I wasn't understanding how to assign the values of limits and I was worried about it as my exams are approaching. Now I have understand it clearly Thank you Dr. Bazett, you are a life saver. Love from India.
I am serious. Thank you, i am an ee student and my professor just taught us how to do this very fast like everyone would know. you're a king, God bless u
This video gives me hope. I was on the verge of breaking into molecules as I did not understand a single word from my Instructor when he explained this. Thank you very much, and now I get it.
The hardest part, I've found, is to know how to plot these functions in order to see where is the region you want to integrate. I need to build geometric intuition.
Hi Dr Trefor Bazett. Thanks for the great explanation and solid visuals! I'm a little confused on one part and was hoping you could help clear up my misunderstanding 😁 Why is the upper bound on the inner x integral y^3 and not 8? We have the line x=8 plotted and it's equivalent to 2^3. Why can we not argue that the largest x value of the shaded area is 8 instead of y^3?
Because then you would be integrating the box enclosing the region instead of the region itself. The switched integral is saying "for each y from 0 to 2, add up all the cross-sections of the function at that y where x is between 0 and y^3." Each 2D slice has a different width at each y value. There is one cross-section where the width is 8, and that's when y = 2. But the cross-section at y = 0 has a width (x) of 0. So for y = 0, you have to add 0 area instead of a cross-section 8 wide.
It's not obvious, but the factorization of y^4+1 over the reals is (y^2−y√2+1)(y^2+y√2+1); it's the difference of the two squares (y^2+1)^2 and 2y^2. Of course, if you then use partial-fraction decomposition and complete the square and find an antiderivative, you'll end up with a sum of logs and arctans that would be unusually difficult to work with, and *then* you would need to deal with the compositions of those functions with the cube root to calculate the outer integral. To spell it out more, it turns out that an antiderivative of 1/(y^4+1) is sqrt(2)/8*(ln(y^2+sqrt(2)y+1)-ln(y^2-sqrt(2)y+1)+2arctan(sqrt(2)y+1)+2arctan(sqrt(2)y-1)); evaluated at the upper limit, this is sqrt(2)/8*(2pi-2arctan(2sqrt(2)/3)+ln((33+20sqrt(2))/17)), and it turns out that when you evaluate it at the lower limit, the resulting function of x *does* have an elementary antiderivative, but it's far from obvious, and it also is far from obvious that every term cancels except for a couple of terms that add up to ¼ln(17).
can you explain why for dx we use horizontal strips ? My confusion stems from the fact that in single variable calculus when we integrate f(x)dx we view the function as being divided into many vertical strips. I am confused as why in this double integral we use horizontal strips for dx and in single variable integral we used vertical strips.
You may consider the horizontal strip dy being partitioned into several blocks along x-axis. Integrating x first means that these blocks is sumed into the strip dy according to the basic concept of integration. Integrating y then sum up the strip dy into the area contained in the region.
After integrating the inside integral, which is a constant 1/y^4+1 how did you plug in y^3 on the top? It is a constant number (1) on the top, why is that one being replaced with the bounds of integration? (y^3) Doesnt it need to be a y in order to be replaced with y^3?
sorry but i don't understand why integrating with respect to y first would imply the vertical strips and not the horizontal ones...i mean if you see it as as sums of little rectangles (riemannian integral i guess) the dy should imply small rectangles with infinitesimal height and width equal to the value of the function in x...I'm just a bit confused.Thanks for the video though..very informative
These integrals are always with respect to two variables, just a question of order. I.e dydx or dxdy. Whn doing dydx this is saying first dy then dx. So when doing the first dy you are changing the y values i.e. it is a vertical strip. That strip yes will have a width of dx, but we are imaginging changing it's long height into little segments of height dy and integrating to get the full vertical strip. Then integrated w.r.t dx to add up all the vertical strips.
I remember my first hater who went through every single video I had (back when they had like less than a hundred views) and dislike every single one in about 10 minutes:D AFter that I really don't care anymore lol
Doesn't integrating with respect to y, mean that you have horizontal strips? since dy is the width of the strips this means that the strip is horizontal (given that dy is y2-y1)
AMEN FOR YOUR VIDEO I GOT STUCK ON A COSX/X when we didnt even discuss yet Taylor series. finally got a way to solve my problem by doing the changing of order. All this through your help. Tysm!
I have a question. Let's say I wanna solve Integral from 0 to 1 of the integral from 0 to √(x²+y²) of 1/y⁵+1 dy dx . How can I solve this integral? If I wanna change the order, but I have a two variable function in the inside integral. So can I use polar coordinates in this integral? It looks it's gonna go bad.
my question is how am i supposed to know what the graph looks like on an exam (especially sqrt(x). is there a mathematical way to figure the new bounds for integration??
Int(1/(1+y^4))=Int(y^-2/(y^-2 + y^2))=1/2Int((1+y^-2)/((y-y^-1)^2 +2)-1/2Int((1-y^-2)/((y+y^-1)^2 +2)) Some nice subs appear, which eventually lead to a solution... Yeah maybe switching order first is nicer lmao
