Vector Fields & Line/Surface Integrals

Trivium

Given a vector field over either \(\mathbf{R}^2\) or \(\mathbf{R}^3,\) determine whether or not it’s conservative. If it’s conservative, determine its potential function.

Exercises

  1. Consider the vector field \(\bm{F}\colon\mathbf{R}^2\to\mathbf{R}^2\) defined as \(\bm{F}(x,y) = -y\mathbf{i}+x\mathbf{j}.\) Sketch some of the vectors of this vector field until you have a good idea of what it looks like.
  2. Sketch the vector field \(\bm{F}\colon\mathbf{R}^3\to\mathbf{R}^3\) defined as \(\bm{F}(x,y,z) = z\mathbf{k}.\)
  3. Sketch the gradient field and a contour plot of the function \(f\) defined as \(f(x,y) = x^2-y^3\) and notice how they’re related.
  4. \(\displaystyle y = mx+b\)
    \(\displaystyle y = mx+b\)
  5. TK MORE HERE

Problems & Challenges

  1. Schey Instead of using arrows to represent vector functions, we can use families of curves called field lines. In two dimensions, a curve \(\bigl(x(t), y(t)\bigr)\) is a field line of the vector field \(\bm{F}\) if at each point \((x_0, y_0)\) on the curve, \(\bm{F}(x_0,y_0)\) is tangent to the curve.

    1. Realize that the field lines \(\bigl(x(t), y(t)\bigr)\) for a vector field \(\bm{F} = P\mathbf{i} + Q\mathbf{j}\) are solutions to the differential equation \[ \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t} = \frac{Q}{P}\,. \]
    2. Determine the family of field lines for each of these vector fields.
      \(\bm{F} = \bigl\langle y, x \bigr\rangle\)
      \(\bm{F} = \bigl\langle x, -y \bigr\rangle\)
      \(\bm{F} = \bigl\langle 0, x \bigr\rangle\)
      \(\bm{F} = \bigl\langle y, xy \bigr\rangle\)
      \(\bm{F} = \bigl\langle 1, y \bigr\rangle\)
      \(\displaystyle \bm{F} = \biggl\langle \frac{y}{\sqrt{x^2+y^2}}, \frac{x}{\sqrt{x^2+y^2}} \biggr\rangle\)

  2. Schey Sometimes the value of a surface integral can be reasoned out without working through the usual arduous procedure simply by remembering that the surface integral calculates the flux, the “sum of flow”, across the surface. For each of the following vector fields \(\bm{F}\) and surfaces \(S,\) reason out the value of the surface integral \(\iint_S \bm{F}\cdot\mathrm{d}\bm{S}.\)

    1. Consider the cube of side-length \(\ell\) in the first quadrant with diagonally opposite corners located at \((0,0,0)\) and \((\ell,\ell,\ell).\) The surface \(S\) consists of the three sides of this cube that lie in the coordinate planes, and \(\bm{F}(x,y,z) = \langle x,y,z \rangle.\)
    2. The surface \(S\) is the cylinder with height \(h\) and base of radius \(R\) in the \(xy\)-plane, and \(\bm{F}(x,y,z) = \bigl\langle x\ln(x^2+y^2), y\ln(x^2+y^2), 0 \bigr\rangle.\)
    3. The surface \(S\) is the sphere of radius \(R\) centered at the origin, and \[\bm{F}(x,y,z) = \biggl\langle x\mathrm{e}^{-\bigl(x^2+y^2+z^2\bigr)}, y\mathrm{e}^{-\bigl(x^2+y^2+z^2\bigr)}, z\mathrm{e}^{-\bigl(x^2+y^2+z^2\bigr)} \biggr\rangle.\]
    4. The surface \(S\) is the cube of side-length \(\ell\) in the first quadrant with diagonally opposite corners located at \((0,0,0)\) and \((\ell,\ell,\ell),\) and \(\bm{F}(x,y,z) = \langle f(x), 0, 0 \rangle\) where \(f\) is an abritrary scalar function of \(x\).