)-(3)\bigg)dr d\theta$

  $=\int\limits_{0}^{2\pi}\int\limits_{0}^{1} r \bigg(-4\bigg)dr d\theta=-4\pi$

  
• Red

  February 2018

  For $\int_C F\cdot d n$ we have $\int_C F\cdot d n=\int \int \frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y} dx dy$ and $\frac{\partial P}{\partial x}=2 $, $\frac{\partial Q}{\partial y}=1$. Thus, when converting to polar we have $\int_{0}^{2\pi} \int_{0}^1 3 r \text{ } dr d\theta=\int_0^{2\pi} \frac{3}{2} \space d\theta=3\pi$