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1

$begingroup$

Consider the function for $0 leq x leq 2$:

$$y = tan^-1x$$

Using the formula:

$$SA = int 2 pi y L, dx$$

where $y = textradius of the frustum$, and $L$ is the arc length: $ L = int sqrt1 + (fracdydx)^2$ where

beginalign
fracdydx &= frac1 1 + x^2 & &textso that & left( fracdydx right)^2 = frac1x^4 + 2x^2 + 1
endalign

But I'm stuck here. I'm still left with:

$$SA = 2 pi int_0^2 tan^-1x cdot sqrt 1 + frac1x^4 + 2x^2 + 1
,dx$$

Where do I even go from here? What can I do with the inverse tan still in my equation?

Should I just use x actually instead of $tan^-1x$?

integration

share|cite|improve this question

asked Mar 21 at 14:31

Kitty CapitalKitty Capital

1096

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have the feeling that I saw this problem earlier today. In my humble opinion, just use numerical integration.
  $endgroup$
  – Claude Leibovici
  Mar 21 at 14:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by that? How do I get rid of the $tan^-1x$? I dont understand what that means @ClaudeLeibovici
  $endgroup$
  – Kitty Capital
  Mar 21 at 15:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just think that there is no way to get an analytical result for this integral.
  $endgroup$
  – Claude Leibovici
  Mar 21 at 15:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But I want to know if I set this integral up correctly. I'm still leaving a tan^-1x in the setup. I can use a CAD, but I want to know if I did this setup correctly.
  $endgroup$
  – Kitty Capital
  Mar 21 at 15:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ClaudeLeibovici does that make sense? Leaving the $tan^-1x$ feels weird...
  $endgroup$
  – Kitty Capital
  Mar 21 at 15:52
  

  
  
  
  

add a comment | 

1

$begingroup$

Consider the function for $0 leq x leq 2$:

$$y = tan^-1x$$

Using the formula:

$$SA = int 2 pi y L, dx$$

where $y = textradius of the frustum$, and $L$ is the arc length: $ L = int sqrt1 + (fracdydx)^2$ where

beginalign
fracdydx &= frac1 1 + x^2 & &textso that & left( fracdydx right)^2 = frac1x^4 + 2x^2 + 1
endalign

But I'm stuck here. I'm still left with:

$$SA = 2 pi int_0^2 tan^-1x cdot sqrt 1 + frac1x^4 + 2x^2 + 1
,dx$$

Where do I even go from here? What can I do with the inverse tan still in my equation?

Should I just use x actually instead of $tan^-1x$?

integration

share|cite|improve this question

asked Mar 21 at 14:31

Kitty CapitalKitty Capital

1096

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have the feeling that I saw this problem earlier today. In my humble opinion, just use numerical integration.
  $endgroup$
  – Claude Leibovici
  Mar 21 at 14:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by that? How do I get rid of the $tan^-1x$? I dont understand what that means @ClaudeLeibovici
  $endgroup$
  – Kitty Capital
  Mar 21 at 15:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just think that there is no way to get an analytical result for this integral.
  $endgroup$
  – Claude Leibovici
  Mar 21 at 15:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But I want to know if I set this integral up correctly. I'm still leaving a tan^-1x in the setup. I can use a CAD, but I want to know if I did this setup correctly.
  $endgroup$
  – Kitty Capital
  Mar 21 at 15:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ClaudeLeibovici does that make sense? Leaving the $tan^-1x$ feels weird...
  $endgroup$
  – Kitty Capital
  Mar 21 at 15:52
  

  
  
  
  

add a comment | 

$begingroup$

Consider the function for $0 leq x leq 2$:

$$y = tan^-1x$$

Using the formula:

$$SA = int 2 pi y L, dx$$

where $y = textradius of the frustum$, and $L$ is the arc length: $ L = int sqrt1 + (fracdydx)^2$ where

beginalign
fracdydx &= frac1 1 + x^2 & &textso that & left( fracdydx right)^2 = frac1x^4 + 2x^2 + 1
endalign

But I'm stuck here. I'm still left with:

$$SA = 2 pi int_0^2 tan^-1x cdot sqrt 1 + frac1x^4 + 2x^2 + 1
,dx$$

Where do I even go from here? What can I do with the inverse tan still in my equation?

Should I just use x actually instead of $tan^-1x$?

integration

share|cite|improve this question

asked Mar 21 at 14:31

Kitty CapitalKitty Capital

1096

$endgroup$

share|cite|improve this question

asked Mar 21 at 14:31

Kitty CapitalKitty Capital

1096

share|cite|improve this question

asked Mar 21 at 14:31

Kitty CapitalKitty Capital

1096

asked Mar 21 at 14:31

Kitty CapitalKitty Capital

1096

asked Mar 21 at 14:31

Kitty CapitalKitty Capital

1096

1 Answer
1

active

oldest

votes

1

$begingroup$

$$S=2piint_0^2arctan(x)sqrt1-frac1(x^2+1)^2dx$$
Let $x=tan u$ so $dx=sec^2(u)du$:
$$S=2piint_0^ausqrt1-frac1(tan^2(u)+1)^2sec^2(u)du$$
Where $a=arctan2$. We have
$$S=2piint_0^a ufracsqrt1-cos^4(u)cos^2(u)du$$
$$S=2piint_0^a usqrtfrac1-cos^4(u)cos^4(u)du$$
$$S=2piint_0^a usqrtsec^4(u)-1du$$
At this point, Wolfram gives a monstrous result for the antiderivative including a bunch of $log$ functions and a few $mathrmLi_2$ functions. So in theory there is a closed form, but realistically, the best you can do is numerical approximation, and from the Integral Calculator, we have $$Sapprox 7.622399360811291...$$
Although there may be a closed form in terms of a special value of a hypergeometric function... I will update if I find one.

share|cite|improve this answer

answered Mar 21 at 17:02

clathratusclathratus

5,0791439

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I suppose that monstrous is just an understatement ! In any manner $to +1$ (may I cinfess that I gave up before starting ?). Cheers :-)
  $endgroup$
  – Claude Leibovici
  Mar 21 at 17:49
  

  
  
  
  

add a comment | 

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1

$begingroup$

$$S=2piint_0^2arctan(x)sqrt1-frac1(x^2+1)^2dx$$
Let $x=tan u$ so $dx=sec^2(u)du$:
$$S=2piint_0^ausqrt1-frac1(tan^2(u)+1)^2sec^2(u)du$$
Where $a=arctan2$. We have
$$S=2piint_0^a ufracsqrt1-cos^4(u)cos^2(u)du$$
$$S=2piint_0^a usqrtfrac1-cos^4(u)cos^4(u)du$$
$$S=2piint_0^a usqrtsec^4(u)-1du$$
At this point, Wolfram gives a monstrous result for the antiderivative including a bunch of $log$ functions and a few $mathrmLi_2$ functions. So in theory there is a closed form, but realistically, the best you can do is numerical approximation, and from the Integral Calculator, we have $$Sapprox 7.622399360811291...$$
Although there may be a closed form in terms of a special value of a hypergeometric function... I will update if I find one.

share|cite|improve this answer

answered Mar 21 at 17:02

clathratusclathratus

5,0791439

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I suppose that monstrous is just an understatement ! In any manner $to +1$ (may I cinfess that I gave up before starting ?). Cheers :-)
  $endgroup$
  – Claude Leibovici
  Mar 21 at 17:49
  

  
  
  
  

add a comment | 

1

$begingroup$

$$S=2piint_0^2arctan(x)sqrt1-frac1(x^2+1)^2dx$$
Let $x=tan u$ so $dx=sec^2(u)du$:
$$S=2piint_0^ausqrt1-frac1(tan^2(u)+1)^2sec^2(u)du$$
Where $a=arctan2$. We have
$$S=2piint_0^a ufracsqrt1-cos^4(u)cos^2(u)du$$
$$S=2piint_0^a usqrtfrac1-cos^4(u)cos^4(u)du$$
$$S=2piint_0^a usqrtsec^4(u)-1du$$
At this point, Wolfram gives a monstrous result for the antiderivative including a bunch of $log$ functions and a few $mathrmLi_2$ functions. So in theory there is a closed form, but realistically, the best you can do is numerical approximation, and from the Integral Calculator, we have $$Sapprox 7.622399360811291...$$
Although there may be a closed form in terms of a special value of a hypergeometric function... I will update if I find one.

share|cite|improve this answer

answered Mar 21 at 17:02

clathratusclathratus

5,0791439

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I suppose that monstrous is just an understatement ! In any manner $to +1$ (may I cinfess that I gave up before starting ?). Cheers :-)
  $endgroup$
  – Claude Leibovici
  Mar 21 at 17:49
  

  
  
  
  

add a comment | 

$begingroup$

$$S=2piint_0^2arctan(x)sqrt1-frac1(x^2+1)^2dx$$
Let $x=tan u$ so $dx=sec^2(u)du$:
$$S=2piint_0^ausqrt1-frac1(tan^2(u)+1)^2sec^2(u)du$$
Where $a=arctan2$. We have
$$S=2piint_0^a ufracsqrt1-cos^4(u)cos^2(u)du$$
$$S=2piint_0^a usqrtfrac1-cos^4(u)cos^4(u)du$$
$$S=2piint_0^a usqrtsec^4(u)-1du$$
At this point, Wolfram gives a monstrous result for the antiderivative including a bunch of $log$ functions and a few $mathrmLi_2$ functions. So in theory there is a closed form, but realistically, the best you can do is numerical approximation, and from the Integral Calculator, we have $$Sapprox 7.622399360811291...$$
Although there may be a closed form in terms of a special value of a hypergeometric function... I will update if I find one.

share|cite|improve this answer

answered Mar 21 at 17:02

clathratusclathratus

5,0791439

$endgroup$

share|cite|improve this answer

answered Mar 21 at 17:02

clathratusclathratus

5,0791439

answered Mar 21 at 17:02

clathratusclathratus

5,0791439

answered Mar 21 at 17:02

clathratusclathratus

5,0791439