Converting rectangular to polar coordinates
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Why when converting rectangular to polar is theta only $0$ to $pi/2$ and not $0$ to $2pi$??
problem I'm working with
multivariable-calculus
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add a comment |
$begingroup$
Why when converting rectangular to polar is theta only $0$ to $pi/2$ and not $0$ to $2pi$??
problem I'm working with
multivariable-calculus
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$begingroup$
No. $theta$ can be anywhere from $-infty$ to $infty.$ Every point in the plane has can be expresses in infinitely many ways in polar.
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– B. Goddard
Dec 10 '18 at 22:18
add a comment |
$begingroup$
Why when converting rectangular to polar is theta only $0$ to $pi/2$ and not $0$ to $2pi$??
problem I'm working with
multivariable-calculus
$endgroup$
Why when converting rectangular to polar is theta only $0$ to $pi/2$ and not $0$ to $2pi$??
problem I'm working with
multivariable-calculus
multivariable-calculus
edited Dec 10 '18 at 22:18
Andrei
11.8k21026
11.8k21026
asked Dec 10 '18 at 22:13
Norman Sabin Jr.Norman Sabin Jr.
132
132
$begingroup$
No. $theta$ can be anywhere from $-infty$ to $infty.$ Every point in the plane has can be expresses in infinitely many ways in polar.
$endgroup$
– B. Goddard
Dec 10 '18 at 22:18
add a comment |
$begingroup$
No. $theta$ can be anywhere from $-infty$ to $infty.$ Every point in the plane has can be expresses in infinitely many ways in polar.
$endgroup$
– B. Goddard
Dec 10 '18 at 22:18
$begingroup$
No. $theta$ can be anywhere from $-infty$ to $infty.$ Every point in the plane has can be expresses in infinitely many ways in polar.
$endgroup$
– B. Goddard
Dec 10 '18 at 22:18
$begingroup$
No. $theta$ can be anywhere from $-infty$ to $infty.$ Every point in the plane has can be expresses in infinitely many ways in polar.
$endgroup$
– B. Goddard
Dec 10 '18 at 22:18
add a comment |
2 Answers
2
active
oldest
votes
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The region of integration is a quarter disk of radius $2$ centered at the origin, in the first quadrant. In polar coordinates, this means integrating $r$ from $0$ to $2$, and $theta$ from $0$ (the x-axis) through the first quadrant to $pi/2$ (the y-axis).
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ohhhhh so r: -2 to 2 would be theta 0 to pi?
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– Norman Sabin Jr.
Dec 10 '18 at 22:37
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@NormanSabinJr. $r$ is always from $0$ to $2$. $theta$ depends on the limits for $x$ and $y$
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– Andrei
Dec 10 '18 at 22:43
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okay thank you.
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:44
add a comment |
$begingroup$
You already know that the integration limits involve the disk centered on origin and radius $2$. If we look at the $y$ integral, we see that it goes from $0$ to $2$. That means that we deal with only the upper half of the disk, so we must restrict $theta$ from $0$ to $pi$. Now we look at $x$. Similarly, you only integrate from $0$, so no negative $x$, which means no $theta>pi/2$.
To get the integral from $0$ to $pi$, the $x$ integration must go between $-sqrt{4-y^2}$ to $sqrt{4-y^2}$, and to extend $theta$ to $2pi$ you would need to change the $y$ integral limits to be $-2$ to $2$.
$endgroup$
$begingroup$
okay thank you!
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:37
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The region of integration is a quarter disk of radius $2$ centered at the origin, in the first quadrant. In polar coordinates, this means integrating $r$ from $0$ to $2$, and $theta$ from $0$ (the x-axis) through the first quadrant to $pi/2$ (the y-axis).
$endgroup$
$begingroup$
ohhhhh so r: -2 to 2 would be theta 0 to pi?
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:37
$begingroup$
@NormanSabinJr. $r$ is always from $0$ to $2$. $theta$ depends on the limits for $x$ and $y$
$endgroup$
– Andrei
Dec 10 '18 at 22:43
$begingroup$
okay thank you.
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:44
add a comment |
$begingroup$
The region of integration is a quarter disk of radius $2$ centered at the origin, in the first quadrant. In polar coordinates, this means integrating $r$ from $0$ to $2$, and $theta$ from $0$ (the x-axis) through the first quadrant to $pi/2$ (the y-axis).
$endgroup$
$begingroup$
ohhhhh so r: -2 to 2 would be theta 0 to pi?
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:37
$begingroup$
@NormanSabinJr. $r$ is always from $0$ to $2$. $theta$ depends on the limits for $x$ and $y$
$endgroup$
– Andrei
Dec 10 '18 at 22:43
$begingroup$
okay thank you.
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:44
add a comment |
$begingroup$
The region of integration is a quarter disk of radius $2$ centered at the origin, in the first quadrant. In polar coordinates, this means integrating $r$ from $0$ to $2$, and $theta$ from $0$ (the x-axis) through the first quadrant to $pi/2$ (the y-axis).
$endgroup$
The region of integration is a quarter disk of radius $2$ centered at the origin, in the first quadrant. In polar coordinates, this means integrating $r$ from $0$ to $2$, and $theta$ from $0$ (the x-axis) through the first quadrant to $pi/2$ (the y-axis).
answered Dec 10 '18 at 22:25
JohnJohn
22.7k32450
22.7k32450
$begingroup$
ohhhhh so r: -2 to 2 would be theta 0 to pi?
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:37
$begingroup$
@NormanSabinJr. $r$ is always from $0$ to $2$. $theta$ depends on the limits for $x$ and $y$
$endgroup$
– Andrei
Dec 10 '18 at 22:43
$begingroup$
okay thank you.
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:44
add a comment |
$begingroup$
ohhhhh so r: -2 to 2 would be theta 0 to pi?
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:37
$begingroup$
@NormanSabinJr. $r$ is always from $0$ to $2$. $theta$ depends on the limits for $x$ and $y$
$endgroup$
– Andrei
Dec 10 '18 at 22:43
$begingroup$
okay thank you.
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:44
$begingroup$
ohhhhh so r: -2 to 2 would be theta 0 to pi?
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:37
$begingroup$
ohhhhh so r: -2 to 2 would be theta 0 to pi?
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:37
$begingroup$
@NormanSabinJr. $r$ is always from $0$ to $2$. $theta$ depends on the limits for $x$ and $y$
$endgroup$
– Andrei
Dec 10 '18 at 22:43
$begingroup$
@NormanSabinJr. $r$ is always from $0$ to $2$. $theta$ depends on the limits for $x$ and $y$
$endgroup$
– Andrei
Dec 10 '18 at 22:43
$begingroup$
okay thank you.
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:44
$begingroup$
okay thank you.
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:44
add a comment |
$begingroup$
You already know that the integration limits involve the disk centered on origin and radius $2$. If we look at the $y$ integral, we see that it goes from $0$ to $2$. That means that we deal with only the upper half of the disk, so we must restrict $theta$ from $0$ to $pi$. Now we look at $x$. Similarly, you only integrate from $0$, so no negative $x$, which means no $theta>pi/2$.
To get the integral from $0$ to $pi$, the $x$ integration must go between $-sqrt{4-y^2}$ to $sqrt{4-y^2}$, and to extend $theta$ to $2pi$ you would need to change the $y$ integral limits to be $-2$ to $2$.
$endgroup$
$begingroup$
okay thank you!
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:37
add a comment |
$begingroup$
You already know that the integration limits involve the disk centered on origin and radius $2$. If we look at the $y$ integral, we see that it goes from $0$ to $2$. That means that we deal with only the upper half of the disk, so we must restrict $theta$ from $0$ to $pi$. Now we look at $x$. Similarly, you only integrate from $0$, so no negative $x$, which means no $theta>pi/2$.
To get the integral from $0$ to $pi$, the $x$ integration must go between $-sqrt{4-y^2}$ to $sqrt{4-y^2}$, and to extend $theta$ to $2pi$ you would need to change the $y$ integral limits to be $-2$ to $2$.
$endgroup$
$begingroup$
okay thank you!
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:37
add a comment |
$begingroup$
You already know that the integration limits involve the disk centered on origin and radius $2$. If we look at the $y$ integral, we see that it goes from $0$ to $2$. That means that we deal with only the upper half of the disk, so we must restrict $theta$ from $0$ to $pi$. Now we look at $x$. Similarly, you only integrate from $0$, so no negative $x$, which means no $theta>pi/2$.
To get the integral from $0$ to $pi$, the $x$ integration must go between $-sqrt{4-y^2}$ to $sqrt{4-y^2}$, and to extend $theta$ to $2pi$ you would need to change the $y$ integral limits to be $-2$ to $2$.
$endgroup$
You already know that the integration limits involve the disk centered on origin and radius $2$. If we look at the $y$ integral, we see that it goes from $0$ to $2$. That means that we deal with only the upper half of the disk, so we must restrict $theta$ from $0$ to $pi$. Now we look at $x$. Similarly, you only integrate from $0$, so no negative $x$, which means no $theta>pi/2$.
To get the integral from $0$ to $pi$, the $x$ integration must go between $-sqrt{4-y^2}$ to $sqrt{4-y^2}$, and to extend $theta$ to $2pi$ you would need to change the $y$ integral limits to be $-2$ to $2$.
answered Dec 10 '18 at 22:24
AndreiAndrei
11.8k21026
11.8k21026
$begingroup$
okay thank you!
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:37
add a comment |
$begingroup$
okay thank you!
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:37
$begingroup$
okay thank you!
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:37
$begingroup$
okay thank you!
$endgroup$
– Norman Sabin Jr.
Dec 10 '18 at 22:37
add a comment |
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$begingroup$
No. $theta$ can be anywhere from $-infty$ to $infty.$ Every point in the plane has can be expresses in infinitely many ways in polar.
$endgroup$
– B. Goddard
Dec 10 '18 at 22:18