SUNY Geneseo Department of Mathematics
Tuesday, April 18
Math 223
Spring 2023
Prof. Doug Baldwin
(No.)
PRISM’s “Calculus Bee” is one week from today — Tuesday, April 25, 5:00-ish - 7:00, Newton 204.
Try out your calculus skills, or just watch.
Based on “Scalar Line Integrals” in section 5.2 of the textbook.
A scalar line integral is basically a Riemann sum.
The calculation formula — see the derivation below or the textbook for the formula.
Geometrically, interpret a line integral of a 2-variable function as the area of a “sheet” between the graph of that function and a curve in the xy plane.
Derivations related to scalar line integrals?
Imagine some function f with variables x, y, and maybe more. Also imagine a curve r through the domain of f. Finally, imagine that curve divided into many short segments at points (x1, y1, …), (x2, y2, …), etc.
Now we can form a sum of function values at points along the line, multiplied by the lengths of the associated segments. This sum, or more precisely its limit as the number of segments grows to infinity, and so the lengths of the segments shrinks to 0, is the definition of the scalar line integral of f along curve r:
Notice that this is an integral with respect to arc length, so if we wanted a parametric form of curve r, it would need to be an arc length parameterization. But arc length parameterizations can be hard to find and work with, so we’d like a way to do line integrals in terms of an arbitrary parameterization of r. The arc length formula is key to doing this: arc length is an integral of the magnitude of r’; if we differentiate both sides of that equation, we get that differential arc length, ds, can be treated as the magnitude of r’ times dt, for any parameter t we like. This gives rise to the calculation formula:
Using scalar line integrals to find arc length?
The arc length integral is just a special case of the calculation formula for scalar line integrals, namely the special case in which function f is always equal to 1.
For example, here’s that idea applied to finding the length of half a turn of a certain helix. Notice that the integral quickly turns into the integral you would have gotten by starting with the arc length formula — so there’s an interesting connection between arc length and scalar line integrals, but line integrals don’t offer a radically new or better way to find arc lengths.
As a geometric introduction, consider integrating f(x, y) = x2 + y2 along the quarter circle of radius 4 running from the positive x axis to the positive y axis. This integral should be the surface area of the quarter-cylinder whose base is that circle and whose height is given by f(x, y) where x and y are any point that is distance 4 from the origin, e.g., (4, 0). Note that the level curves of f are circles centered at the origin, so all points on the quarter circle give the same value for f, namely 16.
To find this area via a line integral, start with the abstract integral of f along the quarter circle C:
To actually evaluate this integral, put it into a form we can use the calculation formula on. This requires finding a parametric equation for the quarter circle (and bounds on t that make the equation define just a quarter of a full circle), and its derivative and the magnitude of that derivative:
Plug the information from the parametric form into the calculation formula for scalar line integrals to get something we can actually evaluate:
Using the scalar line integral to find the area between the quarter circle and the bottom of the surface of f did indeed produce the same value our original geometric argument did.
More examples of scalar line integrals.
No new reading.