OpenStudy (baru):

need help understanding double integrals in polar co-ordinates

OpenStudy (baru):

this is how i see double integrals in rectangular we wan to find volume under the curve z(x,y) \[\int\limits \int\limits z(x,y) dx~ dy\] the first integral \[\int\limits z(x,y)~dx\] gives us area at any cross section parlell to x-z plane... i.e A(y) and the second integral gives use the summation A(y) dy carried out infinite times, which gives volume.

OpenStudy (baru):

is there a similarly intuitive argument in polar co-ordinates because \(dA=rdrd\theta\) dosent convince me at all

OpenStudy (baru):

* under surface z(x,y)

OpenStudy (zale101):

When doing double integrals, always always draw the region, then label the graphs, see which graph is top and which one is at the bottom. Label them as your limits for dy or dx. However, it would be great to post a question. However, when converting them to polar coordinates, always look at these definitions that r is equal to \(\sqrt{x^2+y^2}\) and theta is the radians from where the region starts and stops..Again, it would be easier to explain if there was a question involved....

OpenStudy (astrophysics):

I mean to answer your question..it's really the same thing not sure if you know about generalized coordinates, but it can be transformed from jacobians

OpenStudy (baru):

i can do the questions... i know the procedure... but i dont understand why it gives us volume

OpenStudy (zale101):

also, \(x=rcos\theta \) and \(y=rsin\theta\)

OpenStudy (baru):

@Astrophysics i dont think i know any of those things :/

OpenStudy (astrophysics):

Haha it's alright, guess that sounded like something from star wars..ok but it seems as you're just having a hard time understanding why you are calculating the volume rather than the coordinate system, is that a safe assumption?

OpenStudy (astrophysics):

I mean you know how to calculate but the theory on why/ how it works

OpenStudy (baru):

yes...what i'm looking for is an intuitive understanding,in rectangular i can understand, but in Polar co-ordinates i'm lost for example in rectangular, the first integral gives us area at a cross section similarly \[\int\limits z ~rdr\] gives us what?

OpenStudy (zale101):

The application of double integrals may vary from area to volumes. The area calculates the region \[A=\int\limits_{}^{}\int\limits_{R}^{}dxdy\] The volume calculates under or beneath the surface f(x,y) over the region R \[V=\int\limits_{}^{}\int\limits_{R}^{}f(x,y) dxdy\] |dw:1450776315571:dw|