surface area of a parametric curve or revolution
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I have the equations $x=9t-3t^3 $ and $ y=9t^2$ , $0le tle 2$ rotated about the x-axis
what I did to try and solve this was:
$ S= int^2_0 2pi(9t^2) sqrt(9-9t^2)^2+(18t)^2dt $
$S= int^2_0 18pi t^2 sqrt81t^4 +162t^2+81dt $
$ S=int^2_0 18pi t^2 (9-9t^2)dt $
$S= int^2_0 162pi t^2 (t^2+1) dt $
$ S= 162pi [fract^55 +fract^33] $
$ S= frac7344pi5$
I'm not sure where I went wrong , any help pointing that out is greatly appreciated.
parametric surface-integrals
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I have the equations $x=9t-3t^3 $ and $ y=9t^2$ , $0le tle 2$ rotated about the x-axis
what I did to try and solve this was:
$ S= int^2_0 2pi(9t^2) sqrt(9-9t^2)^2+(18t)^2dt $
$S= int^2_0 18pi t^2 sqrt81t^4 +162t^2+81dt $
$ S=int^2_0 18pi t^2 (9-9t^2)dt $
$S= int^2_0 162pi t^2 (t^2+1) dt $
$ S= 162pi [fract^55 +fract^33] $
$ S= frac7344pi5$
I'm not sure where I went wrong , any help pointing that out is greatly appreciated.
parametric surface-integrals
Hence you are claiming you went wrong somewhere I guess you know the right solution. Could you please add the right one somewhere? I think it is easier to seach for a mistake by knowing what the outcome should look like.
â mrtaurho
Aug 7 at 21:06
Furthermore I have to say that I cannot spot any mistake beside the typo with the third line where it should be $(9+9t^2)$.
â mrtaurho
Aug 7 at 21:12
Do you understand "How this curve is"?ãÂÂYour calculation is correct when x increase monotonically.
â Takahiro Waki
Aug 7 at 21:28
wolfram alpha show this curve need a few integrals. wolframalpha.com/input/?i=x%3D9t-3t%5E3,y%3D9t%5E2,+0%3Ct%3C2
â Takahiro Waki
Aug 7 at 21:35
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the equations $x=9t-3t^3 $ and $ y=9t^2$ , $0le tle 2$ rotated about the x-axis
what I did to try and solve this was:
$ S= int^2_0 2pi(9t^2) sqrt(9-9t^2)^2+(18t)^2dt $
$S= int^2_0 18pi t^2 sqrt81t^4 +162t^2+81dt $
$ S=int^2_0 18pi t^2 (9-9t^2)dt $
$S= int^2_0 162pi t^2 (t^2+1) dt $
$ S= 162pi [fract^55 +fract^33] $
$ S= frac7344pi5$
I'm not sure where I went wrong , any help pointing that out is greatly appreciated.
parametric surface-integrals
I have the equations $x=9t-3t^3 $ and $ y=9t^2$ , $0le tle 2$ rotated about the x-axis
what I did to try and solve this was:
$ S= int^2_0 2pi(9t^2) sqrt(9-9t^2)^2+(18t)^2dt $
$S= int^2_0 18pi t^2 sqrt81t^4 +162t^2+81dt $
$ S=int^2_0 18pi t^2 (9-9t^2)dt $
$S= int^2_0 162pi t^2 (t^2+1) dt $
$ S= 162pi [fract^55 +fract^33] $
$ S= frac7344pi5$
I'm not sure where I went wrong , any help pointing that out is greatly appreciated.
parametric surface-integrals
asked Aug 7 at 20:58
C_bri
112
112
Hence you are claiming you went wrong somewhere I guess you know the right solution. Could you please add the right one somewhere? I think it is easier to seach for a mistake by knowing what the outcome should look like.
â mrtaurho
Aug 7 at 21:06
Furthermore I have to say that I cannot spot any mistake beside the typo with the third line where it should be $(9+9t^2)$.
â mrtaurho
Aug 7 at 21:12
Do you understand "How this curve is"?ãÂÂYour calculation is correct when x increase monotonically.
â Takahiro Waki
Aug 7 at 21:28
wolfram alpha show this curve need a few integrals. wolframalpha.com/input/?i=x%3D9t-3t%5E3,y%3D9t%5E2,+0%3Ct%3C2
â Takahiro Waki
Aug 7 at 21:35
add a comment |Â
Hence you are claiming you went wrong somewhere I guess you know the right solution. Could you please add the right one somewhere? I think it is easier to seach for a mistake by knowing what the outcome should look like.
â mrtaurho
Aug 7 at 21:06
Furthermore I have to say that I cannot spot any mistake beside the typo with the third line where it should be $(9+9t^2)$.
â mrtaurho
Aug 7 at 21:12
Do you understand "How this curve is"?ãÂÂYour calculation is correct when x increase monotonically.
â Takahiro Waki
Aug 7 at 21:28
wolfram alpha show this curve need a few integrals. wolframalpha.com/input/?i=x%3D9t-3t%5E3,y%3D9t%5E2,+0%3Ct%3C2
â Takahiro Waki
Aug 7 at 21:35
Hence you are claiming you went wrong somewhere I guess you know the right solution. Could you please add the right one somewhere? I think it is easier to seach for a mistake by knowing what the outcome should look like.
â mrtaurho
Aug 7 at 21:06
Hence you are claiming you went wrong somewhere I guess you know the right solution. Could you please add the right one somewhere? I think it is easier to seach for a mistake by knowing what the outcome should look like.
â mrtaurho
Aug 7 at 21:06
Furthermore I have to say that I cannot spot any mistake beside the typo with the third line where it should be $(9+9t^2)$.
â mrtaurho
Aug 7 at 21:12
Furthermore I have to say that I cannot spot any mistake beside the typo with the third line where it should be $(9+9t^2)$.
â mrtaurho
Aug 7 at 21:12
Do you understand "How this curve is"?ãÂÂYour calculation is correct when x increase monotonically.
â Takahiro Waki
Aug 7 at 21:28
Do you understand "How this curve is"?ãÂÂYour calculation is correct when x increase monotonically.
â Takahiro Waki
Aug 7 at 21:28
wolfram alpha show this curve need a few integrals. wolframalpha.com/input/?i=x%3D9t-3t%5E3,y%3D9t%5E2,+0%3Ct%3C2
â Takahiro Waki
Aug 7 at 21:35
wolfram alpha show this curve need a few integrals. wolframalpha.com/input/?i=x%3D9t-3t%5E3,y%3D9t%5E2,+0%3Ct%3C2
â Takahiro Waki
Aug 7 at 21:35
add a comment |Â
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Hence you are claiming you went wrong somewhere I guess you know the right solution. Could you please add the right one somewhere? I think it is easier to seach for a mistake by knowing what the outcome should look like.
â mrtaurho
Aug 7 at 21:06
Furthermore I have to say that I cannot spot any mistake beside the typo with the third line where it should be $(9+9t^2)$.
â mrtaurho
Aug 7 at 21:12
Do you understand "How this curve is"?ãÂÂYour calculation is correct when x increase monotonically.
â Takahiro Waki
Aug 7 at 21:28
wolfram alpha show this curve need a few integrals. wolframalpha.com/input/?i=x%3D9t-3t%5E3,y%3D9t%5E2,+0%3Ct%3C2
â Takahiro Waki
Aug 7 at 21:35