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Integrate $int_-a^a, int_0^sqrta^2-x^2 ,(x^2+y^2),texte^x^4+2x^2y^2+y^4 ,textdy ,textdx$ using polar coordinates.
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From $$int_-a^a ,int_0^sqrta^2-x^2 ,(x^2+y^2),texte^x^4+2x^2y^2+y^4 , textdy ,textdx=int_-a^a, int_0^sqrta^2-x^2 ,(x^2+y^2),texte^(x^2+y^2)^2 , textdy , textdx,,$$
so $-aleq xleq a$ or $-asecthetaleq r leq asectheta$,
and since the region is between $y=0$ and $y=sqrta^2-x^2 $ , so $0leq thetaleq pi$.
I integrate $$int_0^pi ,int_-asectheta^asectheta r^3,texte^r^4 , textdr , textdtheta,.$$
Solving it I get the area to be zero. If my calculations are correct, is it safe to assume that whenever I am integrating over a region that is symmetric against $y$-axis the area should be zero?
calculus integration multivariable-calculus polar-coordinates
edited Jul 25 at 18:57
Batominovski
23.1k22777
asked Jul 25 at 18:40
kronos
909
add a comment |Â
up vote
0
down vote
favorite
From $$int_-a^a ,int_0^sqrta^2-x^2 ,(x^2+y^2),texte^x^4+2x^2y^2+y^4 , textdy ,textdx=int_-a^a, int_0^sqrta^2-x^2 ,(x^2+y^2),texte^(x^2+y^2)^2 , textdy , textdx,,$$
so $-aleq xleq a$ or $-asecthetaleq r leq asectheta$,
and since the region is between $y=0$ and $y=sqrta^2-x^2 $ , so $0leq thetaleq pi$.
I integrate $$int_0^pi ,int_-asectheta^asectheta r^3,texte^r^4 , textdr , textdtheta,.$$
Solving it I get the area to be zero. If my calculations are correct, is it safe to assume that whenever I am integrating over a region that is symmetric against $y$-axis the area should be zero?
calculus integration multivariable-calculus polar-coordinates
edited Jul 25 at 18:57
Batominovski
23.1k22777
asked Jul 25 at 18:40
kronos
909
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
From $$int_-a^a ,int_0^sqrta^2-x^2 ,(x^2+y^2),texte^x^4+2x^2y^2+y^4 , textdy ,textdx=int_-a^a, int_0^sqrta^2-x^2 ,(x^2+y^2),texte^(x^2+y^2)^2 , textdy , textdx,,$$
so $-aleq xleq a$ or $-asecthetaleq r leq asectheta$,
and since the region is between $y=0$ and $y=sqrta^2-x^2 $ , so $0leq thetaleq pi$.
I integrate $$int_0^pi ,int_-asectheta^asectheta r^3,texte^r^4 , textdr , textdtheta,.$$
Solving it I get the area to be zero. If my calculations are correct, is it safe to assume that whenever I am integrating over a region that is symmetric against $y$-axis the area should be zero?
calculus integration multivariable-calculus polar-coordinates
edited Jul 25 at 18:57
Batominovski
23.1k22777
asked Jul 25 at 18:40
kronos
909
From $$int_-a^a ,int_0^sqrta^2-x^2 ,(x^2+y^2),texte^x^4+2x^2y^2+y^4 , textdy ,textdx=int_-a^a, int_0^sqrta^2-x^2 ,(x^2+y^2),texte^(x^2+y^2)^2 , textdy , textdx,,$$
so $-aleq xleq a$ or $-asecthetaleq r leq asectheta$,
and since the region is between $y=0$ and $y=sqrta^2-x^2 $ , so $0leq thetaleq pi$.
I integrate $$int_0^pi ,int_-asectheta^asectheta r^3,texte^r^4 , textdr , textdtheta,.$$
Solving it I get the area to be zero. If my calculations are correct, is it safe to assume that whenever I am integrating over a region that is symmetric against $y$-axis the area should be zero?
calculus integration multivariable-calculus polar-coordinates
edited Jul 25 at 18:57
Batominovski
23.1k22777
asked Jul 25 at 18:40
kronos
909
edited Jul 25 at 18:57
Batominovski
23.1k22777
edited Jul 25 at 18:57
Batominovski
23.1k22777
edited Jul 25 at 18:57
Batominovski
23.1k22777
23.1k22777
asked Jul 25 at 18:40
kronos
909
asked Jul 25 at 18:40
kronos
909
asked Jul 25 at 18:40
kronos
909
909
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
The limits of the integral in $r$ are wrong. The area of integration is half a disk of radius $a$ centered at the origin. The graph of $y=sqrta^2-x^2$ is the semicircle $x^2+y^2=a^2$, $yge0$. The region of integration is $0lethetalepi$, $0le rle a$.
answered Jul 25 at 18:44
Julián Aguirre
64.5k23894
could you please point out the flaw in my understanding about the boundary for r.In the 3rd line, I have written my method for the boundary of r.
– kronos
Jul 25 at 18:47
2
It is certainly true that $-ale xle a$, but there is another condition, $0le ylesqrta^2-x^2$. What happens when $theta=0$? Is $r=infty$? You are even accepting negative values for $r$.
– Julián Aguirre
Jul 25 at 18:59
@JuliánAguirre Is there a consistent way of finding the bounds when you change the integration? I understand you can draw it, but is there an algebraic way?
– Sorfosh
Jul 25 at 19:10
1
Most of the times there is. In this problem, the condition $0le ylesqrta^2-x^2$ transforms into $x^2+y^2le a^2$, which in polar coordinates is $0le rle a$.
– Julián Aguirre
Jul 26 at 11:02
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
The limits of the integral in $r$ are wrong. The area of integration is half a disk of radius $a$ centered at the origin. The graph of $y=sqrta^2-x^2$ is the semicircle $x^2+y^2=a^2$, $yge0$. The region of integration is $0lethetalepi$, $0le rle a$.
answered Jul 25 at 18:44
Julián Aguirre
64.5k23894
could you please point out the flaw in my understanding about the boundary for r.In the 3rd line, I have written my method for the boundary of r.
– kronos
Jul 25 at 18:47
2
It is certainly true that $-ale xle a$, but there is another condition, $0le ylesqrta^2-x^2$. What happens when $theta=0$? Is $r=infty$? You are even accepting negative values for $r$.
– Julián Aguirre
Jul 25 at 18:59
@JuliánAguirre Is there a consistent way of finding the bounds when you change the integration? I understand you can draw it, but is there an algebraic way?
– Sorfosh
Jul 25 at 19:10
1
Most of the times there is. In this problem, the condition $0le ylesqrta^2-x^2$ transforms into $x^2+y^2le a^2$, which in polar coordinates is $0le rle a$.
– Julián Aguirre
Jul 26 at 11:02
add a comment |Â
up vote
5
down vote
accepted
The limits of the integral in $r$ are wrong. The area of integration is half a disk of radius $a$ centered at the origin. The graph of $y=sqrta^2-x^2$ is the semicircle $x^2+y^2=a^2$, $yge0$. The region of integration is $0lethetalepi$, $0le rle a$.
answered Jul 25 at 18:44
Julián Aguirre
64.5k23894
could you please point out the flaw in my understanding about the boundary for r.In the 3rd line, I have written my method for the boundary of r.
– kronos
Jul 25 at 18:47
2
It is certainly true that $-ale xle a$, but there is another condition, $0le ylesqrta^2-x^2$. What happens when $theta=0$? Is $r=infty$? You are even accepting negative values for $r$.
– Julián Aguirre
Jul 25 at 18:59
@JuliánAguirre Is there a consistent way of finding the bounds when you change the integration? I understand you can draw it, but is there an algebraic way?
– Sorfosh
Jul 25 at 19:10
1
Most of the times there is. In this problem, the condition $0le ylesqrta^2-x^2$ transforms into $x^2+y^2le a^2$, which in polar coordinates is $0le rle a$.
– Julián Aguirre
Jul 26 at 11:02
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
The limits of the integral in $r$ are wrong. The area of integration is half a disk of radius $a$ centered at the origin. The graph of $y=sqrta^2-x^2$ is the semicircle $x^2+y^2=a^2$, $yge0$. The region of integration is $0lethetalepi$, $0le rle a$.
answered Jul 25 at 18:44
Julián Aguirre
64.5k23894
The limits of the integral in $r$ are wrong. The area of integration is half a disk of radius $a$ centered at the origin. The graph of $y=sqrta^2-x^2$ is the semicircle $x^2+y^2=a^2$, $yge0$. The region of integration is $0lethetalepi$, $0le rle a$.
answered Jul 25 at 18:44
Julián Aguirre
64.5k23894
answered Jul 25 at 18:44
Julián Aguirre
64.5k23894
answered Jul 25 at 18:44
Julián Aguirre
64.5k23894
answered Jul 25 at 18:44
Julián Aguirre
64.5k23894
64.5k23894
could you please point out the flaw in my understanding about the boundary for r.In the 3rd line, I have written my method for the boundary of r.
– kronos
Jul 25 at 18:47
2
It is certainly true that $-ale xle a$, but there is another condition, $0le ylesqrta^2-x^2$. What happens when $theta=0$? Is $r=infty$? You are even accepting negative values for $r$.
– Julián Aguirre
Jul 25 at 18:59
@JuliánAguirre Is there a consistent way of finding the bounds when you change the integration? I understand you can draw it, but is there an algebraic way?
– Sorfosh
Jul 25 at 19:10
1
Most of the times there is. In this problem, the condition $0le ylesqrta^2-x^2$ transforms into $x^2+y^2le a^2$, which in polar coordinates is $0le rle a$.
– Julián Aguirre
Jul 26 at 11:02
add a comment |Â
could you please point out the flaw in my understanding about the boundary for r.In the 3rd line, I have written my method for the boundary of r.
– kronos
Jul 25 at 18:47
2
It is certainly true that $-ale xle a$, but there is another condition, $0le ylesqrta^2-x^2$. What happens when $theta=0$? Is $r=infty$? You are even accepting negative values for $r$.
– Julián Aguirre
Jul 25 at 18:59
@JuliánAguirre Is there a consistent way of finding the bounds when you change the integration? I understand you can draw it, but is there an algebraic way?
– Sorfosh
Jul 25 at 19:10
1
Most of the times there is. In this problem, the condition $0le ylesqrta^2-x^2$ transforms into $x^2+y^2le a^2$, which in polar coordinates is $0le rle a$.
– Julián Aguirre
Jul 26 at 11:02
could you please point out the flaw in my understanding about the boundary for r.In the 3rd line, I have written my method for the boundary of r.
– kronos
Jul 25 at 18:47
could you please point out the flaw in my understanding about the boundary for r.In the 3rd line, I have written my method for the boundary of r.
– kronos
Jul 25 at 18:47
2
2
It is certainly true that $-ale xle a$, but there is another condition, $0le ylesqrta^2-x^2$. What happens when $theta=0$? Is $r=infty$? You are even accepting negative values for $r$.
– Julián Aguirre
Jul 25 at 18:59
It is certainly true that $-ale xle a$, but there is another condition, $0le ylesqrta^2-x^2$. What happens when $theta=0$? Is $r=infty$? You are even accepting negative values for $r$.
– Julián Aguirre
Jul 25 at 18:59
@JuliánAguirre Is there a consistent way of finding the bounds when you change the integration? I understand you can draw it, but is there an algebraic way?
– Sorfosh
Jul 25 at 19:10
@JuliánAguirre Is there a consistent way of finding the bounds when you change the integration? I understand you can draw it, but is there an algebraic way?
– Sorfosh
Jul 25 at 19:10
1
1
Most of the times there is. In this problem, the condition $0le ylesqrta^2-x^2$ transforms into $x^2+y^2le a^2$, which in polar coordinates is $0le rle a$.
– Julián Aguirre
Jul 26 at 11:02
Most of the times there is. In this problem, the condition $0le ylesqrta^2-x^2$ transforms into $x^2+y^2le a^2$, which in polar coordinates is $0le rle a$.
– Julián Aguirre
Jul 26 at 11:02
add a comment |Â
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