surface integrals — is it “dx” or “ds”?
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I was reading this note about Surface Integrals and came across this paragraph:
Let S be a surface parameterized by $mathbf X : D → mathbf R^3$.
A point $(s_0, t_0) in D$, is mapped to $mathbf X(s_0, t_0) inmathbb R^3$.
An infinitesimal $dx × dt$ parallelogram at $(s_0, t_0) in D$ has area $dx dt$. It’s mapped to an infinitesimal $mathbf T_s(s_0, t_0)ds × mathbf T_t(s_0, t_0)$ rectangle with area $||mathbf T_s × mathbf T_t|| ds dt$, which equals $|mathbf N| ds dt$.
We’ll call this infinitesimal parallelogram the surface area differential, denoted $dS$.
I'm quite new to the topic, and I kind of got confused here. When the note says "An infinitesimal $dx times dt$ parallelogram... has area $dx dt$", are the $dx$'s supposed to be $ds$'s? Otherwise, if $dx$ is right, could someone help me understand why we use $dx$?
calculus integration multivariable-calculus surface-integrals
edited Dec 15 '18 at 17:00
amWhy
1
asked Dec 15 '18 at 15:12
jjhhjjhh
2,09611121
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add a comment |
$begingroup$
I was reading this note about Surface Integrals and came across this paragraph:
Let S be a surface parameterized by $mathbf X : D → mathbf R^3$.
A point $(s_0, t_0) in D$, is mapped to $mathbf X(s_0, t_0) inmathbb R^3$.
An infinitesimal $dx × dt$ parallelogram at $(s_0, t_0) in D$ has area $dx dt$. It’s mapped to an infinitesimal $mathbf T_s(s_0, t_0)ds × mathbf T_t(s_0, t_0)$ rectangle with area $||mathbf T_s × mathbf T_t|| ds dt$, which equals $|mathbf N| ds dt$.
We’ll call this infinitesimal parallelogram the surface area differential, denoted $dS$.
I'm quite new to the topic, and I kind of got confused here. When the note says "An infinitesimal $dx times dt$ parallelogram... has area $dx dt$", are the $dx$'s supposed to be $ds$'s? Otherwise, if $dx$ is right, could someone help me understand why we use $dx$?
calculus integration multivariable-calculus surface-integrals
edited Dec 15 '18 at 17:00
amWhy
1
asked Dec 15 '18 at 15:12
jjhhjjhh
2,09611121
$endgroup$
add a comment |
$begingroup$
I was reading this note about Surface Integrals and came across this paragraph:
Let S be a surface parameterized by $mathbf X : D → mathbf R^3$.
A point $(s_0, t_0) in D$, is mapped to $mathbf X(s_0, t_0) inmathbb R^3$.
An infinitesimal $dx × dt$ parallelogram at $(s_0, t_0) in D$ has area $dx dt$. It’s mapped to an infinitesimal $mathbf T_s(s_0, t_0)ds × mathbf T_t(s_0, t_0)$ rectangle with area $||mathbf T_s × mathbf T_t|| ds dt$, which equals $|mathbf N| ds dt$.
We’ll call this infinitesimal parallelogram the surface area differential, denoted $dS$.
I'm quite new to the topic, and I kind of got confused here. When the note says "An infinitesimal $dx times dt$ parallelogram... has area $dx dt$", are the $dx$'s supposed to be $ds$'s? Otherwise, if $dx$ is right, could someone help me understand why we use $dx$?
calculus integration multivariable-calculus surface-integrals
edited Dec 15 '18 at 17:00
amWhy
1
asked Dec 15 '18 at 15:12
jjhhjjhh
2,09611121
$endgroup$
I was reading this note about Surface Integrals and came across this paragraph:
Let S be a surface parameterized by $mathbf X : D → mathbf R^3$.
A point $(s_0, t_0) in D$, is mapped to $mathbf X(s_0, t_0) inmathbb R^3$.
An infinitesimal $dx × dt$ parallelogram at $(s_0, t_0) in D$ has area $dx dt$. It’s mapped to an infinitesimal $mathbf T_s(s_0, t_0)ds × mathbf T_t(s_0, t_0)$ rectangle with area $||mathbf T_s × mathbf T_t|| ds dt$, which equals $|mathbf N| ds dt$.
We’ll call this infinitesimal parallelogram the surface area differential, denoted $dS$.
I'm quite new to the topic, and I kind of got confused here. When the note says "An infinitesimal $dx times dt$ parallelogram... has area $dx dt$", are the $dx$'s supposed to be $ds$'s? Otherwise, if $dx$ is right, could someone help me understand why we use $dx$?
calculus integration multivariable-calculus surface-integrals
calculus integration multivariable-calculus surface-integrals
edited Dec 15 '18 at 17:00
amWhy
1
asked Dec 15 '18 at 15:12
jjhhjjhh
2,09611121
edited Dec 15 '18 at 17:00
amWhy
1
asked Dec 15 '18 at 15:12
jjhhjjhh
2,09611121
edited Dec 15 '18 at 17:00
amWhy
1
edited Dec 15 '18 at 17:00
amWhy
1
edited Dec 15 '18 at 17:00
amWhy
1
1
asked Dec 15 '18 at 15:12
jjhhjjhh
2,09611121
asked Dec 15 '18 at 15:12
jjhhjjhh
2,09611121
asked Dec 15 '18 at 15:12
jjhhjjhh
2,09611121
2,09611121
add a comment |
add a comment |
1 Answer
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I believe $dx$ should be $ds$. I believe also $T_s(s_0, t_0)ds times T_t(s_0, t_0)$ should be $T_s(s_0, t_0)ds times T_t(s_0, t_0)dt$.
answered Dec 15 '18 at 15:28
Lorenzo B.Lorenzo B.
1,8402520
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1 Answer
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1 Answer
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$begingroup$
I believe $dx$ should be $ds$. I believe also $T_s(s_0, t_0)ds times T_t(s_0, t_0)$ should be $T_s(s_0, t_0)ds times T_t(s_0, t_0)dt$.
answered Dec 15 '18 at 15:28
Lorenzo B.Lorenzo B.
1,8402520
$endgroup$
add a comment |
$begingroup$
I believe $dx$ should be $ds$. I believe also $T_s(s_0, t_0)ds times T_t(s_0, t_0)$ should be $T_s(s_0, t_0)ds times T_t(s_0, t_0)dt$.
answered Dec 15 '18 at 15:28
Lorenzo B.Lorenzo B.
1,8402520
$endgroup$
add a comment |
$begingroup$
I believe $dx$ should be $ds$. I believe also $T_s(s_0, t_0)ds times T_t(s_0, t_0)$ should be $T_s(s_0, t_0)ds times T_t(s_0, t_0)dt$.
answered Dec 15 '18 at 15:28
Lorenzo B.Lorenzo B.
1,8402520
$endgroup$
I believe $dx$ should be $ds$. I believe also $T_s(s_0, t_0)ds times T_t(s_0, t_0)$ should be $T_s(s_0, t_0)ds times T_t(s_0, t_0)dt$.
answered Dec 15 '18 at 15:28
Lorenzo B.Lorenzo B.
1,8402520
answered Dec 15 '18 at 15:28
Lorenzo B.Lorenzo B.
1,8402520
answered Dec 15 '18 at 15:28
Lorenzo B.Lorenzo B.
1,8402520
answered Dec 15 '18 at 15:28
Lorenzo B.Lorenzo B.
1,8402520
1,8402520
add a comment |
add a comment |
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