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Using transformation to evaluate double integral

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1

$begingroup$

Given the transformation $T(x, y) = (x - y, x + y)$, evaluate the double integral
$iint_R (x^2+y^2) dA$, where $R$ is the rectangle in the $xy$-plane with vertices $A(1, 1)$, $B(2, 2)$, $C(-1, 5)$ and $D(-2, 4)$.

Computing the Jacobian of the transformation, I get $det[DT(x,y)] = 2$. Change the coordinates of the vertices respectively, we get

• 
  $A(1,1)$ becomes $(0,2)$
  


• 
  $B(2,2)$ becomes $(0,4)$
  


• 
  $C(-1,5)$ becomes $(-6,4)$
  


• 
  $D(-2,4)$ becomes $(-6,2)$
  

Solving the double integral, I did:
$$
int_2^4 int_-6^0 2left[(x-y)^2-(x+y)^2right]dxdy
$$
which gave me an answer of 864, however, I should be getting $-54$ as given in the answer sheet, may I know what went wrong?

multivariable-calculus proof-verification transformation multiple-integral jacobian

share|cite|improve this question

edited Mar 28 at 15:10

gt6989b

35.2k22557

asked Mar 28 at 14:52

Cheryl Cheryl

477

$endgroup$

This question has an open bounty worth +50
reputation from Cheryl ending ending at 2019-04-07 07:43:30Z">in 4 days.

Looking for an answer drawing from credible and/or official sources.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Since $x^2+y^2geqslant0$ for any $(x,y)in R$, $displaystyleiintlimits_R(x^2+y^2),mathrm dA$ cannot be negative.
  $endgroup$
  – Saad
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are using the same variable names before and after transformation. This creates confusion.
  $endgroup$
  – Yves Daoust
  2 days ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Given the transformation $T(x, y) = (x - y, x + y)$, evaluate the double integral
$iint_R (x^2+y^2) dA$, where $R$ is the rectangle in the $xy$-plane with vertices $A(1, 1)$, $B(2, 2)$, $C(-1, 5)$ and $D(-2, 4)$.

Computing the Jacobian of the transformation, I get $det[DT(x,y)] = 2$. Change the coordinates of the vertices respectively, we get

• 
  $A(1,1)$ becomes $(0,2)$
  


• 
  $B(2,2)$ becomes $(0,4)$
  


• 
  $C(-1,5)$ becomes $(-6,4)$
  


• 
  $D(-2,4)$ becomes $(-6,2)$
  

Solving the double integral, I did:
$$
int_2^4 int_-6^0 2left[(x-y)^2-(x+y)^2right]dxdy
$$
which gave me an answer of 864, however, I should be getting $-54$ as given in the answer sheet, may I know what went wrong?

multivariable-calculus proof-verification transformation multiple-integral jacobian

share|cite|improve this question

edited Mar 28 at 15:10

gt6989b

35.2k22557

asked Mar 28 at 14:52

Cheryl Cheryl

477

$endgroup$

This question has an open bounty worth +50
reputation from Cheryl ending ending at 2019-04-07 07:43:30Z">in 4 days.

Looking for an answer drawing from credible and/or official sources.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Since $x^2+y^2geqslant0$ for any $(x,y)in R$, $displaystyleiintlimits_R(x^2+y^2),mathrm dA$ cannot be negative.
  $endgroup$
  – Saad
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are using the same variable names before and after transformation. This creates confusion.
  $endgroup$
  – Yves Daoust
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Given the transformation $T(x, y) = (x - y, x + y)$, evaluate the double integral
$iint_R (x^2+y^2) dA$, where $R$ is the rectangle in the $xy$-plane with vertices $A(1, 1)$, $B(2, 2)$, $C(-1, 5)$ and $D(-2, 4)$.

Computing the Jacobian of the transformation, I get $det[DT(x,y)] = 2$. Change the coordinates of the vertices respectively, we get

• 
  $A(1,1)$ becomes $(0,2)$
  


• 
  $B(2,2)$ becomes $(0,4)$
  


• 
  $C(-1,5)$ becomes $(-6,4)$
  


• 
  $D(-2,4)$ becomes $(-6,2)$
  

Solving the double integral, I did:
$$
int_2^4 int_-6^0 2left[(x-y)^2-(x+y)^2right]dxdy
$$
which gave me an answer of 864, however, I should be getting $-54$ as given in the answer sheet, may I know what went wrong?

multivariable-calculus proof-verification transformation multiple-integral jacobian

share|cite|improve this question

edited Mar 28 at 15:10

gt6989b

35.2k22557

asked Mar 28 at 14:52

Cheryl Cheryl

477

$endgroup$

share|cite|improve this question

edited Mar 28 at 15:10

gt6989b

35.2k22557

asked Mar 28 at 14:52

Cheryl Cheryl

477

share|cite|improve this question

edited Mar 28 at 15:10

gt6989b

35.2k22557

asked Mar 28 at 14:52

Cheryl Cheryl

477

edited Mar 28 at 15:10

gt6989b

35.2k22557

edited Mar 28 at 15:10

gt6989b

35.2k22557

asked Mar 28 at 14:52

Cheryl Cheryl

477

asked Mar 28 at 14:52

Cheryl Cheryl

477

1 Answer
1

active

oldest

votes

0

$begingroup$

HINT

One mistake -- you are transforming $u = x-y,v=x+y$ so $x = (u+v)/2$ and $y = (v-u)/2$, and the final integral would be
$$
beginsplit
A &= int_v=2^v=4 int_u=-6^u=0 left[left(fracu+v2right)^2 + left(fracv-u2right)^2right] 2 du dv \
&= frac12
int_v=2^v=4 int_u=-6^u=0 left[(u+v)^2 + (v-u)^2right] du dv \
&= int_v=2^v=4 int_u=-6^u=0 left[u^2+v^2right] du dv
endsplit
$$

UPDATE

$$
beginsplit
A &= int_v=2^v=4 int_u=-6^u=0 left[u^2+v^2right] du dv \
&= 2 int_u=-6^u=0 u^2 du
+ 6int_v=2^v=4 v^2 dv \
&= left.frac23 u^3 right|_u=-6^u=0
+ left.2 v^3 right|_v=2^v=4 \
&= -frac23 (-6)^3 + 2left(4^3-2^3right)\
&= 36times 4+2(64 - 8)\
&= 256.
endsplit
$$

The arithmetic agrees with Wolfram Alpha.

As for your hint, you are (originally) integrating
$$
iint_A left(x^2+y^2right)dA,
$$
where the integrand is always non-negative, so the integral should be positive as well.

share|cite|improve this answer

edited 2 days ago

answered Mar 28 at 15:06

gt6989bgt6989b

35.2k22557

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hmmms, I've tried as you've suggested above, but the answer is still incorrect though!!
  $endgroup$
  – Cheryl
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Cheryl see update
  $endgroup$
  – gt6989b
  2 days ago
  

  
  
  
  

add a comment | 

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0

$begingroup$

HINT

One mistake -- you are transforming $u = x-y,v=x+y$ so $x = (u+v)/2$ and $y = (v-u)/2$, and the final integral would be
$$
beginsplit
A &= int_v=2^v=4 int_u=-6^u=0 left[left(fracu+v2right)^2 + left(fracv-u2right)^2right] 2 du dv \
&= frac12
int_v=2^v=4 int_u=-6^u=0 left[(u+v)^2 + (v-u)^2right] du dv \
&= int_v=2^v=4 int_u=-6^u=0 left[u^2+v^2right] du dv
endsplit
$$

UPDATE

$$
beginsplit
A &= int_v=2^v=4 int_u=-6^u=0 left[u^2+v^2right] du dv \
&= 2 int_u=-6^u=0 u^2 du
+ 6int_v=2^v=4 v^2 dv \
&= left.frac23 u^3 right|_u=-6^u=0
+ left.2 v^3 right|_v=2^v=4 \
&= -frac23 (-6)^3 + 2left(4^3-2^3right)\
&= 36times 4+2(64 - 8)\
&= 256.
endsplit
$$

The arithmetic agrees with Wolfram Alpha.

As for your hint, you are (originally) integrating
$$
iint_A left(x^2+y^2right)dA,
$$
where the integrand is always non-negative, so the integral should be positive as well.

share|cite|improve this answer

edited 2 days ago

answered Mar 28 at 15:06

gt6989bgt6989b

35.2k22557

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hmmms, I've tried as you've suggested above, but the answer is still incorrect though!!
  $endgroup$
  – Cheryl
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Cheryl see update
  $endgroup$
  – gt6989b
  2 days ago
  

  
  
  
  

add a comment | 

0

$begingroup$

HINT

One mistake -- you are transforming $u = x-y,v=x+y$ so $x = (u+v)/2$ and $y = (v-u)/2$, and the final integral would be
$$
beginsplit
A &= int_v=2^v=4 int_u=-6^u=0 left[left(fracu+v2right)^2 + left(fracv-u2right)^2right] 2 du dv \
&= frac12
int_v=2^v=4 int_u=-6^u=0 left[(u+v)^2 + (v-u)^2right] du dv \
&= int_v=2^v=4 int_u=-6^u=0 left[u^2+v^2right] du dv
endsplit
$$

UPDATE

$$
beginsplit
A &= int_v=2^v=4 int_u=-6^u=0 left[u^2+v^2right] du dv \
&= 2 int_u=-6^u=0 u^2 du
+ 6int_v=2^v=4 v^2 dv \
&= left.frac23 u^3 right|_u=-6^u=0
+ left.2 v^3 right|_v=2^v=4 \
&= -frac23 (-6)^3 + 2left(4^3-2^3right)\
&= 36times 4+2(64 - 8)\
&= 256.
endsplit
$$

The arithmetic agrees with Wolfram Alpha.

As for your hint, you are (originally) integrating
$$
iint_A left(x^2+y^2right)dA,
$$
where the integrand is always non-negative, so the integral should be positive as well.

share|cite|improve this answer

edited 2 days ago

answered Mar 28 at 15:06

gt6989bgt6989b

35.2k22557

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hmmms, I've tried as you've suggested above, but the answer is still incorrect though!!
  $endgroup$
  – Cheryl
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Cheryl see update
  $endgroup$
  – gt6989b
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

HINT

One mistake -- you are transforming $u = x-y,v=x+y$ so $x = (u+v)/2$ and $y = (v-u)/2$, and the final integral would be
$$
beginsplit
A &= int_v=2^v=4 int_u=-6^u=0 left[left(fracu+v2right)^2 + left(fracv-u2right)^2right] 2 du dv \
&= frac12
int_v=2^v=4 int_u=-6^u=0 left[(u+v)^2 + (v-u)^2right] du dv \
&= int_v=2^v=4 int_u=-6^u=0 left[u^2+v^2right] du dv
endsplit
$$

UPDATE

$$
beginsplit
A &= int_v=2^v=4 int_u=-6^u=0 left[u^2+v^2right] du dv \
&= 2 int_u=-6^u=0 u^2 du
+ 6int_v=2^v=4 v^2 dv \
&= left.frac23 u^3 right|_u=-6^u=0
+ left.2 v^3 right|_v=2^v=4 \
&= -frac23 (-6)^3 + 2left(4^3-2^3right)\
&= 36times 4+2(64 - 8)\
&= 256.
endsplit
$$

The arithmetic agrees with Wolfram Alpha.

As for your hint, you are (originally) integrating
$$
iint_A left(x^2+y^2right)dA,
$$
where the integrand is always non-negative, so the integral should be positive as well.

share|cite|improve this answer

edited 2 days ago

answered Mar 28 at 15:06

gt6989bgt6989b

35.2k22557

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered Mar 28 at 15:06

gt6989bgt6989b

35.2k22557

answered Mar 28 at 15:06

gt6989bgt6989b

35.2k22557

answered Mar 28 at 15:06

gt6989bgt6989b

35.2k22557