# Double Integral Problem

1. Mar 18, 2004

### 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!

2. Mar 18, 2004

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

...?

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

Last edited: Mar 18, 2004
3. Mar 18, 2004

### 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!

4. Mar 18, 2004

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. =]

5. Mar 18, 2004

### 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 [Broken]

Last edited by a moderator: May 1, 2017
6. Mar 18, 2004

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.

7. Mar 18, 2004

### 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.

8. Mar 18, 2004

Guess we all need a vacation sometimes.

9. Mar 18, 2004

### 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.

Last edited: Mar 18, 2004
10. Mar 19, 2004

### 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.

11. Mar 19, 2004

### 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.

12. Mar 19, 2004

### matt grime

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

13. Mar 19, 2004

### 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 wanna 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. ).

14. Mar 19, 2004

### 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.

15. Mar 19, 2004

### 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.

Last edited: Mar 19, 2004
16. Mar 19, 2004

### 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.......

17. Mar 19, 2004

### 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?

18. Mar 19, 2004

### matt grime

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.

19. Mar 19, 2004

### 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?

20. Mar 19, 2004

### matt grime

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