Polar is the method I used immediately. I’m very curious to see a full version of the first method just to know how ridiculous it is by comparison. Not enough to do it myself though.
Your video reminded me of the time when I was a teaching fellow more than thirty years ago. 14:01 At that point, since we're integrating first w.r.t. r and *then* w.r.t. theta, I wouldn't have depicted semi circles ranging from r=0 to r=3 but rays with angles ranging from theta = -Pi/2 to Pi/2. I would have also shown the other order of integration too which is just as easy to do. Then of course the semi circles would have come into play! Great video nonetheless. ... I wish I could have communicated as well as you!
the double integral is essentially calculating the volume of the origin-centered half-cylinder, which is capped by the surface x^3+xy^2. It baffles my mind how this volume can be a rational number, given that a circle is involved.
How can be sure dy integrate from -3 to +3, not from +3 to -3? And theta from -pi/2 to pi/2 instead of pi/2 to -pi/2... Maybe always integration from smaller coordinate to larger cooridate? It looks quite certain though that final integration result is positive... If dx was from 3 to 0, would you use theta pi/2 to -pi/2 or r from 3 to 0? This case does it matter which one of the variables would integrate in the negative direction?
using polar coordinates is unfortunately not always the best way. Hence, if you use the polar coordinates to calculate the area of an ellipse, the integral you need to solve turns out to be more difficult to handle than the one you solve using Cartesian coordinates; you can see ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-I1vRL-5e2lQ.html
There's a whole method of evaluating double integrals by changing the order of integration. However, you have to change the bounds between which they are evaluated as well. You can't simply switch dy and dx and the integral bounds in the front. Hope that helps!
The double integral is essentially calculating the volume of the origin-centered half-cylinder, which is capped by the surface x^3+xy^2. It baffles my mind how this volume can be a rational number, given that a circle is involved.
If the height of the cylinder is 1/π then the volume will be r², if r is integer, the volume will not only be rational but also an integer and a perfect square
(2x/3)(sqrt(9-x^2)^3) is actually relatively simple to integrate, as it fits the formula of the integral of f’(x)*f^n(x) Where f(x)=9-x^2 And n=1.5 Therefore all it is is (1/3)*(((9-x^2)^2.5)/2.5) Idk how to integrate the other part though as my integration knowledge is very limited
@@thaovu-yi5tsWe don't but it's pretty self explanatory that we gotta do the inside integral first. It's kinda like those 10yr old algebra questions where u use bodmas and do inside out ig But yeah being a highschool student myself I only knew how to do the first method and i got stuck afterwards
the second way is so much clearer, however i cannot help but try the first method as well edit: well it was intimidating to integrate at first, but wasn't so bad in the end
using polar coordinates is unfortunately not always the best way. Hence, if you use the polar coordinates to calculate the area of an ellipse, the integral you need to solve turns out to be more difficult to handle than the one you solve using Cartesian coordinates; you can see ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-I1vRL-5e2lQ.html
Surely to integrate with respect to y where there are x's you have to assume that the two functions are independent? Like if you wrote x as a function of y (not treating it as constant) it would look different and you would get a different answer. But then later he connects them by saying x^2 + y^2 = 9...
Great video as always. There's actually a 4th way also, since changing the order of integration in polar form also works. I wrote all 4 methods out in detail and got 486/5 each time.
4:08 > _"represents bottom part of circle"_ holly molly, i entirely forgot that and was thinking about root of inverted parabola. and by the way, never noticed this connection before too: root of a parabola gives a semi-circle. awesome.
I tried the same arrangement, but with function x^2+y^2 (instead of x^3+xy^2). Following method 3 (polar), I got (81 / 4) * pi. But if this is half a circle, then its area should be pi * r^2 / 2, and if r = 3, it should be (9 / 2) * pi. What did I do wrong, or maybe the whole thing is not really the area of half a circle? Please explain. Thanks.
x^2+y^2 in 3d is not a plain circle, it is a parabola rotated on itself in the y axis, so what you calcultae with this double integral is the volume under this shape, very different from the area of a circle :D
i pause this video at 0:50 and i want to solve this integral by original way by myself , it take along time and very complex, then when i solved it i continous see this video, that amazing way to solve it, 2 way is so good.
using polar coordinates is unfortunately not always the best way. Hence, if you use the polar coordinates to calculate the area of an ellipse, the integral you need to solve turns out to be more difficult to handle than the one you solve using Cartesian coordinates; you can see ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-I1vRL-5e2lQ.html
Kurtlane it is a matrix of the partial derivatives of the change of coordinates. in this case, x=rcos(theta) and y=rsin(theta) are the change of coordinates, you takes the partial derivatives of both with respect to r and theta, and you take the determinant of the matrix which gives r. its essentially the multidimensional analogue to dealing with the differential du in u-substitution in the single variable case.
