How to evaluate a double integral by using polar coordinates

[@bprpcalculusbasics]

More double integrals: kzbin.info/www/bejne/fl7HlpmEab-BidE

[@MELONS-s8y]

Sum of n^k/n! From 1 to inf pls where k can be any natural number ❤

[@gdlorenzodm5664]

that one to pi transformation is CRAZY

[@jeanrichonez2022]

Nice integral .Nice shirt.Nice glasses. Nice haircut. Nice pokeball.

[@Omar-qf9rw]

finally got it. I was thinking everytime there was a calcle done with y values, but i've just got that as you only need to know the iinterval of the angle, thank you!

[@Budgeman83030]

That was very easy to follow and it’s been forty two years since I had any calculus

[@major__kong]

If the solution is non-elementary in rectangular coordinates but elementary in polar coordinates, that implies an equivalence between the two solutions. Nature doesn't care about coordinate systems after all. So does that mean non-elementary functions can ultimately be expressed as a function of elementary ones and we've been missing this all along?

[@yaboirequis]

I thought that was the idea behind a Taylor polynomial/series, couldn't nonelementary functions be expressed with mere addition over an interval?

[@stephenbeck7222]

The equivalence of the antiderivative will not correspond to equivalence in the conversion of the definite integral bounds. This one worked out nicely so that the integrand was only in one variable and the double integral ended up being just an iterated integral with no variables in the bounds.

[@Mediterranean81]

Integrate all trig functions

[@user-zy5yi9yy5r]

h..i sir can you make a tutorial about applications of derivatives?

[@jamescollier3]

Did I miss something? The picture is 1/2 a circle of radius 1. Area =π/2, no?

[@robertpearce8394]

I am also puzzled. What does cos(r^2) mean?

[@LordQuixote]

@@robertpearce8394 Draw a circle of radius r, say r= 0.5. On that circle draw an angle of 0.25 radians. It's the cosine of that angle

[@robertpearce8394]

​@LordQuixote but it's the radius not the angle. Maybe I'm overthinking it. I'm comparing this with the Gaussian integral.

[@LordQuixote]

@@robertpearce8394 It's just a value. The radius has a value of 0.5 and the angle has a value of 0.25 radians. It's like saying y=x^2--it's just a relationship between two variables. The integral is just summing up all the cosine functions for each circle of radius r. For r=0.5, you have cos(0.25), add that to the cos(1) for r=1, cos(0.09) for r=0.3, cos(0.04) for r=0.2, etc.

[@arieltabbach4946]

no because your integrating over a function and not just the area

[@broytingaravsol]

a piece of cake

[@bitoty9357]

sine of 1 radians, thats disturbing

[@deltalima6703]

He never actually said it was radians. Sloppy imo.

[@darkmask4767]

​@@deltalima6703 In calculus, it's implicitly understood that the angle is in radians because otherwise the derivative and integral formulas don't work.

[@GustavoMerchan79]

kind of puzzled how the result is not just the half the area of a unit circle (r=1): (π . r^2) / 2 = π/2

[@pseudolullus]

Careful, we are integrating over **cosine** of x^2+y^2, a function. You don't really need a double integral for the area of a semicircle, you can just pull y out of x^2+y^2 = 1

[@GustavoMerchan79]

@@pseudolullusyes! You are right, thank you

[@cdkw2]

bring back the pokeball mic pls!

[@wesleyburghardt7189]

I’m not sure I would write dxdy = rdrdtheta. Each is an expression for a differential area, appropriate (respectively) for rectangular and polar coordinates. They play an analogous role in a 2D integral. But, they are not actually equal to one another.

[@dummyaccount1706]

I would use '≡' instead