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Fubini's Theorem about double integration in polar coordinates
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So I am studying James Stewart's Calculus 8th edition and Fubini's theorem is defined in the following way...
If f is continuous on the rectangle R = a ≤ x ≤ b, c ≤ y ≤ d.
Then the integral over this region R is...
$$intint_Rf(x, y)space dA =int_a^bint_c^df(x,y) space dydx = int_c^dint_a^bf(x,y) space dxdy$$
Which I think means that the double integral over a rectangular region can be calculated using a iterated integral. This makes sense. However, in a later section it derives the method of computing double integrals over regions defined by polar coordinates. The following is where I am struggling to understand their logic.
So they discuss defining circular regions using polar coordinates and splitting it up into m x n different regions referred to as 'polar rectangles'. Similar to this image here,
From there, they take a sample point within each polar rectangle and multiply the value of the function at that sample point by the area of the polar rectangle. They do this for all of the polar rectangles in the region and sum them together thus arriving at the following Riemann sum.
$$sum_i=1^msum_j=1^nf(r_i^*cosθ_j^*,r_i^*sinθ_j^*)ΔA_i=
sum_i=1^msum_j=1^nf(r_i^*cosθ_j^*,r_i^*sinθ_j^*)r_i^*ΔrΔθ$$
Where
$$ΔA_i = r_i^*ΔrΔθ$$ is the area of these polar rectangles. From there they make the substitution,
$$g(r, theta) = rf(rcostheta, rsintheta)$$
So the above Riemann sum becomes,
$$sum_i=1^msum_j=1^ng(r_i^*, theta_j^*) ΔrΔtheta$$
Which is, with no explanation and presumably after taking the limit of this sum as n and m tend to infinity, equated to the double integral,
$$int_alpha^betaint_a^bg(r, theta)space drdtheta$$
Where the region in polar coordinates is defined as
$$R = space alpha ≤ theta ≤ beta , a ≤ r ≤ b $$
So finally my question is how can Fubini's theorem be used to relate the above Riemann sum to the iterated integral of g when the theorem states that the region must be a rectangle. My intuition tells me that even though the polar region is not a rectangle, it is contained between constant numbers (like a rectangle) so perhaps Fubini's theorem will still hold.
multivariable-calculus polar-coordinates multiple-integral
edited 23 hours ago
Julián Aguirre
64.3k23894
asked yesterday
Zachery Kish
253
add a comment |Â
up vote
1
down vote
favorite
So I am studying James Stewart's Calculus 8th edition and Fubini's theorem is defined in the following way...
If f is continuous on the rectangle R = a ≤ x ≤ b, c ≤ y ≤ d.
Then the integral over this region R is...
$$intint_Rf(x, y)space dA =int_a^bint_c^df(x,y) space dydx = int_c^dint_a^bf(x,y) space dxdy$$
Which I think means that the double integral over a rectangular region can be calculated using a iterated integral. This makes sense. However, in a later section it derives the method of computing double integrals over regions defined by polar coordinates. The following is where I am struggling to understand their logic.
So they discuss defining circular regions using polar coordinates and splitting it up into m x n different regions referred to as 'polar rectangles'. Similar to this image here,
From there, they take a sample point within each polar rectangle and multiply the value of the function at that sample point by the area of the polar rectangle. They do this for all of the polar rectangles in the region and sum them together thus arriving at the following Riemann sum.
$$sum_i=1^msum_j=1^nf(r_i^*cosθ_j^*,r_i^*sinθ_j^*)ΔA_i=
sum_i=1^msum_j=1^nf(r_i^*cosθ_j^*,r_i^*sinθ_j^*)r_i^*ΔrΔθ$$
Where
$$ΔA_i = r_i^*ΔrΔθ$$ is the area of these polar rectangles. From there they make the substitution,
$$g(r, theta) = rf(rcostheta, rsintheta)$$
So the above Riemann sum becomes,
$$sum_i=1^msum_j=1^ng(r_i^*, theta_j^*) ΔrΔtheta$$
Which is, with no explanation and presumably after taking the limit of this sum as n and m tend to infinity, equated to the double integral,
$$int_alpha^betaint_a^bg(r, theta)space drdtheta$$
Where the region in polar coordinates is defined as
$$R = space alpha ≤ theta ≤ beta , a ≤ r ≤ b $$
So finally my question is how can Fubini's theorem be used to relate the above Riemann sum to the iterated integral of g when the theorem states that the region must be a rectangle. My intuition tells me that even though the polar region is not a rectangle, it is contained between constant numbers (like a rectangle) so perhaps Fubini's theorem will still hold.
multivariable-calculus polar-coordinates multiple-integral
edited 23 hours ago
Julián Aguirre
64.3k23894
asked yesterday
Zachery Kish
253
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So I am studying James Stewart's Calculus 8th edition and Fubini's theorem is defined in the following way...
If f is continuous on the rectangle R = a ≤ x ≤ b, c ≤ y ≤ d.
Then the integral over this region R is...
$$intint_Rf(x, y)space dA =int_a^bint_c^df(x,y) space dydx = int_c^dint_a^bf(x,y) space dxdy$$
Which I think means that the double integral over a rectangular region can be calculated using a iterated integral. This makes sense. However, in a later section it derives the method of computing double integrals over regions defined by polar coordinates. The following is where I am struggling to understand their logic.
So they discuss defining circular regions using polar coordinates and splitting it up into m x n different regions referred to as 'polar rectangles'. Similar to this image here,
From there, they take a sample point within each polar rectangle and multiply the value of the function at that sample point by the area of the polar rectangle. They do this for all of the polar rectangles in the region and sum them together thus arriving at the following Riemann sum.
$$sum_i=1^msum_j=1^nf(r_i^*cosθ_j^*,r_i^*sinθ_j^*)ΔA_i=
sum_i=1^msum_j=1^nf(r_i^*cosθ_j^*,r_i^*sinθ_j^*)r_i^*ΔrΔθ$$
Where
$$ΔA_i = r_i^*ΔrΔθ$$ is the area of these polar rectangles. From there they make the substitution,
$$g(r, theta) = rf(rcostheta, rsintheta)$$
So the above Riemann sum becomes,
$$sum_i=1^msum_j=1^ng(r_i^*, theta_j^*) ΔrΔtheta$$
Which is, with no explanation and presumably after taking the limit of this sum as n and m tend to infinity, equated to the double integral,
$$int_alpha^betaint_a^bg(r, theta)space drdtheta$$
Where the region in polar coordinates is defined as
$$R = space alpha ≤ theta ≤ beta , a ≤ r ≤ b $$
So finally my question is how can Fubini's theorem be used to relate the above Riemann sum to the iterated integral of g when the theorem states that the region must be a rectangle. My intuition tells me that even though the polar region is not a rectangle, it is contained between constant numbers (like a rectangle) so perhaps Fubini's theorem will still hold.
multivariable-calculus polar-coordinates multiple-integral
edited 23 hours ago
Julián Aguirre
64.3k23894
asked yesterday
Zachery Kish
253
So I am studying James Stewart's Calculus 8th edition and Fubini's theorem is defined in the following way...
If f is continuous on the rectangle R = a ≤ x ≤ b, c ≤ y ≤ d.
Then the integral over this region R is...
$$intint_Rf(x, y)space dA =int_a^bint_c^df(x,y) space dydx = int_c^dint_a^bf(x,y) space dxdy$$
Which I think means that the double integral over a rectangular region can be calculated using a iterated integral. This makes sense. However, in a later section it derives the method of computing double integrals over regions defined by polar coordinates. The following is where I am struggling to understand their logic.
So they discuss defining circular regions using polar coordinates and splitting it up into m x n different regions referred to as 'polar rectangles'. Similar to this image here,
From there, they take a sample point within each polar rectangle and multiply the value of the function at that sample point by the area of the polar rectangle. They do this for all of the polar rectangles in the region and sum them together thus arriving at the following Riemann sum.
$$sum_i=1^msum_j=1^nf(r_i^*cosθ_j^*,r_i^*sinθ_j^*)ΔA_i=
sum_i=1^msum_j=1^nf(r_i^*cosθ_j^*,r_i^*sinθ_j^*)r_i^*ΔrΔθ$$
Where
$$ΔA_i = r_i^*ΔrΔθ$$ is the area of these polar rectangles. From there they make the substitution,
$$g(r, theta) = rf(rcostheta, rsintheta)$$
So the above Riemann sum becomes,
$$sum_i=1^msum_j=1^ng(r_i^*, theta_j^*) ΔrΔtheta$$
Which is, with no explanation and presumably after taking the limit of this sum as n and m tend to infinity, equated to the double integral,
$$int_alpha^betaint_a^bg(r, theta)space drdtheta$$
Where the region in polar coordinates is defined as
$$R = space alpha ≤ theta ≤ beta , a ≤ r ≤ b $$
So finally my question is how can Fubini's theorem be used to relate the above Riemann sum to the iterated integral of g when the theorem states that the region must be a rectangle. My intuition tells me that even though the polar region is not a rectangle, it is contained between constant numbers (like a rectangle) so perhaps Fubini's theorem will still hold.
multivariable-calculus polar-coordinates multiple-integral
edited 23 hours ago
Julián Aguirre
64.3k23894
asked yesterday
Zachery Kish
253
edited 23 hours ago
Julián Aguirre
64.3k23894
edited 23 hours ago
Julián Aguirre
64.3k23894
edited 23 hours ago
Julián Aguirre
64.3k23894
64.3k23894
asked yesterday
Zachery Kish
253
asked yesterday
Zachery Kish
253
asked yesterday
Zachery Kish
253
253
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The region must be a rectangle in the variable of integration. The variables of integration on the circular region are $x,y$, they are not defined on a rectangular region and you cannot apply Fubini's theorem$^*$.
Once you change to polar coordinates, the variables of integration are $r,theta$, they move on a rectangular region and you can apply Fubini's theorem.
$^*$Note. Actually it is possible to apply Fubini's theorem by extending $f$ as zero outside the circular region. Let $C=(x,y):x^2+y^2lealpha^2, 0le x, 0le y$. Then
$$
int_Cf(x,y),dA=int_0^alphaint_0^sqrtalpha^2-x^2f(x,y),dy,dx.
$$
answered yesterday
Julián Aguirre
64.3k23894
So in the case of the Riemann sum shown in my question, Fubni's theorem can applied to it since it is expressed in polar coordinates?
– Zachery Kish
yesterday
Yes. When expressed in polar coordinates, the region of integration becomes a rectangle.
– Julián Aguirre
yesterday
Ok thank you. This answers my questions
– Zachery Kish
yesterday
add a comment |Â
up vote
0
down vote
Note that $$int_alpha^betaint_a^bg(r, theta)space drdtheta=int_alpha^betaint_a^bg(x, y)space dxdy $$ Which is rectangular.The region is $$R = space alpha ≤ y ≤ beta , a ≤x ≤ b$$is a rectangle.
answered 23 hours ago
Mohammad Riazi-Kermani
26.9k41849
So you essentially have renamed the variables from r and θ to x and y? That makes it easier to see how r and θ are actually forming a rectangle.
– Zachery Kish
23 hours ago
1
Correct. The so called dummy variables could be renamed.
– Mohammad Riazi-Kermani
23 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The region must be a rectangle in the variable of integration. The variables of integration on the circular region are $x,y$, they are not defined on a rectangular region and you cannot apply Fubini's theorem$^*$.
Once you change to polar coordinates, the variables of integration are $r,theta$, they move on a rectangular region and you can apply Fubini's theorem.
$^*$Note. Actually it is possible to apply Fubini's theorem by extending $f$ as zero outside the circular region. Let $C=(x,y):x^2+y^2lealpha^2, 0le x, 0le y$. Then
$$
int_Cf(x,y),dA=int_0^alphaint_0^sqrtalpha^2-x^2f(x,y),dy,dx.
$$
answered yesterday
Julián Aguirre
64.3k23894
So in the case of the Riemann sum shown in my question, Fubni's theorem can applied to it since it is expressed in polar coordinates?
– Zachery Kish
yesterday
Yes. When expressed in polar coordinates, the region of integration becomes a rectangle.
– Julián Aguirre
yesterday
Ok thank you. This answers my questions
– Zachery Kish
yesterday
add a comment |Â
up vote
1
down vote
accepted
The region must be a rectangle in the variable of integration. The variables of integration on the circular region are $x,y$, they are not defined on a rectangular region and you cannot apply Fubini's theorem$^*$.
Once you change to polar coordinates, the variables of integration are $r,theta$, they move on a rectangular region and you can apply Fubini's theorem.
$^*$Note. Actually it is possible to apply Fubini's theorem by extending $f$ as zero outside the circular region. Let $C=(x,y):x^2+y^2lealpha^2, 0le x, 0le y$. Then
$$
int_Cf(x,y),dA=int_0^alphaint_0^sqrtalpha^2-x^2f(x,y),dy,dx.
$$
answered yesterday
Julián Aguirre
64.3k23894
So in the case of the Riemann sum shown in my question, Fubni's theorem can applied to it since it is expressed in polar coordinates?
– Zachery Kish
yesterday
Yes. When expressed in polar coordinates, the region of integration becomes a rectangle.
– Julián Aguirre
yesterday
Ok thank you. This answers my questions
– Zachery Kish
yesterday
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The region must be a rectangle in the variable of integration. The variables of integration on the circular region are $x,y$, they are not defined on a rectangular region and you cannot apply Fubini's theorem$^*$.
Once you change to polar coordinates, the variables of integration are $r,theta$, they move on a rectangular region and you can apply Fubini's theorem.
$^*$Note. Actually it is possible to apply Fubini's theorem by extending $f$ as zero outside the circular region. Let $C=(x,y):x^2+y^2lealpha^2, 0le x, 0le y$. Then
$$
int_Cf(x,y),dA=int_0^alphaint_0^sqrtalpha^2-x^2f(x,y),dy,dx.
$$
answered yesterday
Julián Aguirre
64.3k23894
The region must be a rectangle in the variable of integration. The variables of integration on the circular region are $x,y$, they are not defined on a rectangular region and you cannot apply Fubini's theorem$^*$.
Once you change to polar coordinates, the variables of integration are $r,theta$, they move on a rectangular region and you can apply Fubini's theorem.
$^*$Note. Actually it is possible to apply Fubini's theorem by extending $f$ as zero outside the circular region. Let $C=(x,y):x^2+y^2lealpha^2, 0le x, 0le y$. Then
$$
int_Cf(x,y),dA=int_0^alphaint_0^sqrtalpha^2-x^2f(x,y),dy,dx.
$$
answered yesterday
Julián Aguirre
64.3k23894
answered yesterday
Julián Aguirre
64.3k23894
answered yesterday
Julián Aguirre
64.3k23894
answered yesterday
Julián Aguirre
64.3k23894
64.3k23894
So in the case of the Riemann sum shown in my question, Fubni's theorem can applied to it since it is expressed in polar coordinates?
– Zachery Kish
yesterday
Yes. When expressed in polar coordinates, the region of integration becomes a rectangle.
– Julián Aguirre
yesterday
Ok thank you. This answers my questions
– Zachery Kish
yesterday
add a comment |Â
So in the case of the Riemann sum shown in my question, Fubni's theorem can applied to it since it is expressed in polar coordinates?
– Zachery Kish
yesterday
Yes. When expressed in polar coordinates, the region of integration becomes a rectangle.
– Julián Aguirre
yesterday
Ok thank you. This answers my questions
– Zachery Kish
yesterday
So in the case of the Riemann sum shown in my question, Fubni's theorem can applied to it since it is expressed in polar coordinates?
– Zachery Kish
yesterday
So in the case of the Riemann sum shown in my question, Fubni's theorem can applied to it since it is expressed in polar coordinates?
– Zachery Kish
yesterday
Yes. When expressed in polar coordinates, the region of integration becomes a rectangle.
– Julián Aguirre
yesterday
Yes. When expressed in polar coordinates, the region of integration becomes a rectangle.
– Julián Aguirre
yesterday
Ok thank you. This answers my questions
– Zachery Kish
yesterday
Ok thank you. This answers my questions
– Zachery Kish
yesterday
add a comment |Â
up vote
0
down vote
Note that $$int_alpha^betaint_a^bg(r, theta)space drdtheta=int_alpha^betaint_a^bg(x, y)space dxdy $$ Which is rectangular.The region is $$R = space alpha ≤ y ≤ beta , a ≤x ≤ b$$is a rectangle.
answered 23 hours ago
Mohammad Riazi-Kermani
26.9k41849
So you essentially have renamed the variables from r and θ to x and y? That makes it easier to see how r and θ are actually forming a rectangle.
– Zachery Kish
23 hours ago
1
Correct. The so called dummy variables could be renamed.
– Mohammad Riazi-Kermani
23 hours ago
add a comment |Â
up vote
0
down vote
Note that $$int_alpha^betaint_a^bg(r, theta)space drdtheta=int_alpha^betaint_a^bg(x, y)space dxdy $$ Which is rectangular.The region is $$R = space alpha ≤ y ≤ beta , a ≤x ≤ b$$is a rectangle.
answered 23 hours ago
Mohammad Riazi-Kermani
26.9k41849
So you essentially have renamed the variables from r and θ to x and y? That makes it easier to see how r and θ are actually forming a rectangle.
– Zachery Kish
23 hours ago
1
Correct. The so called dummy variables could be renamed.
– Mohammad Riazi-Kermani
23 hours ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Note that $$int_alpha^betaint_a^bg(r, theta)space drdtheta=int_alpha^betaint_a^bg(x, y)space dxdy $$ Which is rectangular.The region is $$R = space alpha ≤ y ≤ beta , a ≤x ≤ b$$is a rectangle.
answered 23 hours ago
Mohammad Riazi-Kermani
26.9k41849
Note that $$int_alpha^betaint_a^bg(r, theta)space drdtheta=int_alpha^betaint_a^bg(x, y)space dxdy $$ Which is rectangular.The region is $$R = space alpha ≤ y ≤ beta , a ≤x ≤ b$$is a rectangle.
answered 23 hours ago
Mohammad Riazi-Kermani
26.9k41849
answered 23 hours ago
Mohammad Riazi-Kermani
26.9k41849
answered 23 hours ago
Mohammad Riazi-Kermani
26.9k41849
answered 23 hours ago
Mohammad Riazi-Kermani
26.9k41849
26.9k41849
So you essentially have renamed the variables from r and θ to x and y? That makes it easier to see how r and θ are actually forming a rectangle.
– Zachery Kish
23 hours ago
1
Correct. The so called dummy variables could be renamed.
– Mohammad Riazi-Kermani
23 hours ago
add a comment |Â
So you essentially have renamed the variables from r and θ to x and y? That makes it easier to see how r and θ are actually forming a rectangle.
– Zachery Kish
23 hours ago
1
Correct. The so called dummy variables could be renamed.
– Mohammad Riazi-Kermani
23 hours ago
So you essentially have renamed the variables from r and θ to x and y? That makes it easier to see how r and θ are actually forming a rectangle.
– Zachery Kish
23 hours ago
So you essentially have renamed the variables from r and θ to x and y? That makes it easier to see how r and θ are actually forming a rectangle.
– Zachery Kish
23 hours ago
1
1
Correct. The so called dummy variables could be renamed.
– Mohammad Riazi-Kermani
23 hours ago
Correct. The so called dummy variables could be renamed.
– Mohammad Riazi-Kermani
23 hours ago
add a comment |Â
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