Question arrising from integration homework (advanced integration i guess?)

In summary, the conversation discusses a problem involving finding the surface area when a function is rotated around the x-axis. The speaker mentions different approaches, such as using a substitution or trigonometric substitution. They also mention using integration by parts and finding the proof for a specific formula. They express their interest in math and their reluctance to memorize formulas.
  • #1
teclo
117
0
Hi, so I've got a simple problem to evaluate the surface area for 1/x when rotated around the x axis. Ok, so that takes a quit bit of the magic and you wind up with the integral of (x^4+1)^1/2 over x^3. now with a substition you could make that into a form, sqrt(x2+1) over x2.

That's what the book says to do. I'm a dork though and I don't want to use a 'form'. I mean I have to remember enough already, I'd rather just work though the problems. So I decide to tackle the bastard with some trigonometric substitution. I end up with sec^3 x over tan^2 x. I work that through into an integration by partial fractions problem, and blamo 1/1-u^2 + 1/u^2.

Ok great on the u^-2, but now I've got another integration problem that I have to use a form for -- (1-u^2)^-1. So I tried to put sin x back into u. That winds up asking me to integrate sec x -- blam that one sucks. I've tried integration by parts and the third generation made we think I did something wrong. Anyone familiar with a proof of that and willing to give me any hints on the proper strategy? (on either the proof for (1-u^2)^-1 or sec x)

So while writing this I got an idea that I could try substituting cos x and doing another substitution instead of going back one. I think I've done the homework problem for the integral of csc x. I'm going to do that but I have to get some housework done, go to work and study for physics exam.

It's still going to bug me, I'd really be great for anyone and their ideas, below is the link to the work I've done in pdf if you want to see. The last line should be u^-2 it's a mistake.

http://mypage.iu.edu/~nlcooper/math!.pdf

thanks so much for anyone who can read through my buzzed mathmatical excursions. is differential stuff as fun? I'm looking forward to those and debating on changing my major from physics to math.
 
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  • #2
oh, and i dropped a pi at the last line, i forgot i even put it into parenthesis so i could add it in.
 
  • #3
wouldent that be the integral from -infinity to infinity of 1/Y dx..
since you are revolving around the x axis...
 
  • #4
well the homework gaves a domain of 1 to 3 or something like that. i did the homework problem using the retarded forms, but want to do it the real way. in that case i simplified, making it an integral where the range can be specified after the entire integration process. i don't care about filling in stuff anyway, I'm not in algebra anymore. i just do it when i have to.
 
  • #5
willib said:
wouldent that be the integral from -infinity to infinity of 1/Y dx..
since you are revolving around the x axis...

No, the formula for the surface area is S=2π∫r(x)[1+(f'(x))2]dx.

Teclo,

The integral is pretty easy if you do it by parts. Start from:

∫(1+x4)1/2/x3dx

and let u=(1+x4)1/2 and dv=dx/x3.
 
  • #6
Tom Mattson said:
No, the formula for the surface area is S=2π∫r(x)[1+(f'(x))2]dx.

Teclo,

The integral is pretty easy if you do it by parts. Start from:

∫(1+x4)1/2/x3dx

and let u=(1+x4)1/2 and dv=dx/x3.

i have the solution manual and this is the correct answer up to the part before integration by parts. maybe its around the y-axis instead, but i don't think so

my substituions were a different style than the solution manual, but at the point sqrt(x^4+1)/x^3 i chose to bypass subsituting u^2 for x^4, which would let me use a form sqrt(a^2+u^2)/u^2

basically i guess i wanted to prove that since i didn't understand the way it works out. my book just 'verifies' it. so i start the integration and i believe its correct to 1/(1-u^2)

that integration hint you gave me is awesome, and i will attempt to work it out and learn a new style for something like that. thank you very much.

but i was able to get very far with trig substitiution, i really like using it and i'd like to finish by finding the proof for (1-x^2) or sec x or hints on those differentiations. unless of course there's an error on my pdf (if anyone who has time would look at it you'd be my hero)

i really like math and want to be able to do very well, grad school is a long way away.
 
  • #7
teclo said:
but i was able to get very far with trig substitiution, i really like using it

Trig substitution is a powerful tool, but it is just that: a tool. And as always, it is important to use the right tool for the right job. I tried to continue your work with the trig substitution, and it was unworkable as a secant-tangent type trig integral. I then converted it to a sine-cosine integral, and it was no better. Since those are the only 2 options I can see for continuing, I conclude that using trig substitution for this integral is as inappropriate as using a hacksaw to tighten a screw.

But if you use integration by parts, the answer comes out almost immediately.
 
  • #8
Tom Mattson said:
I then converted it to a sine-cosine integral, and it was no better.

Whoops, your partial fraction expansion does make this approach doable.

However, it's still a lot more work than integrating by parts.
 
  • #9
Tom Mattson said:
Whoops, your partial fraction expansion does make this approach doable.

However, it's still a lot more work than integrating by parts.

yes i am going to work out that method tommorow after my physics exam. I'm primarily interested in integrating sec x. i am almost 100% i can do csc x because i am certain it was a homework problem from earlier in the semester.

i don't memorize the formula things in the back of my calc book. i try i guess somewhat prove them, and then i know how to tackle that type of problem whenever it may arise. using integration by parts hadn't occurred to me since it looked so ugly. the answer key and my professor used a similar technique to what i used. i will have to start working with that some more to see if i can apply it more often, i seem to prefer those crazy trig functions.

thanks for the idea again
 
  • #10
teclo said:
yes i am going to work out that method tommorow after my physics exam. I'm primarily interested in integrating sec x. i am almost 100% i can do csc x because i am certain it was a homework problem from earlier in the semester.

If you can do csc(x), then you can do sec(x). The trick is nearly identical.

In the case of sec(x), multiply the integrand by [sec(x)+tan(x)]/[sec(x)+tan(x)], then do a u-substitution. In the case of csc(x), multiply by [csc(x)+cot(x)]/[csc(x)+cot(x)], and do a u-substitution.
 

1. What is the purpose of integration in advanced calculus?

Integration is a mathematical process used to find the area under a curve or to solve problems involving accumulation, such as calculating total distance traveled or total mass of an object. In advanced calculus, integration is used to solve more complex and abstract problems, such as finding volumes of irregular shapes or determining the convergence of infinite series.

2. What is the difference between indefinite and definite integration?

Indefinite integration is a process of finding a general solution to a function's integral, while definite integration is used to find a specific numerical value of an integral over a given interval. In other words, indefinite integration results in a function, while definite integration results in a number.

3. How do I know which integration method to use?

There are several methods for integration, including substitution, integration by parts, partial fractions, and trigonometric substitutions. The best method to use depends on the form of the integrand and the techniques you are comfortable with. Practice and experience will help you determine which method is most suitable for a given problem.

4. What is the fundamental theorem of calculus?

The fundamental theorem of calculus states that integration and differentiation are inverse operations. In other words, if a function is integrated and then differentiated, the result will be the original function. This theorem is the basis for many integration techniques and is essential to understanding the concept of integration.

5. How can I check if my answer to an integration problem is correct?

You can check your answer by differentiating it and comparing the result to the original integrand. If the two are equal, then your answer is most likely correct. You can also use online integration calculators or ask a tutor or classmate to verify your solution. It is important to check your work to avoid making mistakes and to gain a better understanding of the integration process.

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