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Switching Iterated Integrals from dxdy to dydx

0

$begingroup$

I am having issues switching an iterated integral from dxdy to dydx:

Switch the integral $int_0^2 int_y^2y6xy$ dx dy to a dy dx integral in the form of $int_0^2 int_...^...6xydydx$ + $int_2^4 int_...^...6xydydx$

What I'm having issues with here is trying to break it up into pieces-- is the first one just from 0 to y, and the next one from y to 2y?

calculus integration multivariable-calculus

share|cite|improve this question

asked Oct 25 '17 at 0:21

Scott BScott B

1

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  How about drawing the region of integration?
  $endgroup$
  – Lord Shark the Unknown
  Oct 25 '17 at 0:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The hard part for me is that it is in 3 dimension-- do I draw the traces, for instance, holding X constant?
  $endgroup$
  – Scott B
  Oct 25 '17 at 0:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  But it's not in three dimensions. The domain is a subset of the plane. Draw the set $$left(x,y) : 0 leq y leq 2, y leq x leq 2yright$$
  $endgroup$
  – Matthew Leingang
  Oct 25 '17 at 0:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, so x goes from 0 to 4. So then, for the y, would it be from x to x/2 ?
  $endgroup$
  – Scott B
  Oct 25 '17 at 0:58
  

  
  
  
  

add a comment | 

0

$begingroup$

I am having issues switching an iterated integral from dxdy to dydx:

Switch the integral $int_0^2 int_y^2y6xy$ dx dy to a dy dx integral in the form of $int_0^2 int_...^...6xydydx$ + $int_2^4 int_...^...6xydydx$

What I'm having issues with here is trying to break it up into pieces-- is the first one just from 0 to y, and the next one from y to 2y?

calculus integration multivariable-calculus

share|cite|improve this question

asked Oct 25 '17 at 0:21

Scott BScott B

1

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  How about drawing the region of integration?
  $endgroup$
  – Lord Shark the Unknown
  Oct 25 '17 at 0:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The hard part for me is that it is in 3 dimension-- do I draw the traces, for instance, holding X constant?
  $endgroup$
  – Scott B
  Oct 25 '17 at 0:24
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  But it's not in three dimensions. The domain is a subset of the plane. Draw the set $$left(x,y) : 0 leq y leq 2, y leq x leq 2yright$$
  $endgroup$
  – Matthew Leingang
  Oct 25 '17 at 0:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, so x goes from 0 to 4. So then, for the y, would it be from x to x/2 ?
  $endgroup$
  – Scott B
  Oct 25 '17 at 0:58
  

  
  
  
  

add a comment | 

$begingroup$

I am having issues switching an iterated integral from dxdy to dydx:

Switch the integral $int_0^2 int_y^2y6xy$ dx dy to a dy dx integral in the form of $int_0^2 int_...^...6xydydx$ + $int_2^4 int_...^...6xydydx$

What I'm having issues with here is trying to break it up into pieces-- is the first one just from 0 to y, and the next one from y to 2y?

calculus integration multivariable-calculus

share|cite|improve this question

asked Oct 25 '17 at 0:21

Scott BScott B

1

$endgroup$

share|cite|improve this question

asked Oct 25 '17 at 0:21

Scott BScott B

1

share|cite|improve this question

asked Oct 25 '17 at 0:21

Scott BScott B

1

asked Oct 25 '17 at 0:21

Scott BScott B

1

asked Oct 25 '17 at 0:21

Scott BScott B

1

1 Answer
1

active

oldest

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0

$begingroup$

The domain of integration is the inside of the triangle with vertices at the points $(0,0)$, $(2,2)$ and $(4,2)$, bounded by the lines:$y=x$, $y=x/2$ and $y=2$. The integral of an arbitrary function $f(x,y)$ in this domain can be done first in $x$ and then in $y$, or first in $y$ and then in $x$:

$$int_0^2dyint_y^2yf(x,y),dx = int_0^2dxint_x/2^xf(x,y),dy+int_2^4dxint_x/2^2f(x,y),dy$$

share|cite|improve this answer

answered Oct 25 '17 at 4:57

Anders BetaAnders Beta

76619

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add a comment | 

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0

$begingroup$

The domain of integration is the inside of the triangle with vertices at the points $(0,0)$, $(2,2)$ and $(4,2)$, bounded by the lines:$y=x$, $y=x/2$ and $y=2$. The integral of an arbitrary function $f(x,y)$ in this domain can be done first in $x$ and then in $y$, or first in $y$ and then in $x$:

$$int_0^2dyint_y^2yf(x,y),dx = int_0^2dxint_x/2^xf(x,y),dy+int_2^4dxint_x/2^2f(x,y),dy$$

share|cite|improve this answer

answered Oct 25 '17 at 4:57

Anders BetaAnders Beta

76619

$endgroup$

add a comment | 

0

$begingroup$

The domain of integration is the inside of the triangle with vertices at the points $(0,0)$, $(2,2)$ and $(4,2)$, bounded by the lines:$y=x$, $y=x/2$ and $y=2$. The integral of an arbitrary function $f(x,y)$ in this domain can be done first in $x$ and then in $y$, or first in $y$ and then in $x$:

$$int_0^2dyint_y^2yf(x,y),dx = int_0^2dxint_x/2^xf(x,y),dy+int_2^4dxint_x/2^2f(x,y),dy$$

share|cite|improve this answer

answered Oct 25 '17 at 4:57

Anders BetaAnders Beta

76619

$endgroup$

add a comment | 

$begingroup$

The domain of integration is the inside of the triangle with vertices at the points $(0,0)$, $(2,2)$ and $(4,2)$, bounded by the lines:$y=x$, $y=x/2$ and $y=2$. The integral of an arbitrary function $f(x,y)$ in this domain can be done first in $x$ and then in $y$, or first in $y$ and then in $x$:

$$int_0^2dyint_y^2yf(x,y),dx = int_0^2dxint_x/2^xf(x,y),dy+int_2^4dxint_x/2^2f(x,y),dy$$

share|cite|improve this answer

answered Oct 25 '17 at 4:57

Anders BetaAnders Beta

76619

$endgroup$

share|cite|improve this answer

answered Oct 25 '17 at 4:57

Anders BetaAnders Beta

76619

answered Oct 25 '17 at 4:57

Anders BetaAnders Beta

76619

answered Oct 25 '17 at 4:57

Anders BetaAnders Beta

76619