In the lecture on double integrals over nonrectangular domains we used to demonstrate the basic idea


 Corey Lewis
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1 Double Integals in Pola Coodinates In the lectue on double integals ove nonectangula domains we used to demonstate the basic idea with gaphics and animations the following: Howeve this paticula example didn't show up in the examples. The function hee is f ( xy ) e y ove the cicle x y 9. Had we set up the integal we would have had: = 9x 9x e y dy d x uh no thanks. Howeve if we efomulate the poblem in tems of pola coodinates we have something much moe manageable. Thee ae issues we must decide. How do we set the integation limits?. What fom does the aeal element da take the answe is not simply ddθ as you might fist expect
2 . Setting the Integation Limits Typically when pola coodinates ae used we have a egion such as:. The Aeal element da Recall in Rectangula coodinates we patition the egion R accoding to x = constant vetical lines and y = constant which ae hoizontal lines.
3 This patition ceates ectangles with Δ A = Δ xδ y which when we let the numbe of ectangles go to We obtain da = dxdy Suppose we patition the egion R accoding to θ = constant which ae lines emanating fom the oigin and = constant which ae concentic cicles centeed at the oigin. Recall the length of a cicula ac is s = θ
4 Speaking without much igo we have a "ectangle" whose length and width ae d and dθ and hence da = ddθ. This is a good way of emembeing da but let's be moe igoous. da is the diffeence between the aea of the cicula secto of adius and the one of +d. Recall the aea of a cicula secto subtended by an angle θ is A. If we let denote the midpoint of the segment fom m to +d We have : da m d d m d d da m d m dd 8 d d m d m dd 8 d d
5 da dd so in geneal we have da = ddθ. m Theefoe in Pola Coodinates The geneal fom of the double Integal is : = g ( ) g ( ) f ( ) d d Example Suppose we have the egion inside the Cadioid cos ( ) but outside the cicle. Suppose f(x,y) = x y is the density. Find the Mass. in pola fom f(,θ ) = cos( ) d d
6 Now α and β ae detemined fom the intesections of the cuves cos ( ) This yields cos ( ) theefoe θ = π / and π / cos( ) d d cos( ) d d cos ( ) d sin( ) sin( ) A couple of questions aise. Could we have simply integated fom to π / and double the esult? In geneal no. Conside the gaph at the beginning of this discussion. Even though the aeas in the st and 4th quadants ae symmetic the suface lying ove these egions isn't. We need both symmety ove the egion of integation and the suface ove that egion. Having said that, in ou example f(,θ ) =/ is symmetic and so in ou example we could have integated half the egion and doubled it.. Could we have set the integation limits to be 5π / to π /? No if we did we'd be integating clockwise coveing the following egion:
7 Example Evaluate y 4y dxd y x y Let's conside a gaph of the egion
8 x 4 y is the ight half of the cicle = and x = y is the line θ = π /4. the intesect when 4 y y o 4 y y o y =. Note since the lowe limit on y is we don't need to conside If we convet to pola coodinates x y y 4y dxdy x y 4 d d using the usub u du d we obtain 4 d d 4 5 du d Example Suppose we have Then we ae looking at a solid whose height has the constant value theefoe the volume is numeically equal to the aea of R. i.e. we can use a double integal to compute the aea of a plane egion.
9 With this in mind Calculate the Aea of the egion inside the cicle = and to the ight of the line x =. Hee we can Compute the Aea in the fis quadant and double the esult. What about x = Recall x = cos(θ ) theefoe we have cos(θ ) = o = sec(θ ) Theefoe vaies fom sec(θ ) to. What about θ? At the uppe limit sec(θ ) = fom which we get θ = π / sec( ) A d d sec ( ) d tan ( ) Example 4 Ok we've been putting it off as long as possible What about 9x 9x e y dy d x?
10 We can evaluate ove the fist and fouth quadant and double the esult. x 9 x 9 x y e y d d e sin ( ) d d Which bings us to a vey impotant point  Thee is no panacea!!!! Even though we now have anothe tool in o Calculus chest we will always un into poblems that can't be solved analytically (you might think integate by patsgood idea  ty it and get back to me in a yea o ). This is the powe of technology  to get a numeical appoximation just highlight the integal and hit equal e sin ( ) d d 74.59
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