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Area of a Rectangle using a Double Integral in Polar Coordinates
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I'm trying to find the area (which we know is equal to $ab$) of the rectangle bounded by $0leq xleq a$ and $0leq yleq b$ where $0<a<b$ using a double-integral in polar coordinates. The double-integral must be of the form $r dtheta dr$.
Here is my attempt:
$$2int_0^sqrta^2+b^2int_0^arccos(a/r) r dtheta dr$$
I've attempted to integrate from $theta=0$ until the corner of the rectangle at $theta=arccos(a/r)$. The first integral is in terms of $r$ and the second integral is just integrating $r$ from $r=0$ to the corner at $r=sqrta^2+b^2$. I then multiply the area of this triangle by $2$ in order to produce the complete area of the rectangle.
The issue is that the solutions to this setup are imaginary!
PS: This was a question on a Calculus 3 test I took and I was only able to get 4/16 on this part of the question. I really want to understand if my answer is correct in any way and what is going on with it. I would also appreciate any suggestions on how I might argue for more points back as well! (assuming I'm not just blatantly wrong)
My main question is about WHY this is incorrect, why does this reorganization of variables in order to satisfy the integration-order result in imaginary solutions? Is this just a consequence of the arbitrary constrain of the domain of arccos? If arccos could be extended somehow to a larger space could this integral be solved correctly?
Is it possible to extract the answer, $ab$, from the imaginary solution?
EDIT: I've solved the original integral that I set up and found an oddity. Evaluation of the integral yields a result with three components that I've identified: an imaginary component, an irrational component (in terms of $pi$), and a rational component. Turns out, the magnitude of the rational component is $ab$. Here is the result of the original integral, from Mathematica:
$$a^2mathbbi+(a^2+b^2)arccosleft(a/sqrta^2+b^2right) -ab$$
Clearly the magnitude of the rational component is $ab$... but why; what is the significance of this... Is it merely a coincidence? Is it reasonable to say that, in the context of rational solutions, my original integral is correct in any sense?
calculus integration complex-numbers polar-coordinates
asked Jul 16 at 23:22
Seth Taddiken
1138
add a comment |Â
up vote
1
down vote
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I'm trying to find the area (which we know is equal to $ab$) of the rectangle bounded by $0leq xleq a$ and $0leq yleq b$ where $0<a<b$ using a double-integral in polar coordinates. The double-integral must be of the form $r dtheta dr$.
Here is my attempt:
$$2int_0^sqrta^2+b^2int_0^arccos(a/r) r dtheta dr$$
I've attempted to integrate from $theta=0$ until the corner of the rectangle at $theta=arccos(a/r)$. The first integral is in terms of $r$ and the second integral is just integrating $r$ from $r=0$ to the corner at $r=sqrta^2+b^2$. I then multiply the area of this triangle by $2$ in order to produce the complete area of the rectangle.
The issue is that the solutions to this setup are imaginary!
PS: This was a question on a Calculus 3 test I took and I was only able to get 4/16 on this part of the question. I really want to understand if my answer is correct in any way and what is going on with it. I would also appreciate any suggestions on how I might argue for more points back as well! (assuming I'm not just blatantly wrong)
My main question is about WHY this is incorrect, why does this reorganization of variables in order to satisfy the integration-order result in imaginary solutions? Is this just a consequence of the arbitrary constrain of the domain of arccos? If arccos could be extended somehow to a larger space could this integral be solved correctly?
Is it possible to extract the answer, $ab$, from the imaginary solution?
EDIT: I've solved the original integral that I set up and found an oddity. Evaluation of the integral yields a result with three components that I've identified: an imaginary component, an irrational component (in terms of $pi$), and a rational component. Turns out, the magnitude of the rational component is $ab$. Here is the result of the original integral, from Mathematica:
$$a^2mathbbi+(a^2+b^2)arccosleft(a/sqrta^2+b^2right) -ab$$
Clearly the magnitude of the rational component is $ab$... but why; what is the significance of this... Is it merely a coincidence? Is it reasonable to say that, in the context of rational solutions, my original integral is correct in any sense?
calculus integration complex-numbers polar-coordinates
asked Jul 16 at 23:22
Seth Taddiken
1138
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to find the area (which we know is equal to $ab$) of the rectangle bounded by $0leq xleq a$ and $0leq yleq b$ where $0<a<b$ using a double-integral in polar coordinates. The double-integral must be of the form $r dtheta dr$.
Here is my attempt:
$$2int_0^sqrta^2+b^2int_0^arccos(a/r) r dtheta dr$$
I've attempted to integrate from $theta=0$ until the corner of the rectangle at $theta=arccos(a/r)$. The first integral is in terms of $r$ and the second integral is just integrating $r$ from $r=0$ to the corner at $r=sqrta^2+b^2$. I then multiply the area of this triangle by $2$ in order to produce the complete area of the rectangle.
The issue is that the solutions to this setup are imaginary!
PS: This was a question on a Calculus 3 test I took and I was only able to get 4/16 on this part of the question. I really want to understand if my answer is correct in any way and what is going on with it. I would also appreciate any suggestions on how I might argue for more points back as well! (assuming I'm not just blatantly wrong)
My main question is about WHY this is incorrect, why does this reorganization of variables in order to satisfy the integration-order result in imaginary solutions? Is this just a consequence of the arbitrary constrain of the domain of arccos? If arccos could be extended somehow to a larger space could this integral be solved correctly?
Is it possible to extract the answer, $ab$, from the imaginary solution?
EDIT: I've solved the original integral that I set up and found an oddity. Evaluation of the integral yields a result with three components that I've identified: an imaginary component, an irrational component (in terms of $pi$), and a rational component. Turns out, the magnitude of the rational component is $ab$. Here is the result of the original integral, from Mathematica:
$$a^2mathbbi+(a^2+b^2)arccosleft(a/sqrta^2+b^2right) -ab$$
Clearly the magnitude of the rational component is $ab$... but why; what is the significance of this... Is it merely a coincidence? Is it reasonable to say that, in the context of rational solutions, my original integral is correct in any sense?
calculus integration complex-numbers polar-coordinates
asked Jul 16 at 23:22
Seth Taddiken
1138
I'm trying to find the area (which we know is equal to $ab$) of the rectangle bounded by $0leq xleq a$ and $0leq yleq b$ where $0<a<b$ using a double-integral in polar coordinates. The double-integral must be of the form $r dtheta dr$.
Here is my attempt:
$$2int_0^sqrta^2+b^2int_0^arccos(a/r) r dtheta dr$$
I've attempted to integrate from $theta=0$ until the corner of the rectangle at $theta=arccos(a/r)$. The first integral is in terms of $r$ and the second integral is just integrating $r$ from $r=0$ to the corner at $r=sqrta^2+b^2$. I then multiply the area of this triangle by $2$ in order to produce the complete area of the rectangle.
The issue is that the solutions to this setup are imaginary!
PS: This was a question on a Calculus 3 test I took and I was only able to get 4/16 on this part of the question. I really want to understand if my answer is correct in any way and what is going on with it. I would also appreciate any suggestions on how I might argue for more points back as well! (assuming I'm not just blatantly wrong)
My main question is about WHY this is incorrect, why does this reorganization of variables in order to satisfy the integration-order result in imaginary solutions? Is this just a consequence of the arbitrary constrain of the domain of arccos? If arccos could be extended somehow to a larger space could this integral be solved correctly?
Is it possible to extract the answer, $ab$, from the imaginary solution?
EDIT: I've solved the original integral that I set up and found an oddity. Evaluation of the integral yields a result with three components that I've identified: an imaginary component, an irrational component (in terms of $pi$), and a rational component. Turns out, the magnitude of the rational component is $ab$. Here is the result of the original integral, from Mathematica:
$$a^2mathbbi+(a^2+b^2)arccosleft(a/sqrta^2+b^2right) -ab$$
Clearly the magnitude of the rational component is $ab$... but why; what is the significance of this... Is it merely a coincidence? Is it reasonable to say that, in the context of rational solutions, my original integral is correct in any sense?
calculus integration complex-numbers polar-coordinates
asked Jul 16 at 23:22
Seth Taddiken
1138
edited Jul 18 at 1:58
asked Jul 16 at 23:22
Seth Taddiken
1138
asked Jul 16 at 23:22
Seth Taddiken
1138
asked Jul 16 at 23:22
Seth Taddiken
1138
1138
add a comment |Â
add a comment |Â
2 Answers
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up vote
1
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As it stands, it looks like you've got your order of integration switched (you're integrating $r$ over the bounds you specified for $theta$, and vice versa). Your idea looks perfectly correct, though.
Edit in response to your edit: I doubt changing the lower bound of $r$ from $r=0$ to $r=a$ is the solution. Since the graph of $r=a$ is a circle of radius $a$, all that change would do is remove the area enclosed by that circle from your final answer.
Edit 2: There are a couple problems. First of all, after integrating with respect to $theta$ in the integral you presented, you're left having to integrate $arccos(a/r)$ with respect to $r$, which is indeed imaginary. That's not good. Furthermore, your upper bound for $r$ currently describes a circle with radius $sqrta^2+b^2$, which is not the figure whose area you want to calculate. You need to express the vertical line $x=a$ in polar coordinates, which would be $r=asectheta$. Thus, changing the order of integration, replacing the upper bound of $r$ with $asectheta$, and replacing $arccos(a/r)$ with $arctan(b/a)$ (you can prove those two terms are equivalent using geometry, and using the latter removes an unnecessary $r$ from your bounds), we arrive at the following integral:
$$2int_0^arctan(b/a)int_0^asecthetar dr dtheta,$$
which does, in fact, evaluate to $ab$ (at least according to Desmos).
Edit 3: Here's how that integral actually evaluates to $ab$. Let's call the value of the integral $A$.
$$beginalign
A&=2int_0^arctan(b/a)int_0^asecthetar dr dtheta \
&=2int_0^arctan(b/a)left[frac12r^2right]_0^asecthetadtheta \
&=a^2int_0^arctan(b/a)sec^2theta dtheta \
&=a^2Big[tanthetaBig]_0^arctan(b/a) \
&=a^2left(frac baright) \
&=ab
endalign$$
Another edit: In answer to your main question, yes, it does have to do with the domain of $arccos$. To show why, let's evaluate part of your integral:
$$2int_0^sqrta^2+b^2int_0^arccos(a/r) r dtheta dr=2int_0^sqrta^2+b^2rarccosleft(frac arright)dr$$
The function $f(x)=xarccosleft(frac axright)$ isn't defined for $-alt xlt a$, so it looks like trying to include $r=0$ and any other $r$ less than $a$ in your region of integration is what's giving you the imaginary answer. I don't think there's a way to "fix" that and extract the correct answer; I think getting the correct answer with that order of integration would require different bounds, probably involving $theta=($some function of $r)$, as suggested by the answer to this question, if it's possible at all.
And another edit: It's very interesting that $ab$ turns up in the result of your original integral, but unless there's some deep truth of mathematics I'm not aware of, I'm willing to chalk that up as a coincidence. We often talk about separating the real part of a number from its imaginary part, so perhaps more of an argument could be made if the magnitude of the entire real part of the answer were $ab$, but that's not the case here.
I would suggest that you talk to your instructor and ask how he/she intended you to do this problem with the requisite order of integration (for your sake and also because I'm really curious now).
answered Jul 16 at 23:27
Robert Howard
1,325620
1
Thanks for the catch, that was a typo but the integral still seems to be giving incorrect results!
– Seth Taddiken
Jul 16 at 23:29
1
Ah this is a good solution, but, I should have mentioned this, the question includes the requirement that the double-integral be of the form $r dtheta dr$. You are right to switch the bounds of integration though, that does seem like the logical thing to do!
– Seth Taddiken
Jul 17 at 1:37
The answer to this question (math.stackexchange.com/questions/2358611/…) suggests that what you're trying to do is indeed possible, but I'm still unsure of exactly how to arrive at the integral you want.
– Robert Howard
Jul 17 at 17:57
Ah, the correct way is to integrate theta in terms of r from the border x=a to pi/2 for r from 0 to a, then do something similar for radius a to b, then do the corner for r from b to sqrt(a^2+b^2) and theta from x=a to y=b, all in the corresponding polar forms obviously.
– Seth Taddiken
Jul 20 at 18:22
1
Ok! I’ll do that!
– Seth Taddiken
Jul 20 at 20:50
 |Â
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0
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The limits of the integral in polar coordinates for the lower half of the rectangle are $$ 0le theta le tan^-1 (b/a)$$
$$ 0le rle a sec (theta)$$
Thus you have $$ A= 2 int _0^ tan^-1 (b/a) int _0 ^a sec (theta)rdr d theta =ab$$
Note that the anti derivative of $sec ^2 theta $ is $tan theta $ and $tan (tan ^-1 b/a)=b/a $
answered Jul 17 at 15:53
Mohammad Riazi-Kermani
27.5k41852
Notice the requisite order of integration
– Seth Taddiken
Jul 17 at 16:45
@SethTaddiken The requisite order of integration does not meet the correct format of a polar double integral . The outermost limits of integrals should be constant.
– Mohammad Riazi-Kermani
Jul 17 at 17:11
That’s true, that is why I believe the only way to do this correctly is to split the problem into three integrals. One from $r=0$ to $r=a$, on from $r=a$ to $r=b$, and the last from $r=b$ to $r=sqrta^2+b^2$ with $theta$ ranging from $theta=arccosa/r$ to $theta=arcsinb/r$
– Seth Taddiken
Jul 17 at 17:17
@SethTaddiken You are covering more than the rectangle that you want to cover. Check the graph again and you will see.
– Mohammad Riazi-Kermani
Jul 17 at 17:32
Well with the bounds on $theta$ implied to be from $theta=0$ to $theta=pi /2$, is this what you mean?
– Seth Taddiken
Jul 17 at 17:44
 |Â
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
As it stands, it looks like you've got your order of integration switched (you're integrating $r$ over the bounds you specified for $theta$, and vice versa). Your idea looks perfectly correct, though.
Edit in response to your edit: I doubt changing the lower bound of $r$ from $r=0$ to $r=a$ is the solution. Since the graph of $r=a$ is a circle of radius $a$, all that change would do is remove the area enclosed by that circle from your final answer.
Edit 2: There are a couple problems. First of all, after integrating with respect to $theta$ in the integral you presented, you're left having to integrate $arccos(a/r)$ with respect to $r$, which is indeed imaginary. That's not good. Furthermore, your upper bound for $r$ currently describes a circle with radius $sqrta^2+b^2$, which is not the figure whose area you want to calculate. You need to express the vertical line $x=a$ in polar coordinates, which would be $r=asectheta$. Thus, changing the order of integration, replacing the upper bound of $r$ with $asectheta$, and replacing $arccos(a/r)$ with $arctan(b/a)$ (you can prove those two terms are equivalent using geometry, and using the latter removes an unnecessary $r$ from your bounds), we arrive at the following integral:
$$2int_0^arctan(b/a)int_0^asecthetar dr dtheta,$$
which does, in fact, evaluate to $ab$ (at least according to Desmos).
Edit 3: Here's how that integral actually evaluates to $ab$. Let's call the value of the integral $A$.
$$beginalign
A&=2int_0^arctan(b/a)int_0^asecthetar dr dtheta \
&=2int_0^arctan(b/a)left[frac12r^2right]_0^asecthetadtheta \
&=a^2int_0^arctan(b/a)sec^2theta dtheta \
&=a^2Big[tanthetaBig]_0^arctan(b/a) \
&=a^2left(frac baright) \
&=ab
endalign$$
Another edit: In answer to your main question, yes, it does have to do with the domain of $arccos$. To show why, let's evaluate part of your integral:
$$2int_0^sqrta^2+b^2int_0^arccos(a/r) r dtheta dr=2int_0^sqrta^2+b^2rarccosleft(frac arright)dr$$
The function $f(x)=xarccosleft(frac axright)$ isn't defined for $-alt xlt a$, so it looks like trying to include $r=0$ and any other $r$ less than $a$ in your region of integration is what's giving you the imaginary answer. I don't think there's a way to "fix" that and extract the correct answer; I think getting the correct answer with that order of integration would require different bounds, probably involving $theta=($some function of $r)$, as suggested by the answer to this question, if it's possible at all.
And another edit: It's very interesting that $ab$ turns up in the result of your original integral, but unless there's some deep truth of mathematics I'm not aware of, I'm willing to chalk that up as a coincidence. We often talk about separating the real part of a number from its imaginary part, so perhaps more of an argument could be made if the magnitude of the entire real part of the answer were $ab$, but that's not the case here.
I would suggest that you talk to your instructor and ask how he/she intended you to do this problem with the requisite order of integration (for your sake and also because I'm really curious now).
answered Jul 16 at 23:27
Robert Howard
1,325620
1
Thanks for the catch, that was a typo but the integral still seems to be giving incorrect results!
– Seth Taddiken
Jul 16 at 23:29
1
Ah this is a good solution, but, I should have mentioned this, the question includes the requirement that the double-integral be of the form $r dtheta dr$. You are right to switch the bounds of integration though, that does seem like the logical thing to do!
– Seth Taddiken
Jul 17 at 1:37
The answer to this question (math.stackexchange.com/questions/2358611/…) suggests that what you're trying to do is indeed possible, but I'm still unsure of exactly how to arrive at the integral you want.
– Robert Howard
Jul 17 at 17:57
Ah, the correct way is to integrate theta in terms of r from the border x=a to pi/2 for r from 0 to a, then do something similar for radius a to b, then do the corner for r from b to sqrt(a^2+b^2) and theta from x=a to y=b, all in the corresponding polar forms obviously.
– Seth Taddiken
Jul 20 at 18:22
1
Ok! I’ll do that!
– Seth Taddiken
Jul 20 at 20:50
 |Â
show 1 more comment
up vote
1
down vote
As it stands, it looks like you've got your order of integration switched (you're integrating $r$ over the bounds you specified for $theta$, and vice versa). Your idea looks perfectly correct, though.
Edit in response to your edit: I doubt changing the lower bound of $r$ from $r=0$ to $r=a$ is the solution. Since the graph of $r=a$ is a circle of radius $a$, all that change would do is remove the area enclosed by that circle from your final answer.
Edit 2: There are a couple problems. First of all, after integrating with respect to $theta$ in the integral you presented, you're left having to integrate $arccos(a/r)$ with respect to $r$, which is indeed imaginary. That's not good. Furthermore, your upper bound for $r$ currently describes a circle with radius $sqrta^2+b^2$, which is not the figure whose area you want to calculate. You need to express the vertical line $x=a$ in polar coordinates, which would be $r=asectheta$. Thus, changing the order of integration, replacing the upper bound of $r$ with $asectheta$, and replacing $arccos(a/r)$ with $arctan(b/a)$ (you can prove those two terms are equivalent using geometry, and using the latter removes an unnecessary $r$ from your bounds), we arrive at the following integral:
$$2int_0^arctan(b/a)int_0^asecthetar dr dtheta,$$
which does, in fact, evaluate to $ab$ (at least according to Desmos).
Edit 3: Here's how that integral actually evaluates to $ab$. Let's call the value of the integral $A$.
$$beginalign
A&=2int_0^arctan(b/a)int_0^asecthetar dr dtheta \
&=2int_0^arctan(b/a)left[frac12r^2right]_0^asecthetadtheta \
&=a^2int_0^arctan(b/a)sec^2theta dtheta \
&=a^2Big[tanthetaBig]_0^arctan(b/a) \
&=a^2left(frac baright) \
&=ab
endalign$$
Another edit: In answer to your main question, yes, it does have to do with the domain of $arccos$. To show why, let's evaluate part of your integral:
$$2int_0^sqrta^2+b^2int_0^arccos(a/r) r dtheta dr=2int_0^sqrta^2+b^2rarccosleft(frac arright)dr$$
The function $f(x)=xarccosleft(frac axright)$ isn't defined for $-alt xlt a$, so it looks like trying to include $r=0$ and any other $r$ less than $a$ in your region of integration is what's giving you the imaginary answer. I don't think there's a way to "fix" that and extract the correct answer; I think getting the correct answer with that order of integration would require different bounds, probably involving $theta=($some function of $r)$, as suggested by the answer to this question, if it's possible at all.
And another edit: It's very interesting that $ab$ turns up in the result of your original integral, but unless there's some deep truth of mathematics I'm not aware of, I'm willing to chalk that up as a coincidence. We often talk about separating the real part of a number from its imaginary part, so perhaps more of an argument could be made if the magnitude of the entire real part of the answer were $ab$, but that's not the case here.
I would suggest that you talk to your instructor and ask how he/she intended you to do this problem with the requisite order of integration (for your sake and also because I'm really curious now).
answered Jul 16 at 23:27
Robert Howard
1,325620
1
Thanks for the catch, that was a typo but the integral still seems to be giving incorrect results!
– Seth Taddiken
Jul 16 at 23:29
1
Ah this is a good solution, but, I should have mentioned this, the question includes the requirement that the double-integral be of the form $r dtheta dr$. You are right to switch the bounds of integration though, that does seem like the logical thing to do!
– Seth Taddiken
Jul 17 at 1:37
The answer to this question (math.stackexchange.com/questions/2358611/…) suggests that what you're trying to do is indeed possible, but I'm still unsure of exactly how to arrive at the integral you want.
– Robert Howard
Jul 17 at 17:57
Ah, the correct way is to integrate theta in terms of r from the border x=a to pi/2 for r from 0 to a, then do something similar for radius a to b, then do the corner for r from b to sqrt(a^2+b^2) and theta from x=a to y=b, all in the corresponding polar forms obviously.
– Seth Taddiken
Jul 20 at 18:22
1
Ok! I’ll do that!
– Seth Taddiken
Jul 20 at 20:50
 |Â
show 1 more comment
up vote
1
down vote
up vote
1
down vote
As it stands, it looks like you've got your order of integration switched (you're integrating $r$ over the bounds you specified for $theta$, and vice versa). Your idea looks perfectly correct, though.
Edit in response to your edit: I doubt changing the lower bound of $r$ from $r=0$ to $r=a$ is the solution. Since the graph of $r=a$ is a circle of radius $a$, all that change would do is remove the area enclosed by that circle from your final answer.
Edit 2: There are a couple problems. First of all, after integrating with respect to $theta$ in the integral you presented, you're left having to integrate $arccos(a/r)$ with respect to $r$, which is indeed imaginary. That's not good. Furthermore, your upper bound for $r$ currently describes a circle with radius $sqrta^2+b^2$, which is not the figure whose area you want to calculate. You need to express the vertical line $x=a$ in polar coordinates, which would be $r=asectheta$. Thus, changing the order of integration, replacing the upper bound of $r$ with $asectheta$, and replacing $arccos(a/r)$ with $arctan(b/a)$ (you can prove those two terms are equivalent using geometry, and using the latter removes an unnecessary $r$ from your bounds), we arrive at the following integral:
$$2int_0^arctan(b/a)int_0^asecthetar dr dtheta,$$
which does, in fact, evaluate to $ab$ (at least according to Desmos).
Edit 3: Here's how that integral actually evaluates to $ab$. Let's call the value of the integral $A$.
$$beginalign
A&=2int_0^arctan(b/a)int_0^asecthetar dr dtheta \
&=2int_0^arctan(b/a)left[frac12r^2right]_0^asecthetadtheta \
&=a^2int_0^arctan(b/a)sec^2theta dtheta \
&=a^2Big[tanthetaBig]_0^arctan(b/a) \
&=a^2left(frac baright) \
&=ab
endalign$$
Another edit: In answer to your main question, yes, it does have to do with the domain of $arccos$. To show why, let's evaluate part of your integral:
$$2int_0^sqrta^2+b^2int_0^arccos(a/r) r dtheta dr=2int_0^sqrta^2+b^2rarccosleft(frac arright)dr$$
The function $f(x)=xarccosleft(frac axright)$ isn't defined for $-alt xlt a$, so it looks like trying to include $r=0$ and any other $r$ less than $a$ in your region of integration is what's giving you the imaginary answer. I don't think there's a way to "fix" that and extract the correct answer; I think getting the correct answer with that order of integration would require different bounds, probably involving $theta=($some function of $r)$, as suggested by the answer to this question, if it's possible at all.
And another edit: It's very interesting that $ab$ turns up in the result of your original integral, but unless there's some deep truth of mathematics I'm not aware of, I'm willing to chalk that up as a coincidence. We often talk about separating the real part of a number from its imaginary part, so perhaps more of an argument could be made if the magnitude of the entire real part of the answer were $ab$, but that's not the case here.
I would suggest that you talk to your instructor and ask how he/she intended you to do this problem with the requisite order of integration (for your sake and also because I'm really curious now).
answered Jul 16 at 23:27
Robert Howard
1,325620
As it stands, it looks like you've got your order of integration switched (you're integrating $r$ over the bounds you specified for $theta$, and vice versa). Your idea looks perfectly correct, though.
Edit in response to your edit: I doubt changing the lower bound of $r$ from $r=0$ to $r=a$ is the solution. Since the graph of $r=a$ is a circle of radius $a$, all that change would do is remove the area enclosed by that circle from your final answer.
Edit 2: There are a couple problems. First of all, after integrating with respect to $theta$ in the integral you presented, you're left having to integrate $arccos(a/r)$ with respect to $r$, which is indeed imaginary. That's not good. Furthermore, your upper bound for $r$ currently describes a circle with radius $sqrta^2+b^2$, which is not the figure whose area you want to calculate. You need to express the vertical line $x=a$ in polar coordinates, which would be $r=asectheta$. Thus, changing the order of integration, replacing the upper bound of $r$ with $asectheta$, and replacing $arccos(a/r)$ with $arctan(b/a)$ (you can prove those two terms are equivalent using geometry, and using the latter removes an unnecessary $r$ from your bounds), we arrive at the following integral:
$$2int_0^arctan(b/a)int_0^asecthetar dr dtheta,$$
which does, in fact, evaluate to $ab$ (at least according to Desmos).
Edit 3: Here's how that integral actually evaluates to $ab$. Let's call the value of the integral $A$.
$$beginalign
A&=2int_0^arctan(b/a)int_0^asecthetar dr dtheta \
&=2int_0^arctan(b/a)left[frac12r^2right]_0^asecthetadtheta \
&=a^2int_0^arctan(b/a)sec^2theta dtheta \
&=a^2Big[tanthetaBig]_0^arctan(b/a) \
&=a^2left(frac baright) \
&=ab
endalign$$
Another edit: In answer to your main question, yes, it does have to do with the domain of $arccos$. To show why, let's evaluate part of your integral:
$$2int_0^sqrta^2+b^2int_0^arccos(a/r) r dtheta dr=2int_0^sqrta^2+b^2rarccosleft(frac arright)dr$$
The function $f(x)=xarccosleft(frac axright)$ isn't defined for $-alt xlt a$, so it looks like trying to include $r=0$ and any other $r$ less than $a$ in your region of integration is what's giving you the imaginary answer. I don't think there's a way to "fix" that and extract the correct answer; I think getting the correct answer with that order of integration would require different bounds, probably involving $theta=($some function of $r)$, as suggested by the answer to this question, if it's possible at all.
And another edit: It's very interesting that $ab$ turns up in the result of your original integral, but unless there's some deep truth of mathematics I'm not aware of, I'm willing to chalk that up as a coincidence. We often talk about separating the real part of a number from its imaginary part, so perhaps more of an argument could be made if the magnitude of the entire real part of the answer were $ab$, but that's not the case here.
I would suggest that you talk to your instructor and ask how he/she intended you to do this problem with the requisite order of integration (for your sake and also because I'm really curious now).
answered Jul 16 at 23:27
Robert Howard
1,325620
edited Jul 20 at 21:41
answered Jul 16 at 23:27
Robert Howard
1,325620
answered Jul 16 at 23:27
Robert Howard
1,325620
answered Jul 16 at 23:27
Robert Howard
1,325620
1,325620
1
Thanks for the catch, that was a typo but the integral still seems to be giving incorrect results!
– Seth Taddiken
Jul 16 at 23:29
1
Ah this is a good solution, but, I should have mentioned this, the question includes the requirement that the double-integral be of the form $r dtheta dr$. You are right to switch the bounds of integration though, that does seem like the logical thing to do!
– Seth Taddiken
Jul 17 at 1:37
The answer to this question (math.stackexchange.com/questions/2358611/…) suggests that what you're trying to do is indeed possible, but I'm still unsure of exactly how to arrive at the integral you want.
– Robert Howard
Jul 17 at 17:57
Ah, the correct way is to integrate theta in terms of r from the border x=a to pi/2 for r from 0 to a, then do something similar for radius a to b, then do the corner for r from b to sqrt(a^2+b^2) and theta from x=a to y=b, all in the corresponding polar forms obviously.
– Seth Taddiken
Jul 20 at 18:22
1
Ok! I’ll do that!
– Seth Taddiken
Jul 20 at 20:50
 |Â
show 1 more comment
1
Thanks for the catch, that was a typo but the integral still seems to be giving incorrect results!
– Seth Taddiken
Jul 16 at 23:29
1
Ah this is a good solution, but, I should have mentioned this, the question includes the requirement that the double-integral be of the form $r dtheta dr$. You are right to switch the bounds of integration though, that does seem like the logical thing to do!
– Seth Taddiken
Jul 17 at 1:37
The answer to this question (math.stackexchange.com/questions/2358611/…) suggests that what you're trying to do is indeed possible, but I'm still unsure of exactly how to arrive at the integral you want.
– Robert Howard
Jul 17 at 17:57
Ah, the correct way is to integrate theta in terms of r from the border x=a to pi/2 for r from 0 to a, then do something similar for radius a to b, then do the corner for r from b to sqrt(a^2+b^2) and theta from x=a to y=b, all in the corresponding polar forms obviously.
– Seth Taddiken
Jul 20 at 18:22
1
Ok! I’ll do that!
– Seth Taddiken
Jul 20 at 20:50
1
1
Thanks for the catch, that was a typo but the integral still seems to be giving incorrect results!
– Seth Taddiken
Jul 16 at 23:29
Thanks for the catch, that was a typo but the integral still seems to be giving incorrect results!
– Seth Taddiken
Jul 16 at 23:29
1
1
Ah this is a good solution, but, I should have mentioned this, the question includes the requirement that the double-integral be of the form $r dtheta dr$. You are right to switch the bounds of integration though, that does seem like the logical thing to do!
– Seth Taddiken
Jul 17 at 1:37
Ah this is a good solution, but, I should have mentioned this, the question includes the requirement that the double-integral be of the form $r dtheta dr$. You are right to switch the bounds of integration though, that does seem like the logical thing to do!
– Seth Taddiken
Jul 17 at 1:37
The answer to this question (math.stackexchange.com/questions/2358611/…) suggests that what you're trying to do is indeed possible, but I'm still unsure of exactly how to arrive at the integral you want.
– Robert Howard
Jul 17 at 17:57
The answer to this question (math.stackexchange.com/questions/2358611/…) suggests that what you're trying to do is indeed possible, but I'm still unsure of exactly how to arrive at the integral you want.
– Robert Howard
Jul 17 at 17:57
Ah, the correct way is to integrate theta in terms of r from the border x=a to pi/2 for r from 0 to a, then do something similar for radius a to b, then do the corner for r from b to sqrt(a^2+b^2) and theta from x=a to y=b, all in the corresponding polar forms obviously.
– Seth Taddiken
Jul 20 at 18:22
Ah, the correct way is to integrate theta in terms of r from the border x=a to pi/2 for r from 0 to a, then do something similar for radius a to b, then do the corner for r from b to sqrt(a^2+b^2) and theta from x=a to y=b, all in the corresponding polar forms obviously.
– Seth Taddiken
Jul 20 at 18:22
1
1
Ok! I’ll do that!
– Seth Taddiken
Jul 20 at 20:50
Ok! I’ll do that!
– Seth Taddiken
Jul 20 at 20:50
 |Â
show 1 more comment
up vote
0
down vote
The limits of the integral in polar coordinates for the lower half of the rectangle are $$ 0le theta le tan^-1 (b/a)$$
$$ 0le rle a sec (theta)$$
Thus you have $$ A= 2 int _0^ tan^-1 (b/a) int _0 ^a sec (theta)rdr d theta =ab$$
Note that the anti derivative of $sec ^2 theta $ is $tan theta $ and $tan (tan ^-1 b/a)=b/a $
answered Jul 17 at 15:53
Mohammad Riazi-Kermani
27.5k41852
Notice the requisite order of integration
– Seth Taddiken
Jul 17 at 16:45
@SethTaddiken The requisite order of integration does not meet the correct format of a polar double integral . The outermost limits of integrals should be constant.
– Mohammad Riazi-Kermani
Jul 17 at 17:11
That’s true, that is why I believe the only way to do this correctly is to split the problem into three integrals. One from $r=0$ to $r=a$, on from $r=a$ to $r=b$, and the last from $r=b$ to $r=sqrta^2+b^2$ with $theta$ ranging from $theta=arccosa/r$ to $theta=arcsinb/r$
– Seth Taddiken
Jul 17 at 17:17
@SethTaddiken You are covering more than the rectangle that you want to cover. Check the graph again and you will see.
– Mohammad Riazi-Kermani
Jul 17 at 17:32
Well with the bounds on $theta$ implied to be from $theta=0$ to $theta=pi /2$, is this what you mean?
– Seth Taddiken
Jul 17 at 17:44
 |Â
show 1 more comment
up vote
0
down vote
The limits of the integral in polar coordinates for the lower half of the rectangle are $$ 0le theta le tan^-1 (b/a)$$
$$ 0le rle a sec (theta)$$
Thus you have $$ A= 2 int _0^ tan^-1 (b/a) int _0 ^a sec (theta)rdr d theta =ab$$
Note that the anti derivative of $sec ^2 theta $ is $tan theta $ and $tan (tan ^-1 b/a)=b/a $
answered Jul 17 at 15:53
Mohammad Riazi-Kermani
27.5k41852
Notice the requisite order of integration
– Seth Taddiken
Jul 17 at 16:45
@SethTaddiken The requisite order of integration does not meet the correct format of a polar double integral . The outermost limits of integrals should be constant.
– Mohammad Riazi-Kermani
Jul 17 at 17:11
That’s true, that is why I believe the only way to do this correctly is to split the problem into three integrals. One from $r=0$ to $r=a$, on from $r=a$ to $r=b$, and the last from $r=b$ to $r=sqrta^2+b^2$ with $theta$ ranging from $theta=arccosa/r$ to $theta=arcsinb/r$
– Seth Taddiken
Jul 17 at 17:17
@SethTaddiken You are covering more than the rectangle that you want to cover. Check the graph again and you will see.
– Mohammad Riazi-Kermani
Jul 17 at 17:32
Well with the bounds on $theta$ implied to be from $theta=0$ to $theta=pi /2$, is this what you mean?
– Seth Taddiken
Jul 17 at 17:44
 |Â
show 1 more comment
up vote
0
down vote
up vote
0
down vote
The limits of the integral in polar coordinates for the lower half of the rectangle are $$ 0le theta le tan^-1 (b/a)$$
$$ 0le rle a sec (theta)$$
Thus you have $$ A= 2 int _0^ tan^-1 (b/a) int _0 ^a sec (theta)rdr d theta =ab$$
Note that the anti derivative of $sec ^2 theta $ is $tan theta $ and $tan (tan ^-1 b/a)=b/a $
answered Jul 17 at 15:53
Mohammad Riazi-Kermani
27.5k41852
The limits of the integral in polar coordinates for the lower half of the rectangle are $$ 0le theta le tan^-1 (b/a)$$
$$ 0le rle a sec (theta)$$
Thus you have $$ A= 2 int _0^ tan^-1 (b/a) int _0 ^a sec (theta)rdr d theta =ab$$
Note that the anti derivative of $sec ^2 theta $ is $tan theta $ and $tan (tan ^-1 b/a)=b/a $
answered Jul 17 at 15:53
Mohammad Riazi-Kermani
27.5k41852
edited Jul 17 at 15:59
answered Jul 17 at 15:53
Mohammad Riazi-Kermani
27.5k41852
answered Jul 17 at 15:53
Mohammad Riazi-Kermani
27.5k41852
answered Jul 17 at 15:53
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
Notice the requisite order of integration
– Seth Taddiken
Jul 17 at 16:45
@SethTaddiken The requisite order of integration does not meet the correct format of a polar double integral . The outermost limits of integrals should be constant.
– Mohammad Riazi-Kermani
Jul 17 at 17:11
That’s true, that is why I believe the only way to do this correctly is to split the problem into three integrals. One from $r=0$ to $r=a$, on from $r=a$ to $r=b$, and the last from $r=b$ to $r=sqrta^2+b^2$ with $theta$ ranging from $theta=arccosa/r$ to $theta=arcsinb/r$
– Seth Taddiken
Jul 17 at 17:17
@SethTaddiken You are covering more than the rectangle that you want to cover. Check the graph again and you will see.
– Mohammad Riazi-Kermani
Jul 17 at 17:32
Well with the bounds on $theta$ implied to be from $theta=0$ to $theta=pi /2$, is this what you mean?
– Seth Taddiken
Jul 17 at 17:44
 |Â
show 1 more comment
Notice the requisite order of integration
– Seth Taddiken
Jul 17 at 16:45
@SethTaddiken The requisite order of integration does not meet the correct format of a polar double integral . The outermost limits of integrals should be constant.
– Mohammad Riazi-Kermani
Jul 17 at 17:11
That’s true, that is why I believe the only way to do this correctly is to split the problem into three integrals. One from $r=0$ to $r=a$, on from $r=a$ to $r=b$, and the last from $r=b$ to $r=sqrta^2+b^2$ with $theta$ ranging from $theta=arccosa/r$ to $theta=arcsinb/r$
– Seth Taddiken
Jul 17 at 17:17
@SethTaddiken You are covering more than the rectangle that you want to cover. Check the graph again and you will see.
– Mohammad Riazi-Kermani
Jul 17 at 17:32
Well with the bounds on $theta$ implied to be from $theta=0$ to $theta=pi /2$, is this what you mean?
– Seth Taddiken
Jul 17 at 17:44
Notice the requisite order of integration
– Seth Taddiken
Jul 17 at 16:45
Notice the requisite order of integration
– Seth Taddiken
Jul 17 at 16:45
@SethTaddiken The requisite order of integration does not meet the correct format of a polar double integral . The outermost limits of integrals should be constant.
– Mohammad Riazi-Kermani
Jul 17 at 17:11
@SethTaddiken The requisite order of integration does not meet the correct format of a polar double integral . The outermost limits of integrals should be constant.
– Mohammad Riazi-Kermani
Jul 17 at 17:11
That’s true, that is why I believe the only way to do this correctly is to split the problem into three integrals. One from $r=0$ to $r=a$, on from $r=a$ to $r=b$, and the last from $r=b$ to $r=sqrta^2+b^2$ with $theta$ ranging from $theta=arccosa/r$ to $theta=arcsinb/r$
– Seth Taddiken
Jul 17 at 17:17
That’s true, that is why I believe the only way to do this correctly is to split the problem into three integrals. One from $r=0$ to $r=a$, on from $r=a$ to $r=b$, and the last from $r=b$ to $r=sqrta^2+b^2$ with $theta$ ranging from $theta=arccosa/r$ to $theta=arcsinb/r$
– Seth Taddiken
Jul 17 at 17:17
@SethTaddiken You are covering more than the rectangle that you want to cover. Check the graph again and you will see.
– Mohammad Riazi-Kermani
Jul 17 at 17:32
@SethTaddiken You are covering more than the rectangle that you want to cover. Check the graph again and you will see.
– Mohammad Riazi-Kermani
Jul 17 at 17:32
Well with the bounds on $theta$ implied to be from $theta=0$ to $theta=pi /2$, is this what you mean?
– Seth Taddiken
Jul 17 at 17:44
Well with the bounds on $theta$ implied to be from $theta=0$ to $theta=pi /2$, is this what you mean?
– Seth Taddiken
Jul 17 at 17:44
 |Â
show 1 more comment
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