# Thread: Double Integration in Polar and its Applications

1. ## Double Integration in Polar and its Applications

Hello, math wizzes! I am in need of your assistance. In the spoilers you'll find pictures of problems that I'm having issues getting the answers to and would love if you could either point me in the right direction or identify whatever error I may be making, or possibly even both.

Converting to polar coordinates:

Spoiler

Now, for this one it looks like I'm going to be doing a trig substitution, but I can't seem to get past that last line since I can't do a u-sub with what I'm given. I don't think I made a mistake anywhere, but me making a mistake is definitely not unheard of.

Finding mass and center of mass of a lamina over a region:

1:
Spoiler

For this one, I'm not entirely sure what my limits are supposed to be, at least for theta. We have an example from class that used the hypotenuse of one triangle as the upper limit of theta, and the x-axis was the lower limit (i.e. 0 <= theta <= upper limit). I'm also somewhat certain that the limit for r is supposed to go from 0 to 2sec(theta) (by setting 2 = rcos(theta)) but I may be wrong for that, too. At first I didn't convert to polar and just plugged and chugged by integrating w.r.t y and then x, but got no where near the answer of 54.

2:
Spoiler

For this one, I mat just plain have the limits wrong or am doing something wrong in the integration, and possible both. I've done it twice and have gotten the same answer, which is not the answer I'm supposed to be getting.

If anyone could help me out with these issues and possibly teach me some general tricks I could apply to other problems I would greatly appreciate it.

Oh, and hopefully you can read my handwriting. I know that sometimes it can get a bit messy.

2. ## Re: Double Integration in Polar and its Applications

Well for sec^3 x, you use d/dx tan x = sec^2 x, so let dv = sec^2 x dx
Anyway I'm not clear on how you figured out the limits, I'd go about it something like
Spoiler

which doesn't match up with what you got. That has the integral
Spoiler

#1 really should be easier with x,y coordinates - putting it in as written in Wolfram Alpha gives 54, it's just a matter of doing the integral correctly I think. (antiderivative of x+y with respect to y is xy+1/2y^2 | y=x/2 to 3-x, which should give you a polynomial of x)

#2 I would also not do with polar coordinates unless you have to.

3. ## Re: Double Integration in Polar and its Applications

For the conversion I was going off an example from class that I thought was similar, though your method makes more sense. Still a little bit confused as to what you would use for the u-sub (or v, in your case), but I guess that won't have any relevance since it's wrong to begin with. Also confused as to why the theta integral only goes from 0 to pi/2. :\

Was able to get 32/3 as the answer, ha. Looks like I'm missing a half somewhere and I think it might come from a trig property, though I'm not sure.

Subbed 1+sin^2(theta) instead of 1 - sin^2(theta) for cos^2(theta). Got 16/3.

I'll go ahead and give number 1 another try and take it slow to see if I can actually get to 54 this time. I got the polynomial but I might be simplifying wrong.

Got 54 after giving it another go. Blah.

I'll give number 2 a go without converting.

Got 16/5. Guess I was too anxious to try converting to polar that I made it harder than it should have been.

I'll report shortly.

4. ## Re: Double Integration in Polar and its Applications

That's just observation, in the diagram the function's only positive in the first quadrant, which is r>0, 0<theta<pi/2.

5. ## Re: Double Integration in Polar and its Applications

Oh. That makes sense.

Polar coordinates may just be the death of me in calculus, ugh. Thanks for your help!

6. ## Re: Double Integration in Polar and its Applications

Just a note that for the first qns you were looking at the diagram centred at x = 1, in which your integral would have made a little more sense if you instead transformed the graph to the left by 1 unit. It is a common mistake to forget that the origin for polar should still be at the same origin as the cartesian axes, so be more careful next time. You should do the transformation trick occasionally to exploit symmetries in integrals (and more obvious trigo integrals), but do be careful when doing that.

Polar coordinates are supposed to make things easier, because you can pull all kinds of trigo tricks here and there, and lots of shortcuts too. Makes life a lot easier if you're not dealing with flat stuff. Also sometimes used for DEs, and heavily needed for Quantum. Anything touching e^(_), be prepared to pull some trigo... <---- those will kill me one day too.

As a note, please always label your integrals i.e. y=0, sqrt(2x - x^2). The pendantic annoyance is little compared to the strands of hair you pull out in frustration trying to debug your working.

7. ## Re: Double Integration in Polar and its Applications

Oops, misspoke. I mean integration by parts. sec^3x dx, u=secx, dv=sec^2x dx. That has du = secxtanx dx, v = tanx. So you get int udv = uv - int vdu = secxtanx - int (tan^2x secx)
Which by sec^2x = 1 + tan^2x is int (sec^3 x - sec x)
so 2 int (sec^3 x dx) = secxtanx + int(secx)
The integral of secant x is not particularly easy to figure out so I'd just expect to memorize it or look it up.

8. ## Re: Double Integration in Polar and its Applications

also, for as long as youre in calculus, remember you (probably) wont be asked to solve pathological integrals... the point is to teach you not frustrate you, even if trig subs or [insert here] frustrate you a lot. the problem is you have to learn math backwards because you wouldnt understand or need things like where calculus comes from without the gross oversimplifications demonstrated to you in calc 1-3.

integrating things you have to look up outside your book means youre probably doing it wrong if your book has tables of integrals. at least until higher level undergrad physics and grad math.

9. ## Re: Double Integration in Polar and its Applications

Just another couple of pennies to keep you interested. The Gaussian integral for a cartesian coordinate system can be derived/solved (sorry don't know which is the correct word) by changing from cartesian to polar coordinates, and invoking a very nice equality of product of integrals. Once you understand how it works, you will never need to flip to the data book to find out whether to put your constants below or on top of or with 2pi.

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