I am new to vector analysis and have attempted the questions below, I dont have any answers or person that can solve them for me. So if anyone could tell me wether my working and answers are correct or not it would be much appreciated. I havent put in all steps but i think its fairly obvious what \i have done.
- $\nabla \cdot(\nabla \times A)=0$, $\nabla = i\frac {d} {dx}+j\frac {d} {dy}+k\frac {d} {dz}$
the cross product of $(\nabla \times A)$ end up minussing out giving 0 meaning we're left with $(\nabla \cdot 0)$ which equals 0.
- $\nabla \times(\nabla \times A)=\nabla(\nabla \cdot A)-\nabla^2A$
weve proven $(\nabla \times A)= 0$ and therefore $(\nabla \cdot 0) =0$
Looking at the other side $\nabla(\nabla \cdot A) = \nabla(i\frac {dA} {dx}+j\frac {dA} {dy}+k\frac {dA} {dz})= \nabla^2A$
leaving $\nabla^2A-\nabla^2A=0$ giving $0=0$.
- evaluate line integral for vector field $A=yzi-xzj$ along curve $r(t)=cost\textbf{i}+sint\textbf{j}+3t^2k , (0\leq t\leq \Pi )$
Using $\int^\Pi_0A\cdot dr$
$A\cdot dr=(sint(3t^2)-cost(3t^2)) \cdot (-sint+cost+6t)$ which results in, $\int^\Pi_0 -3t^2 + 6t$
which can be solved to $-\Pi^3+3\Pi^2$