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Week 6 Discussion Notes

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Line Integrals

Line integrals are usually of the form

∫Cf(x,y,z) dsor∫CF⃗(x,y,z)⋅dr⃗or∫CF1 dx+F2 dy+F3 dz,\int_{\mathscr{C}} f\p{x, y, z} \,\diff{s} \quad\text{or}\quad \int_{\mathscr{C}} \vec{F}\p{x, y, z} \cdot \diff{\vec{r}} \quad\text{or}\quad \int_{\mathscr{C}} F_1 \,\diff{x} + F_2 \,\diff{y} + F_3 \,\diff{z},

where C\mathscr{C} is a curve. To calculate these, there are three main steps:

  1. Parametrize C\mathscr{C}, which gives you a parametrization r⃗(t)=⟨x(t),y(t),z(t)⟩\vec{r}\p{t} = \ang{x\p{t}, y\p{t}, z\p{t}}.

  2. Plug in r⃗(t)\vec{r}\p{t} into the integrand, differential, and bounds.

    To plug in r⃗(t)\vec{r}\p{t} into the integrand, you replace xx with x(t)x\p{t}, yy with y(t)y\p{t}, and zz with z(t)z\p{t} to get f(r⃗(t))f\p{\vec{r}\p{t}} or F⃗(r⃗(t))\vec{F}\p{\vec{r}\p{t}}.

    For the differential, it's basically like a uu-substitution from 1D: u=u(x)  ⟹  du=u′(x) dxu = u\p{x} \implies \diff{u} = u'\p{x} \,\diff{x}, but instead, you can have vector quantities:

    r⃗=r⃗(t)  ⟹  dr⃗=r⃗′(t) dtandds=∥r⃗′(t)∥ dt.\vec{r} = \vec{r}\p{t} \implies \diff{\vec{r}} = \vec{r}'\p{t} \,\diff{t} \quad\text{and}\quad \diff{s} = \norm{\vec{r}'\p{t}} \,\diff{t}.

    For the third line integral above, though, it will look slightly different:

    x=x(t)  ⟹  dx=x′(t) dty=y(t)  ⟹  dy=y′(t) dtz=z(t)  ⟹  dz=z′(t) dt.\begin{aligned} x = x\p{t} &\implies \diff{x} = x'\p{t} \,\diff{t} \\ y = y\p{t} &\implies \diff{y} = y'\p{t} \,\diff{t} \\ z = z\p{t} &\implies \diff{z} = z'\p{t} \,\diff{t}. \end{aligned}

    The bounds will depend on the parametrization, too.

  3. Integrate like normal.

Example 1.

Compute

∫C⟨1+x2,xy2⟩⋅dr⃗,\int_{\mathscr{C}} \ang{1 + x^2, xy^2} \cdot \diff{\vec{r}},

where C\mathscr{C} is the line segment from (0,0)\p{0, 0} to (1,3)\p{1, 3}.
Solution.

You can always parametrize line segments like so:

r⃗(t)=(1−t)⟨0,0⟩+t⟨1,3⟩=⟨t,3t⟩,0≤t≤1.\vec{r}\p{t} = \p{1 - t}\ang{0, 0} + t\ang{1, 3} = \ang{t, 3t}, \quad 0 \leq t \leq 1.

Note that r⃗(0)=⟨0,0⟩\vec{r}\p{0} = \ang{0, 0} and r⃗(1)=⟨1,3⟩\vec{r}\p{1} = \ang{1, 3}, so this is the correct direction (C\mathscr{C} starts at (0,0)\p{0, 0} and ends at (1,3)\p{1, 3}).

Now we just plug everything in:

F⃗(r⃗(t))=⟨1+t2,t⋅(3t)2⟩=⟨1+t2,9t3⟩dr⃗=⟨1,3⟩ dt,\begin{gathered} \vec{F}\p{\vec{r}\p{t}} = \ang{1 + t^2, t \cdot \p{3t}^2} = \ang{1 + t^2, 9t^3} \\ \diff\vec{r} = \ang{1, 3} \,\diff{t}, \end{gathered}

which gives

∫C⟨1+x2,xy2⟩⋅dr⃗=∫01⟨1+t2,9t3⟩⋅⟨1,3⟩ dt=∫011+t2+27t3 dt=9712.\begin{aligned} \int_{\mathscr{C}} \ang{1 + x^2, xy^2} \cdot \diff{\vec{r}} &= \int_0^1 \ang{1 + t^2, 9t^3} \cdot \ang{1, 3} \,\diff{t} \\ &= \int_0^1 1 + t^2 + 27t^3 \,\diff{t} \\ &= \boxed{\frac{97}{12}}. \end{aligned}

(Like with double/triple integrals, the hardest part is almost always just setting it up, i.e., the first equals sign above was the hardest step.)