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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
edited Jul 26 at 2:23
Nosrati
19.3k41544
asked Jul 24 at 1:31
johny
515
add a comment |Â
up vote
3
down vote
favorite
$$int_0^ ae^pi/4 int_2 log(r/a)^pi/2 f(r, theta) r dr d theta$$
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
edited Jul 26 at 2:23
Nosrati
19.3k41544
asked Jul 24 at 1:31
johny
515
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$$int_0^ ae^pi/4 int_2 log(r/a)^pi/2 f(r, theta) r dr d theta$$
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
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
edited Jul 26 at 2:23
Nosrati
19.3k41544
asked Jul 24 at 1:31
johny
515
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 |Â
add a comment |Â
1 Answer
1
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
$$colorblueint_-infty^fracpi2int_0^ae^fractheta2f(r,theta)r mathrmdr mathrmdtheta$$
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 À/2À/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 |Â
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
$$colorblueint_-infty^fracpi2int_0^ae^fractheta2f(r,theta)r mathrmdr mathrmdtheta$$
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 À/2À/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 |Â
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
$$colorblueint_-infty^fracpi2int_0^ae^fractheta2f(r,theta)r mathrmdr mathrmdtheta$$
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 À/2À/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 |Â
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
$$colorblueint_-infty^fracpi2int_0^ae^fractheta2f(r,theta)r mathrmdr mathrmdtheta$$
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
$$colorblueint_-infty^fracpi2int_0^ae^fractheta2f(r,theta)r mathrmdr mathrmdtheta$$
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
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 À/2À/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 |Â
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 À/2À/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 À/2À/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 À/2À/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 |Â
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