wow, what a pita

[DrPizza]

Go for it,

Integrate [Pi * 1/x * sqrt(1 + 1/x^4)] dx
(surface area of the function 1/x rotated about the x-axis.)

Anyone out there with Mathematica want to confirm:

Pi * ln [ sqrt(1 + x^2) + x^2 ] - [ Pi * sqrt(1 + x^2) ]/x^2

Calc class and I did it on the back of napkins at an awards banquet last night. Just wanted to verify the answer is correct.

I checked 1 interval vs. TI83+'s numerical integration and the answers were close, but not quite close enough for me to be comfortable.

 

[K1052]

I though this thread would be about Gyros.

 

[Lonyo]

I would try, but I'm not totally sure what the question is
Not very easy to show integrations in a forum. Any limits, or is it indefinate?

 

[maziwanka]

The Integrator

edit: this is definitely not fun by hand

 

[DrPizza]

no, it wasn't fun by hand... we ended up with an x^3 in the denominator, made it x^4 with an x in the numberator, then used the substitution x^2 = tan(theta). That resulted in the wonderful integral of 1/(cos(theta)sin^2(theta)) which was another pita

 

[DrPizza]

Originally posted by: maziwanka
The Integrator

edit: this is definitely not fun by hand

wow, the answer it gave is really odd:

what is

sqrt
-----
2x^4

sqrt of what?! (that was one of the terms)

 

[Legendary]

Gross but I'll give it a shot.

 

[Yossarian]

Originally posted by: DrPizza

Calc class and I did it on the back of napkins at an awards banquet last night. Just wanted to verify the answer is correct.

good times, good times

 

[Legendary]

Hmmm I didn't get very far but methinks maybe 1/arctan(1/x^2) is involved? - I don't know though, the math I've been taking recently hasn't had me integrating so I'm a bit rusty.

 

[loup garou]

Originally posted by: K1052
I though this thread would be about Gyros.

 

[Random Variable]

Try to integrate x^3/(e^x-1) from zero to infinity, an integral that you encounter in quantum statistics. The answer is pi^4/15. :evil:

 

[MrDudeMan]

i got...

pi*[x^2*ln(sqrt(x^4+1)+x^2)-sqrt(x^4+1)]/(2x^2)

 

[dullard]

MathCad spits out:

[pi/2] * [asinh(x^2)-(1+1/x^4)^0.5]

 

[MrChad]

Originally posted by: K1052
I though this thread would be about Gyros.

Man, am I hungry.

 

[MrDudeMan]

Originally posted by: dullard
MathCad spits out:

[pi/2] * [asinh(x^2)-(1+1/x^4)^0.5]

:shocked: when did hyperbolic sine get introduced?

 

[dullard]

Originally posted by: Bigsm00th
:shocked: when did hyperbolic sine get introduced?

I didn't do the work, nor check the work. I just spent 30 seconds typing in the equation and asked for the integral. I don't think I've seen asinh more than once in my life anyways, so I don't really know how it behaves. But it does always appear in cylindrical coordinate problems. So if it is the real answer, I'm not surprized that it appears in a rotated area calculation.

 

[Brutuskend]

My solution

 

[dullard]

Ok I looked up asinh. It is the same as log[x+(x^2+1)^0.5]. That is nealy what DrPizza had in his admittedly nearly correct solution.

Originally posted by: Brutuskend
My solution

 

[ElMonoDelMar]

I <3 Wolfram.

 

[maziwanka]

hey drpizza, you gotta input the formula exactly as required by mathematica:

for sqrt it needs to be "Sqrt[..]"

note the square brackets and the capitalized S....

 

[ArmenK]

TI-89 says

Difference is the x^4 vs x^2 in the square roots and a factor of 1/2

 

[DrPizza]

Originally posted by: maziwanka
hey drpizza, you gotta input the formula exactly as required by mathematica:

for sqrt it needs to be "Sqrt[..]"

note the square brackets and the capitalized S....

Thanks... I scratched my head and realized I had lowercase S.
I also ended up with arcsinh when I used the integrator online. Not a complete surprise, but quite different from what I had, making me wonder how it's so different from what I had (or rather, creating one helluva tough identity to understand)