SUNY Geneseo Department of Mathematics
Wednesday, November 2
Math 223 01
Fall 2022
Prof. Doug Baldwin
Finding a parametric form for the level curve in problem set 8 question 1?
The best way to start is to plot, or imagine a plot of, the function and level curve in question (the level curve being technically a curve in the xy-plane). Then ask yourself what shape the level curve is, and what parametric or vector equation would produce that shape. (The shape is a circle, and the simplest vector equations for circles look like s(t) = ⟨ r cos t, r sin t ⟩where r is the radius of the circle.) Once you have the vector form of the level curve, its derivative is tangent to the level curve. Evaluate that derivative at a t value that corresponds to s(t) = ⟨ √2, -√2 ⟩, and see if the result is a scalar multiple of what you got in part B of the question.
A colloquium talk on the history of calculus, aimed specifically at students in the calculus courses, is an annual event in our math department.
This year the colloquium is planned for the week after Thanksgiving break, but exact times are still being worked out. The colloquium organizers are interested in when students in the target audience (i.e., you) would like to have it. Possibilities include
If you have any thoughts, including for completely different times, let me know and I’ll pass them on to the right people.
Based on “General Regions of Integration,” “Double Integrals over Non-Rectangular Regions,” and “Changing the Order of Integration” in section 4.2 of the textbook.
The way the bounds for the integrals work isn’t very clear.
True. Finding formulas for the bounds can be tricky at first, and while what you do with those bounds (plug them into antiderivatives just like you would do with numeric bounds in any definite integral) sounds simple, the details of carrying it out can be confusing. But both of these become clearer with practice, so the best way to handle this question is probably to practice some integrals and watch and talk about where their bounds come from and how they work.
Type I (x bounds are fixed and y bounds are functions of x) and type II (y bounds are fixed and x bounds are functions of y) regions.
The strong version of Fubini’s Theorem: you can evaluate integrals over irregular regions by letting the bounds of inner integrals be functions of variables from outer integrals. Then evaluate the integrals much like you would any other definite integral, by finding an antiderivative, plugging the upper and lower bounds into it, and subtracting the second result from the first.
Integrate xy2 over the triangle bounded by the x and y axes and the line y = 1 - x.
Everyone started by treating the region to integrate over as a Type I region, i.e., it has fixed bounds of x = 0 and x = 1, and between those bounds y ranges from 0 up to the line y = 1 - x.
Now integrate, using 1 - x as the upper bound for the inner integral. Substitute 1 - x for y in the antiderivative. The result is a function of x, rather than a constant, but at least it’s a function you can integrate:
Finally, integrate that function. This time, since all the bounds are constants, all the variables go away and the result is a constant.
More practice with integration over general regions.
And integration in Mathematica (bring Mathematica if you want to try it out).
No new reading.