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2

$begingroup$

This is in my lecture notes, I understand that $x=rcostheta$ so therefore its the integral of $x^2=(rcostheta)^2$ but why is there an r in $rdrdtheta$ in there?

Any help would be appreciated.

integration definite-integrals circles

share|cite|improve this question

edited May 11 at 18:07

user21820

40.7k545165

asked May 11 at 15:10

kingking

698

$endgroup$

add a comment | 

2

$begingroup$

This is in my lecture notes, I understand that $x=rcostheta$ so therefore its the integral of $x^2=(rcostheta)^2$ but why is there an r in $rdrdtheta$ in there?

Any help would be appreciated.

integration definite-integrals circles

share|cite|improve this question

edited May 11 at 18:07

user21820

40.7k545165

asked May 11 at 15:10

kingking

698

$endgroup$

add a comment | 

$begingroup$

This is in my lecture notes, I understand that $x=rcostheta$ so therefore its the integral of $x^2=(rcostheta)^2$ but why is there an r in $rdrdtheta$ in there?

Any help would be appreciated.

integration definite-integrals circles

share|cite|improve this question

edited May 11 at 18:07

user21820

40.7k545165

asked May 11 at 15:10

kingking

698

$endgroup$

share|cite|improve this question

edited May 11 at 18:07

user21820

40.7k545165

asked May 11 at 15:10

kingking

698

share|cite|improve this question

edited May 11 at 18:07

user21820

40.7k545165

asked May 11 at 15:10

kingking

698

edited May 11 at 18:07

user21820

40.7k545165

edited May 11 at 18:07

user21820

40.7k545165

asked May 11 at 15:10

kingking

698

asked May 11 at 15:10

kingking

698

3 Answers
3

active

oldest

votes

4

$begingroup$

When we convert co-ordinates $(x,y)$ to $(r,theta)$,

$dx dy = |J| dr dtheta$

where $J$ is the Jacobian of transformation.

$x = r costheta , y = r sintheta$

Taking partial derivatives,

$x_r = costheta , y_r = sintheta$

$x_theta = - r sintheta , y_theta = r costheta $

$J = beginvmatrix
costheta & sintheta \
- r sintheta & r costheta endvmatrix$

$J = r(costheta)(costheta) + r(sintheta)(sintheta) = r(cos^2theta + sin^2theta) = r$

$$|J| = r$$
Thus, $$dx dy = r dr dtheta$$

share|cite|improve this answer

edited May 11 at 17:55

answered May 11 at 15:24

Ak19Ak19

1,581110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very clearly laid out, thank you!
  $endgroup$
  – king
  May 11 at 15:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome.
  $endgroup$
  – Ak19
  May 11 at 15:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  good answer, but it would look better and be easier to read if you put the backslash in front of all the "sin" and "cos" in it instead of just two of the "cos".
  $endgroup$
  – Paul Sinclair
  May 11 at 17:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Paul Sinclair Thanks for your advice, I've edited it :)
  $endgroup$
  – Ak19
  May 11 at 17:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But you didn't put in the backslashes: rsin theta produces this: $$rsintheta$$ while r sin theta produces this: $$r sin theta$$ Most common functions include a special operator name form like this: sin cos tan log ln arcsin, min, max, sup, inf, lim, limsup, etc. Putting them in helps the TEX engine figure out how better to format the text (more generally you can use operatornamemyoperator to do the same with a "myoperator" function.
  $endgroup$
  – Paul Sinclair
  May 11 at 17:15
  
  

  
  
  
  

 | 
show 1 more comment

5

$begingroup$

Here's the underlying geometric intuition (not a rigorous argument).

In Cartesian coordinates the size of an infinitesimal rectangle with sides $dx$ and $dy$ is just the product $dxdy$, independent of where it is in the plane.

In polar coordinates, changing the angle changes the area more when you're farther from the origin. The area of a small not quite rectangular patch determined by $dtheta$ and $dr$ is $ dr times r dtheta = r dr dtheta$.

There's a picture here: http://citadel.sjfc.edu/faculty/kgreen/vector/block3/jacob/node4.html

share|cite|improve this answer

edited May 11 at 15:22

answered May 11 at 15:16

Ethan BolkerEthan Bolker

48.1k556123

$endgroup$

add a comment | 

3

$begingroup$

That is the formula for changing variables from cartesian coordinates to polar cordinates in double integrals $mathrm d x,mathrm d y$ is replaced with $r,mathrm d r,mathrm dtheta$.

Intuitively, this come from the fact that the area of a small circular sector corresponding to a small increment $mathrm d r$ of the radius and a small increment of the angle $mathrm dtheta$ is approximately $;mathrm drcdot r,mathrm dtheta$

share|cite|improve this answer

edited May 11 at 16:35

Ethan Bolker

48.1k556123

answered May 11 at 15:22

BernardBernard

126k743120

$endgroup$

add a comment | 

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4

$begingroup$

When we convert co-ordinates $(x,y)$ to $(r,theta)$,

$dx dy = |J| dr dtheta$

where $J$ is the Jacobian of transformation.

$x = r costheta , y = r sintheta$

Taking partial derivatives,

$x_r = costheta , y_r = sintheta$

$x_theta = - r sintheta , y_theta = r costheta $

$J = beginvmatrix
costheta & sintheta \
- r sintheta & r costheta endvmatrix$

$J = r(costheta)(costheta) + r(sintheta)(sintheta) = r(cos^2theta + sin^2theta) = r$

$$|J| = r$$
Thus, $$dx dy = r dr dtheta$$

share|cite|improve this answer

edited May 11 at 17:55

answered May 11 at 15:24

Ak19Ak19

1,581110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very clearly laid out, thank you!
  $endgroup$
  – king
  May 11 at 15:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome.
  $endgroup$
  – Ak19
  May 11 at 15:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  good answer, but it would look better and be easier to read if you put the backslash in front of all the "sin" and "cos" in it instead of just two of the "cos".
  $endgroup$
  – Paul Sinclair
  May 11 at 17:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Paul Sinclair Thanks for your advice, I've edited it :)
  $endgroup$
  – Ak19
  May 11 at 17:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But you didn't put in the backslashes: rsin theta produces this: $$rsintheta$$ while r sin theta produces this: $$r sin theta$$ Most common functions include a special operator name form like this: sin cos tan log ln arcsin, min, max, sup, inf, lim, limsup, etc. Putting them in helps the TEX engine figure out how better to format the text (more generally you can use operatornamemyoperator to do the same with a "myoperator" function.
  $endgroup$
  – Paul Sinclair
  May 11 at 17:15
  
  

  
  
  
  

 | 
show 1 more comment

4

$begingroup$

When we convert co-ordinates $(x,y)$ to $(r,theta)$,

$dx dy = |J| dr dtheta$

where $J$ is the Jacobian of transformation.

$x = r costheta , y = r sintheta$

Taking partial derivatives,

$x_r = costheta , y_r = sintheta$

$x_theta = - r sintheta , y_theta = r costheta $

$J = beginvmatrix
costheta & sintheta \
- r sintheta & r costheta endvmatrix$

$J = r(costheta)(costheta) + r(sintheta)(sintheta) = r(cos^2theta + sin^2theta) = r$

$$|J| = r$$
Thus, $$dx dy = r dr dtheta$$

share|cite|improve this answer

edited May 11 at 17:55

answered May 11 at 15:24

Ak19Ak19

1,581110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very clearly laid out, thank you!
  $endgroup$
  – king
  May 11 at 15:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome.
  $endgroup$
  – Ak19
  May 11 at 15:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  good answer, but it would look better and be easier to read if you put the backslash in front of all the "sin" and "cos" in it instead of just two of the "cos".
  $endgroup$
  – Paul Sinclair
  May 11 at 17:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Paul Sinclair Thanks for your advice, I've edited it :)
  $endgroup$
  – Ak19
  May 11 at 17:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But you didn't put in the backslashes: rsin theta produces this: $$rsintheta$$ while r sin theta produces this: $$r sin theta$$ Most common functions include a special operator name form like this: sin cos tan log ln arcsin, min, max, sup, inf, lim, limsup, etc. Putting them in helps the TEX engine figure out how better to format the text (more generally you can use operatornamemyoperator to do the same with a "myoperator" function.
  $endgroup$
  – Paul Sinclair
  May 11 at 17:15
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

When we convert co-ordinates $(x,y)$ to $(r,theta)$,

$dx dy = |J| dr dtheta$

where $J$ is the Jacobian of transformation.

$x = r costheta , y = r sintheta$

Taking partial derivatives,

$x_r = costheta , y_r = sintheta$

$x_theta = - r sintheta , y_theta = r costheta $

$J = beginvmatrix
costheta & sintheta \
- r sintheta & r costheta endvmatrix$

$J = r(costheta)(costheta) + r(sintheta)(sintheta) = r(cos^2theta + sin^2theta) = r$

$$|J| = r$$
Thus, $$dx dy = r dr dtheta$$

share|cite|improve this answer

edited May 11 at 17:55

answered May 11 at 15:24

Ak19Ak19

1,581110

$endgroup$

share|cite|improve this answer

edited May 11 at 17:55

answered May 11 at 15:24

Ak19Ak19

1,581110

answered May 11 at 15:24

Ak19Ak19

1,581110

answered May 11 at 15:24

Ak19Ak19

1,581110

5

$begingroup$

Here's the underlying geometric intuition (not a rigorous argument).

In Cartesian coordinates the size of an infinitesimal rectangle with sides $dx$ and $dy$ is just the product $dxdy$, independent of where it is in the plane.

In polar coordinates, changing the angle changes the area more when you're farther from the origin. The area of a small not quite rectangular patch determined by $dtheta$ and $dr$ is $ dr times r dtheta = r dr dtheta$.

There's a picture here: http://citadel.sjfc.edu/faculty/kgreen/vector/block3/jacob/node4.html

share|cite|improve this answer

edited May 11 at 15:22

answered May 11 at 15:16

Ethan BolkerEthan Bolker

48.1k556123

$endgroup$

add a comment | 

5

$begingroup$

Here's the underlying geometric intuition (not a rigorous argument).

In Cartesian coordinates the size of an infinitesimal rectangle with sides $dx$ and $dy$ is just the product $dxdy$, independent of where it is in the plane.

In polar coordinates, changing the angle changes the area more when you're farther from the origin. The area of a small not quite rectangular patch determined by $dtheta$ and $dr$ is $ dr times r dtheta = r dr dtheta$.

There's a picture here: http://citadel.sjfc.edu/faculty/kgreen/vector/block3/jacob/node4.html

share|cite|improve this answer

edited May 11 at 15:22

answered May 11 at 15:16

Ethan BolkerEthan Bolker

48.1k556123

$endgroup$

add a comment | 

$begingroup$

Here's the underlying geometric intuition (not a rigorous argument).

In Cartesian coordinates the size of an infinitesimal rectangle with sides $dx$ and $dy$ is just the product $dxdy$, independent of where it is in the plane.

In polar coordinates, changing the angle changes the area more when you're farther from the origin. The area of a small not quite rectangular patch determined by $dtheta$ and $dr$ is $ dr times r dtheta = r dr dtheta$.

There's a picture here: http://citadel.sjfc.edu/faculty/kgreen/vector/block3/jacob/node4.html

share|cite|improve this answer

edited May 11 at 15:22

answered May 11 at 15:16

Ethan BolkerEthan Bolker

48.1k556123

$endgroup$

share|cite|improve this answer

edited May 11 at 15:22

answered May 11 at 15:16

Ethan BolkerEthan Bolker

48.1k556123

answered May 11 at 15:16

Ethan BolkerEthan Bolker

48.1k556123

answered May 11 at 15:16

Ethan BolkerEthan Bolker

48.1k556123

3

$begingroup$

That is the formula for changing variables from cartesian coordinates to polar cordinates in double integrals $mathrm d x,mathrm d y$ is replaced with $r,mathrm d r,mathrm dtheta$.

Intuitively, this come from the fact that the area of a small circular sector corresponding to a small increment $mathrm d r$ of the radius and a small increment of the angle $mathrm dtheta$ is approximately $;mathrm drcdot r,mathrm dtheta$

share|cite|improve this answer

edited May 11 at 16:35

Ethan Bolker

48.1k556123

answered May 11 at 15:22

BernardBernard

126k743120

$endgroup$

add a comment | 

3

$begingroup$

That is the formula for changing variables from cartesian coordinates to polar cordinates in double integrals $mathrm d x,mathrm d y$ is replaced with $r,mathrm d r,mathrm dtheta$.

Intuitively, this come from the fact that the area of a small circular sector corresponding to a small increment $mathrm d r$ of the radius and a small increment of the angle $mathrm dtheta$ is approximately $;mathrm drcdot r,mathrm dtheta$

share|cite|improve this answer

edited May 11 at 16:35

Ethan Bolker

48.1k556123

answered May 11 at 15:22

BernardBernard

126k743120

$endgroup$

add a comment | 

$begingroup$

That is the formula for changing variables from cartesian coordinates to polar cordinates in double integrals $mathrm d x,mathrm d y$ is replaced with $r,mathrm d r,mathrm dtheta$.

Intuitively, this come from the fact that the area of a small circular sector corresponding to a small increment $mathrm d r$ of the radius and a small increment of the angle $mathrm dtheta$ is approximately $;mathrm drcdot r,mathrm dtheta$

share|cite|improve this answer

edited May 11 at 16:35

Ethan Bolker

48.1k556123

answered May 11 at 15:22

BernardBernard

126k743120

$endgroup$

share|cite|improve this answer

edited May 11 at 16:35

Ethan Bolker

48.1k556123

answered May 11 at 15:22

BernardBernard

126k743120

edited May 11 at 16:35

Ethan Bolker

48.1k556123

edited May 11 at 16:35

Ethan Bolker

48.1k556123

answered May 11 at 15:22

BernardBernard

126k743120

answered May 11 at 15:22

BernardBernard

126k743120