1. Prove the identity, assuming that the appropriate partial derivatives exist and are continuous. If $f$ is a scalar field and $F$, $G$ are vector fields, then $fF$, $F\cdot G$, and $F\times G$ are defined by

    \[\begin{align*} (fF)(x,y,z)=f(x,y,z)F(x,y,z)\\ (F\cdot G)(x,y,z)=F(x,y,z)\cdot G(x,y,z)\\ (F\times G)(x,y,z)=F(x,y,z)\times G(x,y,z) \end{align*}\]
    1. $\text{div}(F+G)=\text{div}(F)+\text{div}(G)$
    2. $\text{div}(fF)=f\text{div}(F)+F\cdot\nabla f$
    3. $\text{div}(F\times G) = G\cdot \text{curl}(F)-F\cdot \text{curl}(G)$
    4. $\text{div}(\nabla f\times \nabla g)=0$
    5. $\text{curl}(F+G)=\text{curl}(F)+\text{curl}(G)$
    6. $\text{curl}(fF)=f\text{curl}(F)+\nabla f\times F$
    7. $\text{curl}(\text{curl}(F))=\nabla(\text{div} F)-\nabla^2F$

  2. Use Green’s Theorem in the flux form to prove Green’s first identity:

    \[\iint_Df\nabla^2gdA=\int_Cf(\nabla g)\cdot \mathbf{n}ds-\iint_D\nabla f\cdot\nabla gdA\]

  3. Use the definition of an oriented surface to explain why a Mobius strip is not orientable. You can follow this guide to answer the question:

    1. Find a parametric surface $r$ that represents a Mobius strip (you can Google this or refer to the answer provided in class).
    2. Find its unit normal vector $r_u\times r_v$.
    3. Note that $r_u\times r_v$ points in the opposite direction near a certain point on the surface.