Vector Calculus II: Summer-2021
03 : Due 21: M 8/9

  1. Around or through; Stokes in action S0 4268S Let \(\boldsymbol{\vec{F}}=-y^2\,\boldsymbol{\hat{x}}\).
    • Compute \(\nabla \times \boldsymbol{\vec{F}}\).
    • Is \(\boldsymbol{\vec{F}}\) conservative?
    • Let \(S\) be the portion of the surface defined by \(z=2-y^2\) inside the cylinder \(x^2+y^2=3\). Construct \(d\boldsymbol{\vec{S}}\) for this surface.
    • Deduce the value of \(\int\!\!\int_S\nabla\times\boldsymbol{\vec{F}}\cdot d\boldsymbol{\vec{S}}\).
    • Let \(C\) be the boundary of \(S\). Construct \(d\boldsymbol{\vec{r}}\) for \(C\).
    • Calculate \(\int_C\boldsymbol{\vec{F}}\cdot d\boldsymbol{\vec{r}}\).
    • Are your answers to the integrals the same? Should they be?
  2. $d\boldsymbol{\vec{S}}$ and surfaces S0 4268S
    Let \(\boldsymbol{\vec{F}}=x\,\boldsymbol{\hat{y}}\) and \(\boldsymbol{\vec{G}}=-2x\,\boldsymbol{\hat{x}}-2y\,\boldsymbol{\hat{y}}-2z\boldsymbol{\hat{z}}\). Compute the flux of each vector field through the following surfaces:
    • \(z=5-x^2\) where \(-2\leq x \leq 2\) and \(-1 \leq y \leq 2\).
    • The upper hemisphere of radius 4 centimeters.
  3. Flux Fooling S0 4268S Let \(D\) be the cylinder with diameter of 10 inches, height of 13 inches including the bottom but not the top. Presume that the \(z\)-axis is through the center of \(D\) parallel to the height.
    • If \(\boldsymbol{\vec{F}}=-3z\boldsymbol{\hat{z}}\), use geometry to deduce whether the value of \(\int\!\!\int_D\boldsymbol{\vec{F}}\cdot d\boldsymbol{\vec{S}}\) is positive, negative or zero.
    • Support your deduction with explicit calculations.
    • How would your deduction change if the top were included?
    • Use the Divergence Theorem to write down an equation relating the flux through \(D\) and the flux through the top.
    • Create a visual equation expressing the same relationship you wrote down for the previous part. That means: draw pictures. Include some color.
    • Repeat the above steps using the vector field \(\boldsymbol{\vec{G}}=-4x\,\boldsymbol{\hat{x}}+-4y\,\boldsymbol{\hat{y}}\).