Converting rectangular to polar coordinates












2












$begingroup$


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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$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















0












$begingroup$

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).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















0












$begingroup$

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 $2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    okay thank you!
    $endgroup$
    – Norman Sabin Jr.
    Dec 10 '18 at 22:37











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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).






share|cite|improve this answer









$endgroup$



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).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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











0












$begingroup$

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 $2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    okay thank you!
    $endgroup$
    – Norman Sabin Jr.
    Dec 10 '18 at 22:37
















0












$begingroup$

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 $2$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    okay thank you!
    $endgroup$
    – Norman Sabin Jr.
    Dec 10 '18 at 22:37














0












0








0





$begingroup$

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 $2$.






share|cite|improve this answer









$endgroup$



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 $2$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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