I have tried solving this question for some time. I would like to know if my solution is correct.
Let $f(r)=y\cdot{\hat{z}}$ be a vector field and a surface $S$ inside the triangle with the vertices $(a,0,0),(0,2a,0),(0,0,a)\text{ where }a>0$.
- Compute $\displaystyle \Phi=\int\limits_S \left(\nabla\times f\right)\cdot{\text{d}\textbf{S}}\ ; \ \text{d}\textbf{S}=\hat{n}\text{d}S$ where $\hat{n}$ is the normal to the triangle plane.
My Try:
First $f(x,y,z)=(0,0,y)$, thus $\nabla\times f=(1,0,0)$.
The triangle plane is given by $2x+y+2z=2a$ or $\displaystyle z(x,y)=a-x-\frac{y}{2}$. The normal to the plane at some point $(x_0,y_0)$ is $$\vec{N}=(-z_x(x_0,y_0),-z_y(x_0,y_0),1)=\left(1,\frac{1}{2},1\right)\Rightarrow \hat{n}=\frac{\vec{N}}{|N|}=\left(\frac{2}{3},\frac{1}{3},\frac{2}{3}\right)$$ Eventually $$\int\limits_S \left(\nabla\times f\right)\text{d}\textbf{S}=\int\limits_{S}\left(1,0,0\right)\cdot\left(\frac{2}{3},\frac{1}{3},\frac{2}{3}\right)\text{d}S=\frac{2}{3}\int\limits_S \text{d}S$$ The area of the triangle is $\displaystyle \frac{3}{2}a^2$, hence the required result is $a^2$.
- Compute the line integral $\displaystyle \int\limits_C f\cdot{\text{d}\ell}$ where $C$ is the perimeter of the triangle.
My Try:
Let $C=C_1\cup C_2\cup C_3$ where$$C_1\text{ is the line between }(a,0,0)\text{ and }(0,2a,0) \\ C_2\text{ is the line between }(0,2a,0)\text{ and }(0,0,a) \\ C_3\text{ is the line between }(0,0,a)\text{ and }(a,0,0)$$ Parameterizing each line$$C_1:\vec{\ell}(t)=(a-at,2at,0),\ t:0\to 1 \\ C_2:\vec{\ell}(t)=(0,2a-2at,at),\ t:0\to 1\\ C_3:\vec{\ell}(t)=(at,0,a-at),\ t:0\to 1$$ Now,\begin{align*}\int\limits_C \vec{F}\cdot \text{d}\vec{\ell}&=\int\limits_{C_1}\vec{F}\cdot \text{d}\vec{\ell}+\int\limits_{C_2}\vec{F}\cdot \text{d}\vec{\ell}+\int\limits_{C_3}\vec{F}\cdot \text{d}\vec{\ell} \\&=\int\limits_{0}^{1}(0,0,2at)\cdot{(-a,2a,0)}\text{d}t+\int\limits_{0}^{1}(0,0,2a-2at)\cdot{(0,-2,a)}\text{d}t+\int\limits_{0}^{1}(0,0,0)\cdot{(a,0,-a)}\text{d}t\\&=0+\int\limits_{0}^{1}\left(2a^2-2a^2t\right)\text{d}t+0=a^2\end{align*}
- Make sure that Stokes' law holds.
By Stokes' law the surface integral of the curl of a vector field $f$ over surface $S$ equals the line integral of $f$ over the boundary of $S$, i.e $\displaystyle \int\limits_{S}\left(\nabla\times f\right)\text{d}S=\int\limits_{\partial S}f\cdot{\text{d}}\ell$. By results (1) and (2), Stokes' law holds.
Please share your thoughts regarding my solution. Thanks!