Calculus Small Changes, Big Ideas

Section 19.1 Line integrals

• If \(C\) is a curve, then the line integral of \(f\) along \(C\) is:

  \begin{equation*} \int_{C} f\dd{s}=\int_{a}^{b}f(x(t),y(t))\sqrt{\qty(\dv{x}{t})^{2}+\qty(\dv{y}{t})^{2}}\dd{s} \end{equation*}

  Similar works for 3D.

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• Let \(\bv r(t)\) be a parametrization of \(C\text{;}\) then we can write

  \begin{equation*} \int_{C} f\dd{s}=\int_{a}^{b}f(\bv r(t))\norm{\bv r'(t)}\dd{t} \end{equation*}

  Similar works for 3D.

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• Integrating \(f(x,y)=1\) just returns the arc length, so the arc length was really a line integral all along!

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• If we integrate a density, we get the mass, as usual.

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• Always independent of parametrization!

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• For piecewise parametrizations, add the individual integrals.

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• Consider the Jacobian of our transformation \(\bv r\) from \(t\) to \((x,y)\text{:}\)

  \begin{equation*} \bv J=\begin{bmatrix}\DS\pdv{x}{t}\\[2 ex] \DS\pdv{y}{t}\end{bmatrix} \end{equation*}

  This matrix is not square so it doesn’t have a determinant; however we can calculate a related quantity (called the metric tensor, I think?):

  \begin{equation*} \sqrt{\abs{\det\bv{J}^T\bv{J}}}=\sqrt{\qty(\dv{x}{t})^{2}+\qty(\dv{y}{t})^{2}} \end{equation*}

  This is precisely the ‘fudge factor’ that shows up when we convert our variables of integration from \((x,y)\) to \(t\text{.}\)

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