Imagine that you have a field of snowdrifts like the one pictured here. That will be our three-dimensional function $z = f(x,y)$ where $(x,y)$ are coordinates on the ground and $z=f(x,y)$ gives the depth of the snow at each location. Now imagine we want to walk through that snowfield, shoveling a path as we go. Clearly, the total amount of snow we shovel will depend on the path we take. We could go straight through, take a longer, loopier path that avoids the deepest snow, or choose to meander through the peaks to shovel the most snow possible. There are endless possibilities.

In this section we'll learn how to "walk" along or traverse any path and add up the function above (or below) that path as we go.

Often the direction vector is normalized, but that won't be crucial in evaluating line integrals.

Curved paths

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As a simple example, suppose we wanted to integrate over the path formed by $y = x^2$ between $(-1,1)$ and $(1,1)$. One parameterization could be to set $x=t$, then $y=t^2$, and the limits of $t$ would be $-1 \lt t \lt 1$. We could also set $y=t$, then $x=\sqrt{t}$, and so on. In this graph we've used the first parameterization. Notice that as the parameter (often interpreted as time) grows, we move along the curve from right to left

Paramaterizations are directional

Other coordinate systems

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Some paths have circular symmetry, so it will be handy to be able to parameterize a function $f(x,y)$ in terms of a radial parameter, $r$, and an angle $\theta$. For example, if our path is the unit circle

$$x^2+y^2=1$$

we can use the parameterization

$$ \begin{align} x &= r \, \text{cos}(\theta) \\[5pt] y &= r \, \text{sin}(\theta) \\[5pt] r^2 &= x^2 + y^2 \end{align}$$

We'll do some examples using this kind of parameterization later. There are all kinds of ways to parameterize a function, and there is no unique parameterization of any function or curve.

And here's a zoomed in version of one of those trapezoids:

The dimensions of our rectangle are the height of the function $f(x,y)$ on one side (doesn't matter which), and the width is the arc length, $\Delta S$, which we can approximate by a line. Take a look at the zoomed-in picture of the bottom of the rectangle:

We can approximate $\Delta S$ by the straight-line distance across, $\Delta S \approx \sqrt{(\Delta x)^2 + (\Delta y)^2}$. So the approximate area of the i^th rectangle is

$$\Delta A_i = f(x_i,y_i) \sqrt{(\Delta x)^2 + (\Delta y)^2}$$

Now we employ a trick: Divide and multiply each term under the radical by $\Delta t$:

$$\Delta A_i = f(g(t),h(t)) \sqrt{ \left(\frac{\Delta x}{\Delta t}\right)^2 + \left(\frac{\Delta y}{\Delta t}\right)^2 } \, \Delta t$$

Now recognize that in the limit of infinitely-narrow rectangles,

$\frac{\Delta x}{\Delta t}$ becomes $\frac{dx}{dt}$. Then if we parameterize $f(x,y)$ in terms of $t$ we get

$$dA = f(g(t),h(t)) \sqrt{(g'(t))^2 + (h'(t))^2} \, dt$$

Here $x=g(t)$ and $y=h(t)$. The part under the radical is just our expression for arc length of a parameterized function. Integration of $dA$ with respect to the parameter gives us our line integral:

$$ \begin{align} f(x) &= x^2+y^2 \\[5pt] &= (1+2t)^2 + (2+3t^2) \\[5pt] &= 1 + 4t + 4t^2 + 4 + 12t + 9t^2 \\[5pt] &= 13t^2 + 16t + 5 \end{align}$$

Now we an put the integral together:

$$ \begin{align} A &= \int \limits_C f(g(t),h(t)) \, \sqrt{(g'(t))^2 + (h'(t))^2} \, dt \\[5pt] &= \sqrt{13} \int_0^1 [13t^2+16t+5] \, dt \\[5pt] &= \sqrt{13} \left[\frac{13t^3}{3}+8t^2+5t \right]_0^1 \\[5pt] &= \sqrt{13} \left[ \frac{13}{3}+8+5 \right] \\[5pt] &= \sqrt{13} \left[\frac{13+24+15}{3} \right] = \frac{52 \sqrt{13}}{3} \end{align}$$

In equation (1), we used the power-reduction formula for $\text{cos}^2(\theta)$,

$$\text{cos}^2(\theta) = \frac{1+\text{cos}(2\theta)}{2}$$

The second integral can be done with u-substitution:

$$ \begin{align} A_2 &= 2 \int_0^{2\pi} \text{cos}(\theta)\, \text{sin}^2(\theta) \, d\theta \\[5pt] \text{Let } & u = \text{sin}(\theta), \; \rightarrow \; du = \text{cos}(\theta) \, d\theta \\[5pt] &\rightarrow 2\int u^2 \, du = \frac{1}{3}u^3 \\[5pt] &= 2 \left[ \frac{1}{3} \text{sin}^3(\theta) \right]_0^{2\pi} \\[5pt] &= 0 \end{align}$$

So the total value of the integral is $3\pi$.

Note: Although our path was circular, this was not a polar integral, thus we didn't need the whole $r \, dr \, d\theta$ thing. We recognized that our path had the form of a circle with a fixed radius, and therefore could be parameterized with the single parameter $\theta$.

Path $C_2$

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The parameterization of the line segment is

$$ \begin{align} L &= (1,1) + t(0-1,0-1) \\[5pt] &= (1,1) + t(-1,-1) \\[5pt] x &= 1-t \phantom{000} y = 1-t \end{align}$$

The limits on $t$ are $0 \le t \le 1$. The derivatives are

$$ (x'(t))^2 = 1 \phantom{000} (y'(t))^2 = 1$$

So the radical term in the integral is $\sqrt{2}$.

The function $f(x,y) = 2+xy^2$ transforms to

$$ \begin{align} f(g(t),h(t)) &= 2 + (1-t)(1-t)^2 \\[5pt] &= 2+ (1-t)(1-2t+t^2) \\[5pt] &= 2+ 1-2t+t^2-t+2t^2-t^3 \\[5pt] &= -t^3 + 3t^2 -3t +3 \end{align}$$

The line integral is

$$ \begin{align} A &= \sqrt{2} \int_0^1 (-t^3 + 3t^2 -3t +3) \, dt \\[5pt] &= \sqrt{2} \left[ \frac{-t^4}{4} + t^3 - \frac{3}{2}t^2 + 3t \right]_0^1 \\[5pt] &= \sqrt{2} \left[ -\frac{1}{4} + 1 - \frac{3}{2} + 3 \right] \\[5pt] &= \frac{9 \sqrt{2}}{4} = 3.18 \end{align}$$

Total area $\approx -0.11$

It may be a bit unnerving to have a negative area, but remember that just like in 1-D integration, some of the function may lie below the plane of any path we choose. A negative result is OK.

$$\sqrt{\left( \frac{dx(t)}{dt} \right)^2 + \left( \frac{dy(t)}{dt} \right)^2 }$$

In a line integral strictly along the x-axis, $\frac{dy(t)}{dt}$ is zero. Likewise for integrals along the y-direction, thus we end up with the formulas above. Given that for any integral,

We can also encounter integrals of the form

$$\int \limits_C f(x,y) \, dx + g(x,y) \, dy,$$

which can just be split into the relevant $dx$ and $dy$ line integrals and added back together.

Now just for drill, let's say we went the other way along our path, from $(1,2)$ to $(0,0)$. Then our parameterization would be $(1,2)+t(-1,-2)$ or $x = 1-t$ and $y=2-2t$ for $0 \le t \le 1$.

The integrand would be $\frac{1}{2}(1-t)(2-2t) = (1-t)^2 = 1-2t+t^2$. Multiplying by the derivative of $y$ gives

$$ \begin{align} -2\int_0^1 (1-2t+t^2) &= t -t^2+\frac{t^3}{3} \, \bigg|_0^1 \\[5pt] &= -2\left( 1-1-\frac{1}{3} \right) \\[5pt] &= \frac{-2}{3} \end{align}$$

This is just the negative of the first integral. If we switch directions along a path, the path integral switches signs.

4. Evaluate the line integral $\int \limits_C (xy+z^3) \, ds$, where $C$ is the piece of the helix $\vec r(t) = (\text{cos}(t), \text{sin}(t), t)$ from $t=0$ to $t=\pi$.

   Solution

   

   
   

   Here's the path for this integral.

   It's a helix with a circular profile in the $xy$ plane as we look down the $z$-axis, and the $z$ component of $\vec r(t)$ gives the amount of rise or pitch of the helix as it turns.

   Our path for this exercise is just the red portion.

   We already have the parameterization of this path, so the squares of the derivatives are

   $$ \begin{align} (x'(t))^2 &= \text{sin}^2(t) \\[5pt] (y'(t))^2 &= \text{cos}^2(t) \\[5pt] (z'(t))^2 &= 1 \end{align}$$

   So the root term of our line integral is

   $$\sqrt{\text{cos}^2(t) + \text{sin}^2(t) + 1} = \sqrt{2}$$

   Our integrand is

   $$xy+z^3 = \text{cos}(t) \text{sin}(t) + t^3$$

   The integral is

   $$ \begin{align} \sqrt{2}\int_0^{\pi} &\text{cos}(t) \text{sin}(t) + t^3 \, dt \\[5pt] &= \sqrt{2} \left[ \frac{1}{2} \text{sin}^2(t) + \frac{t^4}{4} \right]_0^{\pi} \\[5pt] &= \frac{\pi^4 \sqrt{2}}{4} \approx 34.4 \end{align}$$




5. The mass of a fine wire of variable density can be found by performing a line integral of the density function, $\delta$, over the curve of the wire. If the wire has the shape of the top part of a circle: $\sqrt{9-x^2}$ with $0 \le x \le 3$and its density function is $\delta(x,y) = x \sqrt{y}$, calculate the mass of the wire.

   Solution

   The path is the quarter of the circle of radius $r = 3$ centered at the origin and in the first quadrant, so we can parameterize it in terms of angle $\theta$, with $0 \le \theta \le \pi/2$. Then the integrand (the density function) transforms to

   $$x \sqrt(t) = 3 \, \text{cos}(\theta) \sqrt{3 \text{sin}(\theta)}$$

   The coordinates are $x(t) = 3 \, \text{cos}(\theta)$ and $y(t) = 3 \, \text{sin}(\theta)$. So the root term will just turn out to be $3$. The integral is:

   $$ \begin{align} 9\sqrt{3} \int_0^{\pi/2} &\text{cos}(\theta) \sqrt{\text{sin}(\theta)} \, dt \\[5pt] \text{Let } u &= \text{sin}(\theta), \; du = \text{cos}(\theta) \, d\theta \\[5pt] 9\sqrt{3} &\int u^{1/2} du \\[5pt] &= 9\sqrt{3} \frac{2}{3} \text{sin}^{3/2}(\theta) \bigg|_0^{\pi/2} \\[5pt] &= 9\sqrt{3} \frac{2}{3} [1-0] = 6 \sqrt{3} \end{align}$$




6. Evaluate $\int \limits_C (y-x) \, dx + (xy) \, dy$ where $C$ is the line segment from $(2,4)$ to $(3,1)$.

   Solution

   Think of this as the sum of line integrals along the $x$- and $y$-axes, respectively. First parameterize the line segment:

   $$ \begin{align} C(t) &= (2,4) + t(3-2, \; 1-4) \\[5pt] &= (2, 4) + t(1, -3) \\[5pt] x &= 2+t \phantom{000} y = 4-3t \end{align}$$

   Now the two parts of our integrand are

   $$ \begin{align} y - x &= 4-3t - (2+t) \\[5pt] &= 2 - 4t \\[5pt] xy &= (4-3t)(2+t) \\[5pt] &= 8+4t-6t-3t^2 \\[5pt] &= -3t^2-2t+8 \end{align}$$

   Our derivatives are

   $$\frac{dx}{dt} = 1 \phantom{0000} \frac{dy}{dt} = -3$$

   Finally, the integral is

   $$ \begin{align} \int_0^1 &(2-4t)(1) + (-3t^2-2t+8)(-3) \, dt \\[5pt] &= \int_0^1 2-4t+9t^2+6t-24 \, dt \\[5pt] &= \int_0^1 9t^2 +2t-22 \, dt \\[5pt] &= \left[ 3t^3 + t^2 -22t \right]_0^1 \\[5pt] &= 3+1-22 = -18 \end{align}$$