Need Help with Double Integral Problem?

[Claire84]

Double Integral Problem...

We've been given a question about double integrals and I'm confued by the integration needed and I figure I'm doing something really dozey because all the others have worked out with the exception of this one-

(sorry I don't know how to do the integration signs!)

integral between 0 and 1 then integral between x^2 and x(the limits mentioned first being the bottom limits) of x/y dydx

I've worked out the 'inner' integral as being the integral between the limits 0 and 1 as (xln(2-x) - xlnx) dx

I'm not really sure where to go from here though because integration by parts doesn't seem to be getting me anywhere. Is there some trick I should use or have I just gone the wrong way about this? Any help would be great, even just to get me going in the right direction. Thanks!

 

[cookiemonster]

$$\int_0^1\int_{x^2}^x\frac{x}{y}\,dy\,dx$$
$$\int_0^1x\int_{x^2}^x\frac{1}{y}\,dy\,dx$$
$$\int_0^1x(\ln{y}\Big|_{x^2}^x)\,dx$$
$$\int_0^1x(\ln{x} - 2\ln{x})\,dx$$
$$-\int_0^1x\ln{x}\,dx$$

...?

Edit: By the way, you can click on the pretty pictures of integrals to see how to write them.

cookiemonster

 

[Claire84]

Damn, I'm such a numpty (there goes that word again) the limits are between 2-x (upper limit) and x (lower limit) for the integral with respect to y.

Thanks for letting me know about the integral signs btw!

 

[cookiemonster]

So the problem is:

$$\int_0^1\int_{x}^{2-x}\frac{x}{y}\,dy\,dx$$

?

Don't want to write another thing up for the wrong problem again. =]

cookiemonster

 

[Claire84]

Here's the link to the question sheet to save any confusion- it's number 4.

http://www.am.qub.ac.uk/users/j.mccann/teaching/ama102/2003/assignments/assign_6.pdf

 

[cookiemonster]

Ah, okay.

From,

$$\int_0^1x(\ln{(2-x)} - \ln{x})\,dx$$

just split up the integral into xln(2-x) and xln(x). The second integral is easy to evaluate and the first can be evaluated using a u = 2-x substitution.

cookiemonster

 

[Claire84]

Thank you! I'm so stupid. I just did a question like that for pure and didn't even think twice about it, I really need my head seen to. Thanks again. I'm so embarrassed.

 

[cookiemonster]

Guess we all need a vacation sometimes.

cookiemonster

 

[Claire84]

Tell me about it. [zz)] I did try substitution initially but I stupidly let u=ln(2-x) Only 2 weeks til Easter hols.

 

[Claire84]

For question 5 of that sheet, should the limits of integration for the dr integral be 0 as the lower limit and 2(1-costheta) as the upper limit? I've carried out the integral with respect to r and have ended up with 8/3 costheta(1-costheta)^3 which I'm now not sure how to integrate with respect to theta. Probably just me doing some dumb thing as before and not realising what was going on. I'm just checking to make sure I've done the initially stuff right because we've only done the double integral with polar coordinates for circles so far in class and they're a bit different to this in terms of the limits of itegration. Thanks.

 

[matt grime]

Been in the pub too long to figure out the details, but to integrate powers of cos and sin remember that

cos(2A)= 2cos^2(A)-1=1-2sin^2(A)

so cos^4(A)= 1/4 * (cos2A+1)^2, and you can write cos^2(2A) in terms of cos4A... there may be a better way of doing it.

 

[matt grime]

Oh me Miss, me. Erm, int cos and int of cos^3 between 0 and 2pi are zero too!

 

[Claire84]

Take it you had a good time down the pub then? Just kidding.

Anyhoo, with that last bit you popsted, what does that mean for my answer? I don't wan be getting any zeros! I need 10pi at the end of it all!

Also I'm not sure where to bring in the double angle formulae stuff. Bloody question. The poblem with half of this guy's questions is that nearly every homework sheet has errors all through so so that's why I don't want to go spending my wole weekend faffing about over it to find it's wrong. However, I'm pretty sure the problem is me and my hatred of integration(a recent occurrence, at one stage in my life I did actually have a brain. ).

 

[matt grime]

expand the bracket multiply be the remaining theta. now you've got 4 terms to integrate. my last hint tells you two of them can be ignored cos you've only got cos and cos cubed, if you'll forgive me for putting it like that. that leaves two to do. integrating cos^2 is moderately hard (remember your Fourier analyis) and gives you some mulitple of pi. putting cos^4(-) in terms of cos^2(2-), and then that in terms of cos4(-) etc gives you something you can integrate.

I'm expecting the numpty post any moment.

and yes the pub was ok, been there since 3 watching the cricket, or the rain as it turned out to be for about an hour or so. ever tried explaining the axiom of choice to an engineer after 4 pints? it's not very easy, or interesting.

 

[Claire84]

Numpty.

Is cos^4theta equal to ((2cos2theta +1)/1)^2? Just checking before I become an even bigger numpty because we haven't used that trig identity like that this year and I can't remember it from my a levels (hence the numptiness).

Sounds like a fun time down the pub- not sure about the engineering talk though! My nights down there tend to consist of me telling everyone every embarrassing secret and then hiding under my duvet for the next week.

What's with cricket anyway? Every guy I know loves it and I just don't see the appeal. Do you playyourself or just take my approach to sport and just sit and watch? I'm more of an F1 girly myself, although it does involve getting up at ridicuolous hours to watch the damn races.

 

[Claire84]

Okay, I'm now getting the answer but it's negative so I'm guessing I'm having a numpty moment somewhere but am on the right track...

 

[matt grime]

Not sure you've got exactly the right trig identity:

$$cos^4\theta = (cos^2\theta)^2 = (\frac{cos2\theta + 1}{2})^2$$

multiply that out and you get some cos two thetas some constant, and the $$cos^22\theta$$ term, which you can use more double angle formulas on.

As for sport, I'm afraid I'm on the participation side of things (a lot; 4 badminton games this week already, plus 2 hours of cricket and 3 of softball, but no tennis or squash sadly) though I'm not averse to watching stuff it means i avoid doing any work. Snooker is good for that kind of thing.

I watched the start (ie theonly interesting bit these days) of the Aussie grand prix. It's Malaysia this weekend isn't it?

 

[matt grime]

Originally posted by Claire84
Okay, I'm now getting the answer but it's negative so I'm guessing I'm having a numpty moment somewhere but am on the right track...

dunno about that cos most things in sight are positive.

don't worry about duvet inducing embarrassment in pubs. Hell, I got asked if I'm always display the lack of tact and diplomacy that I did this evening (opening question: X reckons you hate them, do you?). Oh, and I usually hate talking maths in the pub, but there was no one there to talk to about ordinary stuff - mathematicians are very one dimensional, and predominantly dull men too. Bah.

 

[Claire84]

Exercise freak. I used to do a load of athletics and tennis, but then laziness kicked in and all that went out the window. My trek to uni everday is counted as my exercise!

Oh and yeah, it's Malaysia. Shall have pleasant dark circles under my eyes on Monday me thinks. This race had better be better than the last one because I'm not wasting valuable sleep time to watch Ferrai strut their stuff.

Okay, and for the formula for cos^2 (2theta), don't you get (cos4theta +1)/2?

 

[matt grime]

Originally posted by Claire84
Exercise freak. I used to do a load of athletics and tennis, but then laziness kicked in and all that went out the window. My trek to uni everday is counted as my exercise!

Oh and yeah, it's Malaysia. Shall have pleasant dark circles under my eyes on Monday me thinks. This race had better be better than the last one because I'm not wasting valuable sleep time to watch Ferrai strut their stuff.

Okay, and for the formula for cos^2 (2theta), don't you get (cos4theta +1)/2?

aye that last bit looks about right.

please, this is a quiet time for sport too. i used to do 3 hours of tennis practice a day when i was supposed to be revising for GCSEs

 

[Claire84]

Okay, then never classify yourself as a mathematician cos from what I can tel you don't seem too 1-D and boring. Unless you're hiding something and actually intend to turn out dull. Then you'll start having lunch with other mathematicians everyday and eventually you'll grow mad fuzzy hair and develop a liking for very young girls (it's true- one of the lecturers at uni is slightly odd and all the pure mathematicians hang out with him at lunch- it's the hihglight of my day watching their eating habits).

Anyhoo, as I said, I'll go through and chec for numpty mistakes in this. The expansion before I integrate is 8/3(costheta - 3cos^2theta +3cos^3theta - cos^4theta) so maybe that's wrong

So you only saw the start of the race then? It was probably the only bit that got any adrenaline pumping so you truly didn't miss much.

 

[Claire84]

Why'd you quit then? You could have made it as a tennis pro(you would have had to have been better than 'Tiger' Tim, who I had the 'pleasure' of meeting at Wimbledon last year. ).

 

[matt grime]

I think you're on the right tracks with the integral and it should all drop out.

As for tennis, I had to got to university to do Maths! And terms got in the way of the season. That and the fact that all the people I played against at university were county or better, and I didn't get in the first team.

Grand Prix could be fun if Ralf Shumie takes out his team mate as he's promised to do.

 

[Claire84]

If Ralf does that I shall personally fly out to Malaysia with a samurai sword and chop off something he might just happen to miss! I'm keeping my fingers crossed for a Juan Pablo win thank you very much!

And it seems to be if you do something like Maths there isn't actually much time for anything else. All I'm in is the Applied Maths and Physics society which is just an excuse for free booze really.

Anyway, thanks for your help and I'll let you know if this doofer works out!

 

[Claire84]

Still negative. If I put my limits of integration the other way around for dr I think I'd get it as a positive, but I don't really see why you'd do that. I mean, r doesn't matter if you go into the negtive axis, does it? I mean for a circle it's the radius so can't be negative. I'm confused.

 

[matt grime]

yeah, -ve looks right really doesn't it? just look at the region you've sketched - you're adding up the x coordinate at each point, most of the points have -v x value so the sum will be -ve

 

[Claire84]

Okies, so do you think it'd be okay to leave it like that? Or shoudl I go faff about with the limits to make it positive? Please say the former. ;)

I honestly didn't know that the x-coordinates could have an influence on the sign btw (see, a numpty moment) because I just thought that it was the region contained within the curve and that it would alway be positive. Like for a circle with centre at the origin and say radius 2, how do you not end up with an answer of zero if you're adding positive and negatives?

 

[Claire84]

Oh, and one thin else (not related to this, I just want to make sur I'm on the righttrak with lurvely Analysis). we have e>0 and we've to prove that the serquence ((8n +5)/2n) is eventually confined to the interval (4-e,4+e)

I can get a proof for it being less than 4+e (well, it looks okay ), but I can't get it for being greater than 4-e. I mean is it okay just to say that it's obviously going to be greater than 4-e since the sequence can be written as 4 +5/2n and is therefore always going to be greater than 4? I mean the n's here are the natural numbers, right? Since a sequence doesn't have a term lower than the first term. Hope I'm on the right track here. We'd a similar example in the nots but that had a sin term in it which made the proof a bit different.

 

[matt grime]

adding up bit: whether things cancel out or not depends on what you're integrating: if you're integrating 1 over the disk you get the area, if you're integrating x over the disk you get zero - everything is perfectly symmetric about the origin, and sending x to -x doesn't alter the addition and simultaneously it must be minus 1 times the sum as well! so the answer msut be zero.

second bit:
as 4-e <4<4 + 5/2n <4+e
for n greater than 5/2e the sequence eventaully lies in the interval {4-e,4+e} eventually, yes.

also remember that a-e<b<a+e in general just means

|b-a|<e, that's all,

so here you need b=4, a=4+5/2n,so you just need |-5/2n|<e

ie 5/2n<e

 

[Claire84]

Ah that's good, that last bit you put down was the start of my proof. Okay, not exactly a big deal but we're talkin about me here. I used that for the 4+ e bit but it got a bit stuck for being greater than 4-e so that's why I was hoping you could just state the answer. Just trying to make life easy.

And one final thing (I'm like 20 questions at the moment). If we have a<x<b and c<y<d isn't it okay to multiply these two together provided all the terms are >0 so you'd end up with ac<xy<bd?

 

[matt grime]

yes:

a<x and c positive means

ac<xc

c<y and x positive means

cx<xyso putting them together,

ac<cx<xy

the other bit is left as an easy exercise for the reader

 

[Claire84]

Okies, thanks for that. means one of the questions was a lot easier than was expeted.

I thought the next question was going to be something similar but it wasn't. I'll give you the question and what I've done of it, and any input would be fab.

Let 0<r<=s and take some number e with 0<e<3rs. Suppose that (xn) is a b approximation to r and (yn) is a b approximation to s, where b=e/3s. Prove that (xn,yn) is an e approximation to rs.

I started out by just showing that eventually r-b<xn<r+b and s-b<yn<s+b

I then just thought you could multiply the 2 together and do a bit of subbing in, but that has all gone very pear-shaped and I'm guessing that's not what you're supposed to do since the question says to prove this rather than to show (as the previous question did that I thought would be similar to this one). To get an idea for this proof do I need to have rs-e<xnyn<rs+e and then work backwards from that using the things they've given us?

 

[matt grime]

What does it mean to be a "b approximation" exactly?

here's how i'd start it:

|x(n)y(n)-rs| = |x(n)y(n) - sx(n) + sx(n) - rs|


so you get this is less than or equal to

|x(n)[y(n)-s]| + |s[x(n)-r]|

and possibly by this approx answer you get this is less than or equal to

|x(n)||y(n)-s| +sb

and sb=e/3

I'd need to know exactly what the terminology means though to do more

 

[Claire84]

If b>0 we can say that (xn) is a b approximation to r if (xn) is eventually confined to the interval (r-b,r+b), according to the notes anyway.

 

[matt grime]

Ok, so in my last post subs in the word 'eventually'

now if we can show that |x(n)||(y(n)-s)| is at most 2e/3 we are done, right?

well, we know y(n) is a b approx so this quantity is eventually no more than

|x(n)|b



but we know that |x(n)| < r+b eventually as well since -r-b < r-b < x(n) < r+b by definition.

so this is no more than |r+b||b|

The question is how big is r+b?

well, b is less than s, as is r, so r+b is less than 2s

so |r+b||b| < 2sb = 2e/3

so we've shown that

|x(n)y(n)-xy| < 2e/3 +e/3 = e

as required.