Approach to Polar Change of Variable in $int_0^ ae^pi/4 int_2 log(r/a)^pi/2 f(r, theta) r dr d theta$.

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$$int_0^ ae^pi/4 int_2 log(r/a)^pi/2 f(r, theta) r dr d theta$$

In the above integral, for evaluating and sketching it, I have to change the order of integral, but I am not able to get the right approach. Is there any general method to handle change of order of integration in polar coordinates?

The answer is $theta$ varies from $0$ to $pi/2$ and $r$ varies from $a$ to $ae^theta/2$.

calculus integration multivariable-calculus polar-coordinates

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edited Jul 26 at 2:23

Nosrati

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asked Jul 24 at 1:31

johny

515

add a comment | 

up vote
3
down vote

favorite

2

$$int_0^ ae^pi/4 int_2 log(r/a)^pi/2 f(r, theta) r dr d theta$$

In the above integral, for evaluating and sketching it, I have to change the order of integral, but I am not able to get the right approach. Is there any general method to handle change of order of integration in polar coordinates?

The answer is $theta$ varies from $0$ to $pi/2$ and $r$ varies from $a$ to $ae^theta/2$.

calculus integration multivariable-calculus polar-coordinates

share|cite|improve this question

edited Jul 26 at 2:23

Nosrati

19.3k41544

asked Jul 24 at 1:31

johny

515

add a comment | 

up vote
3
down vote

favorite

2

up vote
3
down vote

favorite

2

2

$$int_0^ ae^pi/4 int_2 log(r/a)^pi/2 f(r, theta) r dr d theta$$

In the above integral, for evaluating and sketching it, I have to change the order of integral, but I am not able to get the right approach. Is there any general method to handle change of order of integration in polar coordinates?

The answer is $theta$ varies from $0$ to $pi/2$ and $r$ varies from $a$ to $ae^theta/2$.

calculus integration multivariable-calculus polar-coordinates

share|cite|improve this question

edited Jul 26 at 2:23

Nosrati

19.3k41544

asked Jul 24 at 1:31

johny

515

$$int_0^ ae^pi/4 int_2 log(r/a)^pi/2 f(r, theta) r dr d theta$$

In the above integral, for evaluating and sketching it, I have to change the order of integral, but I am not able to get the right approach. Is there any general method to handle change of order of integration in polar coordinates?

The answer is $theta$ varies from $0$ to $pi/2$ and $r$ varies from $a$ to $ae^theta/2$.

calculus integration multivariable-calculus polar-coordinates

share|cite|improve this question

edited Jul 26 at 2:23

Nosrati

19.3k41544

asked Jul 24 at 1:31

johny

515

share|cite|improve this question

share|cite|improve this question

edited Jul 26 at 2:23

Nosrati

19.3k41544

edited Jul 26 at 2:23

Nosrati

19.3k41544

edited Jul 26 at 2:23

Nosrati

19.3k41544

19.3k41544

asked Jul 24 at 1:31

johny

515

asked Jul 24 at 1:31

johny

515

asked Jul 24 at 1:31

johny

515

515

add a comment | 

add a comment | 

1 Answer
1

active

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votes

up vote
0
down vote

I haven't ever seen a general method to handle change of order in such integrals in (between) any coordinates. In this example we have $2lndfracraleq thetaleqdfracpi2$ and $rgeq0$ hence
$$colorred0leq rleq ae^fractheta2$$
and also
$$0leq rleq ae^pi/4$$
this makes an annulus and gives restriction $0leq dfracraleq e^pi/4$, taking $log$ of sides, then
$$-inftyleq lndfracraleq dfracpi4$$
or $-inftyleq 2lndfracraleq dfracpi2$ gives bound for $theta$
$$colorred-inftyleq thetaleq dfracpi2$$
finally your integral is
$$colorblueint_-infty^fracpi2int_0^ae^fractheta2f(r,theta)r mathrmdr mathrmdtheta$$

share|cite|improve this answer

answered Jul 24 at 2:48

Nosrati

19.3k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry the answer given is different. How did u get area in negative region. Can you plot the two graphs.
  – johny
  Jul 24 at 4:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The answer "The answer is θθ varies from 0 to π/2π/2 and rr varies from a to aeθ" is false. I'll give you an example
  – Nosrati
  Jul 24 at 4:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $a=1$, $f=1$, then $$int_0^ ae^pi/4 int_2 log(r/a)^pi/2 f(r, theta) r dr d theta=dfrac12e^fracpi2$$
  – Nosrati
  Jul 24 at 4:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  while $$int_0^pi/2int_a^ae^t/2 f(r, theta) r dr d theta=-frac12pi-frac12+e^fracpi2$$
  – Nosrati
  Jul 24 at 4:41
  
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

I haven't ever seen a general method to handle change of order in such integrals in (between) any coordinates. In this example we have $2lndfracraleq thetaleqdfracpi2$ and $rgeq0$ hence
$$colorred0leq rleq ae^fractheta2$$
and also
$$0leq rleq ae^pi/4$$
this makes an annulus and gives restriction $0leq dfracraleq e^pi/4$, taking $log$ of sides, then
$$-inftyleq lndfracraleq dfracpi4$$
or $-inftyleq 2lndfracraleq dfracpi2$ gives bound for $theta$
$$colorred-inftyleq thetaleq dfracpi2$$
finally your integral is
$$colorblueint_-infty^fracpi2int_0^ae^fractheta2f(r,theta)r mathrmdr mathrmdtheta$$

share|cite|improve this answer

answered Jul 24 at 2:48

Nosrati

19.3k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry the answer given is different. How did u get area in negative region. Can you plot the two graphs.
  – johny
  Jul 24 at 4:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The answer "The answer is θθ varies from 0 to π/2π/2 and rr varies from a to aeθ" is false. I'll give you an example
  – Nosrati
  Jul 24 at 4:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $a=1$, $f=1$, then $$int_0^ ae^pi/4 int_2 log(r/a)^pi/2 f(r, theta) r dr d theta=dfrac12e^fracpi2$$
  – Nosrati
  Jul 24 at 4:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  while $$int_0^pi/2int_a^ae^t/2 f(r, theta) r dr d theta=-frac12pi-frac12+e^fracpi2$$
  – Nosrati
  Jul 24 at 4:41
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

I haven't ever seen a general method to handle change of order in such integrals in (between) any coordinates. In this example we have $2lndfracraleq thetaleqdfracpi2$ and $rgeq0$ hence
$$colorred0leq rleq ae^fractheta2$$
and also
$$0leq rleq ae^pi/4$$
this makes an annulus and gives restriction $0leq dfracraleq e^pi/4$, taking $log$ of sides, then
$$-inftyleq lndfracraleq dfracpi4$$
or $-inftyleq 2lndfracraleq dfracpi2$ gives bound for $theta$
$$colorred-inftyleq thetaleq dfracpi2$$
finally your integral is
$$colorblueint_-infty^fracpi2int_0^ae^fractheta2f(r,theta)r mathrmdr mathrmdtheta$$

share|cite|improve this answer

answered Jul 24 at 2:48

Nosrati

19.3k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry the answer given is different. How did u get area in negative region. Can you plot the two graphs.
  – johny
  Jul 24 at 4:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The answer "The answer is θθ varies from 0 to π/2π/2 and rr varies from a to aeθ" is false. I'll give you an example
  – Nosrati
  Jul 24 at 4:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $a=1$, $f=1$, then $$int_0^ ae^pi/4 int_2 log(r/a)^pi/2 f(r, theta) r dr d theta=dfrac12e^fracpi2$$
  – Nosrati
  Jul 24 at 4:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  while $$int_0^pi/2int_a^ae^t/2 f(r, theta) r dr d theta=-frac12pi-frac12+e^fracpi2$$
  – Nosrati
  Jul 24 at 4:41
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

I haven't ever seen a general method to handle change of order in such integrals in (between) any coordinates. In this example we have $2lndfracraleq thetaleqdfracpi2$ and $rgeq0$ hence
$$colorred0leq rleq ae^fractheta2$$
and also
$$0leq rleq ae^pi/4$$
this makes an annulus and gives restriction $0leq dfracraleq e^pi/4$, taking $log$ of sides, then
$$-inftyleq lndfracraleq dfracpi4$$
or $-inftyleq 2lndfracraleq dfracpi2$ gives bound for $theta$
$$colorred-inftyleq thetaleq dfracpi2$$
finally your integral is
$$colorblueint_-infty^fracpi2int_0^ae^fractheta2f(r,theta)r mathrmdr mathrmdtheta$$

share|cite|improve this answer

answered Jul 24 at 2:48

Nosrati

19.3k41544

I haven't ever seen a general method to handle change of order in such integrals in (between) any coordinates. In this example we have $2lndfracraleq thetaleqdfracpi2$ and $rgeq0$ hence
$$colorred0leq rleq ae^fractheta2$$
and also
$$0leq rleq ae^pi/4$$
this makes an annulus and gives restriction $0leq dfracraleq e^pi/4$, taking $log$ of sides, then
$$-inftyleq lndfracraleq dfracpi4$$
or $-inftyleq 2lndfracraleq dfracpi2$ gives bound for $theta$
$$colorred-inftyleq thetaleq dfracpi2$$
finally your integral is
$$colorblueint_-infty^fracpi2int_0^ae^fractheta2f(r,theta)r mathrmdr mathrmdtheta$$

share|cite|improve this answer

answered Jul 24 at 2:48

Nosrati

19.3k41544

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 2:48

Nosrati

19.3k41544

answered Jul 24 at 2:48

Nosrati

19.3k41544

answered Jul 24 at 2:48

Nosrati

19.3k41544

19.3k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry the answer given is different. How did u get area in negative region. Can you plot the two graphs.
  – johny
  Jul 24 at 4:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The answer "The answer is θθ varies from 0 to π/2π/2 and rr varies from a to aeθ" is false. I'll give you an example
  – Nosrati
  Jul 24 at 4:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $a=1$, $f=1$, then $$int_0^ ae^pi/4 int_2 log(r/a)^pi/2 f(r, theta) r dr d theta=dfrac12e^fracpi2$$
  – Nosrati
  Jul 24 at 4:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  while $$int_0^pi/2int_a^ae^t/2 f(r, theta) r dr d theta=-frac12pi-frac12+e^fracpi2$$
  – Nosrati
  Jul 24 at 4:41
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry the answer given is different. How did u get area in negative region. Can you plot the two graphs.
  – johny
  Jul 24 at 4:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The answer "The answer is θθ varies from 0 to π/2π/2 and rr varies from a to aeθ" is false. I'll give you an example
  – Nosrati
  Jul 24 at 4:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $a=1$, $f=1$, then $$int_0^ ae^pi/4 int_2 log(r/a)^pi/2 f(r, theta) r dr d theta=dfrac12e^fracpi2$$
  – Nosrati
  Jul 24 at 4:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  while $$int_0^pi/2int_a^ae^t/2 f(r, theta) r dr d theta=-frac12pi-frac12+e^fracpi2$$
  – Nosrati
  Jul 24 at 4:41
  
  

  
  
  
  

sorry the answer given is different. How did u get area in negative region. Can you plot the two graphs.
– johny
Jul 24 at 4:30

sorry the answer given is different. How did u get area in negative region. Can you plot the two graphs.
– johny
Jul 24 at 4:30

The answer "The answer is θθ varies from 0 to π/2π/2 and rr varies from a to aeθ" is false. I'll give you an example
– Nosrati
Jul 24 at 4:34

The answer "The answer is θθ varies from 0 to π/2π/2 and rr varies from a to aeθ" is false. I'll give you an example
– Nosrati
Jul 24 at 4:34

$a=1$, $f=1$, then $$int_0^ ae^pi/4 int_2 log(r/a)^pi/2 f(r, theta) r dr d theta=dfrac12e^fracpi2$$
– Nosrati
Jul 24 at 4:38

$a=1$, $f=1$, then $$int_0^ ae^pi/4 int_2 log(r/a)^pi/2 f(r, theta) r dr d theta=dfrac12e^fracpi2$$
– Nosrati
Jul 24 at 4:38

while $$int_0^pi/2int_a^ae^t/2 f(r, theta) r dr d theta=-frac12pi-frac12+e^fracpi2$$
– Nosrati
Jul 24 at 4:41

while $$int_0^pi/2int_a^ae^t/2 f(r, theta) r dr d theta=-frac12pi-frac12+e^fracpi2$$
– Nosrati
Jul 24 at 4:41

add a comment | 

 

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