SUNY Geneseo Department of Mathematics
Friday, April 21
Math 223
Spring 2023
Prof. Doug Baldwin
How to set up the integrals for problem 1 in problem set 11?
Since the region to integrate over isn’t rectangular, one of the integrals is going to have bounds that are functions of the other integral’s variable. I’d start by deciding which integral should play which role, i.e., whether I want to express y in terms of x, or x in terms of y. In this case, expressing y in terms of x seems easy because the question already does it.
So our inner integral will be with respect to y, and will use the functions from the question as its bounds. Use the function that marks the bottom of the region as the lower bound, the function that marks the top as the upper. Looking at the plot of the region can be helpful for deciding which is which.
Finally, we need to find bounds for x. Since x has to range over the entire region between the curves, the most reliable way to find x bounds is to set the functions for those curves equal to each other and solve for x.
These ideas lead to this iterated integral:
Part B of problem 1 in problem set 11, finding the area of the region between the curves?
Start with the idea that the “volume” of a region is the integral of 1 over that region. This generalizes to any number of dimensions, including 2, in which case the generalization of volume is area.
Since you’re integrating over the same region as in Part A, you can use the same bounds, leading to this integral:
How does this apply to the 4-dimensional integral in problem 3 of problem set 11, especially to the shape it’s the volume of?
One way to think about the shape is to work from the inner integrals out, graphing their regions as you go. The inner integral integrates over line segments of length z, and the next one defines z as ranging from 0 to 1. So between them, these two integrals find the area of a triangle. The third integral then adds another dimension, y, in which coordinates range from 0 to 2. This pulls the triangle out in the third dimension to form a prism. The outermost integral does something similar in the 4th dimension, i.e., pulls the prism out into a 4-dimensional “prism” shape.
Actuary Club is having an informational meeting Tuesday, May 2, at 6:30 PM in South 328.
A good chance to learn more about actuarial careers, Geneseo’s actuary courses, etc.
Problem set 12 is ready, on integration in polar coordinates and scalar line integrals.
Work on it next week; grade it in the first part of the following week.
There will be one more problem set, to work on roughly May 1 to 4 and then to grade May 5 to 10 (i.e., during the last days of classes).
Based on “Examples of Vector Fields,” “Vector Fields in R2,” “Drawing a Vector Field,” and “Vector Fields in R3” in section 5.1.
A vector field is a vector-valued function with multivariable component functions; usually the dimension of the vectors is the same as the number of variables in the component functions.
Plotting a vector field amounts to drawing the vectors at each point; this is generally best done by a computer or calculator, as demonstrated below with Mathematica.
Some common kinds of vector fields include…
Continuity of vector fields is defined very similarly to continuity for other kinds of function.
Mathematica plots vector fields with the VectorPlot (fields in 2 dimensions) and VectorPlot3D (3 dimensions) functions.
Arrows show the directions of selected vectors, while colors encode magnitude (darker/cooler colors are lower magnitudes; brighter/hotter colors are higher magnitudes).
You can download a notebook we developed in class with several examples of 2- and 3-dimensional vector fields and their plots.
Line integrals of vector fields.
Please read “Vector Line Integrals” in section 5.2 of the textbook.