SUNY Geneseo Department of Mathematics

Line Integrals in Vector Fields, Part 2

Monday, April 16

Math 223 04
Spring 2018
Prof. Doug Baldwin

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Previous Lecture

Misc

GROW STEM Talk

“Status of Women and Under-Represented Groups in various fields of STEM”

Led by Provost Stacey Robertson.

Thursday, April 19, 5:00 pm.

Newton 204.

Questions?

Line Integrals in Vector Fields

Theorem 6.5.

There’s another typo (in the PDF version, it’s fixed online) — try to find it and suggest a correction.

The third equation

Integral over C of F dot dr equals integral over negative C of F dot dr

should be

Integral over C of F dot dr equals negative integral over negative C of F dot dr

This result makes intuitive sense based on the idea that a line integral of a vector field is the integral of the components of the vectors in the direction of motion along the path. If you reverse that direction, the vector components negate.

Curve passing through vectors with each vector projected onto curve

Examples

Space Rock, Continued

Recall the space rock and star from Friday. Consider another rock that is in orbit instead of falling towards the star. This rock moves from (2,0,0) to (0,2,0) along a quarter circle of radius 2, under gravitational field

Gravity is a force directed in the minus x minus y minus z direction with magnitude 1 over distance from origin squared

Coordinate system with rock at (2,0,0) moving on circle to (0,2,0)

Calculate the work done on the rock by the star’s gravity.

We first need to come up with a parametric description of the path the rock follows. One is based on the standard parametric form for a circle, namely

r(t) is the vector of 2 cos(t) and 2 sin(t) and 0

Another possibility is to describe the quarter circle by the equation y = √(4-x2) and parameterize accordingly, with x ranging from 2 down to 0:

r(t) equals vector of t and square root of 4 minus t squared and 0

This works better in subsequent integrations than it might seem (try it), but I’ll finish this example using the first parameterization.

To start, find the derivative of r(t)

r-prime equals vector of -2 sin(t), 2 cos(t), and 0

and then calculate the denominators from the gravitational field so it can be plugged into later expressions.

Sum of squares of x(t), y(t), and z(t) equals 4

Now we can put these pieces together in an integral for work:

Work equals integral from 0 to pi/2 of G(r(t)) dot r-prime(t) which equals 0

This result makes physical sense, because gravity hasn’t moved the rock closer to or farther from the star, so no work has been done.

A Line Integral with No Physical Significance

Let f(x,y) = 〈 x2, y2 〉 and r(t) = 〈 cos t, sin t 〉, 0 ≤ t ≤ π . Find the integral of F along r.

Semicircle passing through vectors pointing generally up and right

Start by finding the derivative of r(t)

r-prime(t) equals vector of -sin(t) and cos(t)

and f(r(t))

f(r(t) equals vector of cosine squared of t and sin squared of t

Now we can integrate:

Expand f(r(t)) and r-prime(t) in integral from 0 to pi of f(r(t)) dot r-prime(t)

Integrate each part of the sum separately, using two different substitutions: u = cos t, du = -sin t dt and v = sin t, dv = cos t dt. Doing this turns the integral into

Cosine cubed of t plus sine cubed of t all over 3 and evaluated from 0 to pi, which equals -2/3

Next

Flux and circulation.

Read the “Flux and Circulation” subsection of section 6.2.

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