Multiple Integrals 6: Limits of Region R (Type II)
Uploader Comments (donylee)
All Comments (22)
-
weeeeeeeeeeeeeeeeeeeeeeeeee
-
hooray
-
Thats wt I means...:)
-
@Losuol yeah, western high school is so far behind Asian countries.
-
@anujjuneja1 It is easier to realise that for a dy integral, the limits must be y=f(x) and similarly for a dx integral the limits must be x=g(y). That is how my maths professor taught me. Then just make sure you have the constants on the outer integral to ensure that the final value is a scalar, not a function of some other variable.
-
Dude how to sketch the function and find the region? =( that's crucial before determining the limits...help me please =(
-
Just check out for the limits...if limits are in term of x, it will be type 2, which i think he mention in 4th lecture if i m right...
Just simply see, if the variable in limits(limits are in terms of x or y), then integrate 1st wrt other variable...I hope its helpful...
-
thank you, that was such a help!! x
i wonder what happens when R hasnt got any flat sides... i'll have to keep watching...
jimmyshitbags 3 years ago 2
Yup, good question. I see that you are thinking ahead.
A simple answer is that our regions or R are limited to type I and type II, when either the top and bottom, or left and right are bounded by straight lines. I did mention earlier is that plane regions can take a variety of forms and this is why FOR NOW, we will restrict ourselves to simple ones.
The most complicated region we'll deal with is a circle.
donylee 3 years ago
Hey Donny, excellent videos, you're a great instructor. However, I have a question over the example in this video. You give the answer to the inside integral as 2y^2 - 3y^3, but after several tries, I keep getting it as 2y^2 - 2y^3, and the final answer as -26/3. I'm giving you the benefit of the doubt, since you're much (much much much) better at math than I am, but I'm still unsure where the mistake could be. Would you be willing to do a quick re-check to make sure you had it right?
imafknninja 3 years ago
Hello imafknninja,
Yup, you are indeed right. The result of the inner integral is 2y^2 - 2y^3. However, the final answer is still -68/3 after integrating 2y^2 - 2y^3 wrt to y from 1 to 3.
I think you missed out either one of the coefficients from y^2 or y^3 term.
donylee 3 years ago
wow, you sure like to move around a lot.
facevokk 3 years ago
Haha, I'll do whatever it takes to grab the viewer's attention!
donylee 3 years ago