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Changing order of integration

I have no idea how to do this because I got the wrong answer. What I did made sense to me but its completely wrong according to the solutions.

Q7

Posted from TSR Mobile

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Reply 1
Original post by cooldudeman
I have no idea how to do this because I got the wrong answer. What I did made sense to me but its completely wrong according to the solutions.

Q7

Posted from TSR Mobile


stupid question

firstly the region is the first two bounded regions of sinx, between x = 0 and x =2pi
Original post by TeeEm
stupid question

firstly the region is the first two bounded regions of sinx, between x = 0 and x =2pi


Oh yeah my bad. My answer would change to -1<y<1

So my answer is finding all if the sinx regions from 0 to infinity. I dont know how to limit it to 0 to 2pi

Posted from TSR Mobile
(edited 8 years ago)
Reply 3
Original post by cooldudeman
Oh yeah my bad. My answer would change to -1<y<1

So my answer is finding all if the sinx regions from 0 to infinity. I dont know how to limit it to 0 to 2pi

Posted from TSR Mobile


I think also it is not arcsin for the whole region
Original post by cooldudeman
I have no idea how to do this because I got the wrong answer. What I did made sense to me but its completely wrong according to the solutions.

Q7

Posted from TSR Mobile


It looks to me that you are integrating over all of the area defined by y=sinxy=\sin x for x[0,2π]x \in [0,2\pi] so when you swap order the limits on x for the first chunk will look like:

x=arcsinyx=?f(x,y)dx\int_{x=\arcsin y}^{x=?} f(x,y) dx

where you need to choose ? to make a line segment parallel to the x-axis inside the sine curve, bordered by the left hand limit in [0,90] and the right hand limit in [90,180]. (Yeah, I know - poor description - sorry) I'll leave the rest to you.

I don't think this can be done without splitting up the integral; maybe that's what TeeEm means by "stupid"?
(edited 8 years ago)
Reply 5
Original post by cooldudeman
Oh yeah my bad. My answer would change to -1<y<1

So my answer is finding all if the sinx regions from 0 to infinity. I dont know how to limit it to 0 to 2pi

Posted from TSR Mobile


I made a question for my own use, to explain your one...

I have not typed it yet so you are looking at the solution.

IMG.pdf

part (a)
Integrate 4y over the the region bounded by y = sinx, from 0 to π
this is very easy

next part (b)
verify the answer of part (a) by reversing the limits


your integral has to be split 4 ways!!!

from x = arcsiny to x = π/2, y = 0 to y=1
from x= π/2 to x = π - arcsiny, y = 0 to y=1
from x = π - arcsiny to x =3π/2, y = 0 to y=-1
from x =3π/2 to x = + arcsiny, y = 0 to y=-1
(edited 8 years ago)
Original post by TeeEm


your integral has to be split 4 ways!!!



I thought it can be done in 2, but it's too late for me to work through: arcsiny\arcsin y to πarcsiny\pi - \arcsin y for the +ve chunk etc?
Reply 7
Original post by atsruser
I thought it can be done in 2, but it's too late for me to work through: arcsiny\arcsin y to πarcsiny\pi - \arcsin y for the +ve chunk etc?


possibly it can be done (depending on the integrand) by using some symmetries which at the moment my brain cannot handle ...

in all cases this question with an actual integrand (even as simple as mine, and with half the region) is a joke.

I guess all it is required in the exercise is to show the general expression for 4 integrals of f(x,y) with the limits I am giving (I hope no mistakes on my part)
Reply 8
Original post by cooldudeman
Oh yeah my bad. My answer would change to -1<y<1

So my answer is finding all if the sinx regions from 0 to infinity. I dont know how to limit it to 0 to 2pi

Posted from TSR Mobile



PDF.pdf

and look at this link for other stuff

http://madasmaths.com/archive_maths_booklets_advanced_topics.html


Thanks.

Have you got questions like these? Where finding the limits are tricky.

In this one I still can't make sense out of it. Its just that z is between r and sqrt (1-r^2) which is so tricky for me.

Posted from TSR Mobile
Reply 10
Original post by cooldudeman
Thanks.

Have you got questions like these? Where finding the limits are tricky.

In this one I still can't make sense out of it. Its just that z is between r and sqrt (1-r^2) which is so tricky for me.

Posted from TSR Mobile


look at
the multivariable integration files
Original post by TeeEm
look at
the multivariable integration files


Remember when you told me about symmetry calculations in integrals that have no contribution. You said if there's any odd powers of x or y then it has no contribution. Is this technique only for double integrals?

Posted from TSR Mobile
Reply 12
Original post by cooldudeman
Remember when you told me about symmetry calculations in integrals that have no contribution. You said if there's any odd powers of x or y then it has no contribution. Is this technique only for double integrals?

Posted from TSR Mobile


any odd function integrated in a symmetrical domain equals zero.

it applies to any integral, single, double triple, parametric, vector, surface etc.


trig functions integrated over certain ranges have no contribution either but this is harder to explain ...
Original post by TeeEm
any odd function integrated in a symmetrical domain equals zero.

it applies to any integral, single, double triple, parametric, vector, surface etc.


trig functions integrated over certain ranges have no contribution either but this is harder to explain ...


I have learned about odd and even functions and and if it is odd then on a symmertrical domain, the integral would be zero.

And we were told that sinx is odd. That means sin(f(x)) is also odd right? Where f id any function.

Also how does this hold here. We have x which when integrated is zero but how would you know if x is an odd function?

Posted from TSR Mobile
Reply 14
Original post by cooldudeman
I have learned about odd and even functions and and if it is odd then on a symmertrical domain, the integral would be zero.

And we were told that sinx is odd. That means sin(f(x)) is also odd right? Where f id any function.

Also how does this hold here. We have x which when integrated is zero but how would you know if x is an odd function?

Posted from TSR Mobile


I am not sure what exactly you are asking me.

what holds where?
Original post by TeeEm
I am not sure what exactly you are asking me.

what holds where?


Never mind I just figured it out.

Posted from TSR Mobile
Reply 16
Original post by cooldudeman
Never mind I just figured it out.

Posted from TSR Mobile


here are 2 questions with a lot of simplifications due to "symmetries"


the first has Cartesian simplifications
VI.pdf


the second one has trigonometric simplifications.
BIT.pdf


See if they help
Original post by TeeEm
I am not sure what exactly you are asking me.

what holds where?


When we normally parametrise we almost always give the range of theta for eq as 0 to 2pi. Can we purposely put something like -pi to pi so we can consider symmerty cancellations?

Posted from TSR Mobile
Reply 18
Original post by cooldudeman
When we normally parametrise we almost always give the range of theta for eq as 0 to 2pi. Can we purposely put something like -pi to pi so we can consider symmerty cancellations?

Posted from TSR Mobile


sometimes is possible.
but if you have trig functions there more symmetries so you may not need that.
Graph knowledge very important too


e.g.

sinx, cosx, sin3x, cos5x, sin7x, sinxcosx, sinxcos2x, cosxsin4x etc yield zero from 0 to 2pi

others do not need a full range 0 to 2pi

cosx, sin2x, cos4x etc yields zero from 0 to pi


I also use gamma and beta functions so use more results
Original post by TeeEm
sometimes is possible.
but if you have trig functions there more symmetries so you may not need that.
Graph knowledge very important too


e.g.

sinx, cosx, sin3x, cos5x, sin7x, sinxcosx, sinxcos2x, cosxsin4x etc yield zero from 0 to 2pi

others do not need a full range 0 to 2pi

cosx, sin2x, cos4x etc yields zero from 0 to pi


I also use gamma and beta functions so use more results


That seems like every trig function lol.

So this is not the case for ones like (sinx)^2 right? Because that is what usually what I have to integrate in our questions.

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