15.3: Double Integrals in Polar Coordinates

Objectives:
4. Express a double integral as an iterated integral in polar coordinates.
5. Find area and volume using an iterated integral in polar coordinates.

[@andrewmaksimovich2432]

I really appeciate teachers with great handwriting. It may be a small thing but really great for clarity as a student, unlike some other professors

[@moisesdevs]

Hands down one of the best channels on here. I do not much about who you are but thank you so much.

[@vihaankedia3606]

I quite literally take a second every video to just admire the pattern of your ring.

[@stormixgaming8389]

12:36 it should be 1/4 r_k^2 not r_k* right? for both expanded brackets and I'm guessing that the r_k*Deltar_k^2 is also a mistake since it was -r_k*Deltar_k in the second term

[@oxord1111]

23:09, when you found the bound for X I don’t get how you found these values for X, like did you take the bounds of x from the 3D or did you take it on the circle on the XY plane?

[@aditigali9293]

Is there any place I could go to memorize reoccurring graphs and their functions in multivariable?

[@aakaashks3520]

Who is ur maths inspiration?

[@alexandraniedden5337]

We use Stewart's 9th edition Multivariable Calculus.

[@benjaminandrew7940]

14:02 When you say don't forget you need an additional r, does that mean the r becomes 2r or r^2?

[@zombieprincess12]

is the link to the schoology open for general public or just students in the class?

[@alexandraniedden5337]

No, sorry! It's not a public page. Good luck!

[@geraldcalero4179]

2:04 you say practicing converting on your own which section according to Stewart help on practicing on them?

[@alexandraniedden5337]

If you want some extra practice with polar coordinates, check out section 10.3 in the Stewart book. There is practice converting, plotting, and graphing.

[@geraldcalero4179]

@@alexandraniedden5337 thank you so much your such great help I appreciate it.

[@zacharyquinn2144]

at about 24:00 why does r go between 0 and 1, and not -1 and 1? I must be confused, I thought this was the radius bounds of the circle.

[@alexandraniedden5337]

You are correct in that r represents the radius bounds of the circle. Remember that in polar form, a a negative radius sends the point in the opposite direction. For example, the point with r = -1 and theta = 0 is really the point x = -1, y = 0. When writing the bounds for r, we use the domain at r >= 0.

[@Cruzylife]

can you link to the schoology for polar graphing, please and thank you

[@alexandraniedden5337]

No, sorry! It's not a public page. Good luck!

[@hussainaligopang203]

Have you apload 13.1 polar coordinate

[@alexandraniedden5337]

Section 13.1 in my book covers Vector Functions: kzbin.info/www/bejne/naLPh5mOps2Eg8k I don't think that's what you are looking for though...

[@hussainaligopang203]

@@alexandraniedden5337 ma'am I need Laplace equation in polar form and in spherical form

[@adamkhan5378]

When do you know if you can split up the equation or not?

[@alexandraniedden5337]

If the integrand is a product of function f(x)g(y)h(z), then you can split up the integrand. For example 3xyz can be split into 3x*y*z. On the other hand, 3x - 4y cannot be split into a product of functions, so we could not split up that integral.

[@adamkhan5378]

@@alexandraniedden5337 do the limits of integration matter for splitting it up? Like if the upper bound for the dr integral was in terms of theta but the integrand was 3rcos(theta), could we split it up?

[@alexandraniedden5337]

@@adamkhan5378 Ah good question! I neglected to mention that. Yes, the limits of integration matter. In your example, no, that integrand could not be split up because one set of limits relies on another variable. So not only do you need a product of functions, but you also need rectangular limits of integration in order to split the integral. Sorry for the omission!

[@squirrel8121]

fk this class imma just hope linear algebra is easier

[@anonymous-is5uz]

I don't trust u solve for me

[@listlessviewer153]

This is where my brain fails me. It’s been a good run 🫡