Polar Form of the Unit circle
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I just want to ask if the polar form of the unit circle is
$cos^2theta+sin^2theta=1$ so if I were to try and find the area of it using an integral I would get
$int_0^2pi(cos^2theta+sin^2theta-1)dtheta$ which equals $0$ and therefore you can't find the area using polar coordinates? Or am I wrong
polar-coordinates
asked Mar 10 '17 at 18:54
Heavenly96
279215
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show 1 more comment
up vote
0
down vote
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I just want to ask if the polar form of the unit circle is
$cos^2theta+sin^2theta=1$ so if I were to try and find the area of it using an integral I would get
$int_0^2pi(cos^2theta+sin^2theta-1)dtheta$ which equals $0$ and therefore you can't find the area using polar coordinates? Or am I wrong
polar-coordinates
asked Mar 10 '17 at 18:54
Heavenly96
279215
3
The polar form for the unit circle is $r=1$. Thus the area is $int_theta = 0^2piint_r=0^1 rdr dtheta=int_theta = 0^2pi frac 12 dtheta = pi$.
– lulu
Mar 10 '17 at 18:58
There is a slight subtlety when you create the change of coordinates from rectangular to polar. You must involve a concept called the Jacobian matrix of the change of variables: en.wikipedia.org/wiki/Jacobian_matrix_and_determinant Moreover, the unit disk which you can use to describe the region you're trying to find the area of is the region such that $0le r le 1$ and $0 le theta le 2pi$.
– Decaf-Math
Mar 10 '17 at 19:01
I can't use a Jacobian here, the person is in Calc 2 that I'm trying to show this to. @pyrazolam
– Heavenly96
Mar 10 '17 at 19:01
Why use polar coordinates then? Just write the upper part of the circle as the graph of $y=sqrt 1-x^2$ and integrate that from $-1$ to $1$ (and then multiply by $2$ to get the lower half).
– lulu
Mar 10 '17 at 19:03
Well, I know that the area is just $pi r^2$ but the professor requires polar for this question @lulu
– Heavenly96
Mar 10 '17 at 19:04
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I just want to ask if the polar form of the unit circle is
$cos^2theta+sin^2theta=1$ so if I were to try and find the area of it using an integral I would get
$int_0^2pi(cos^2theta+sin^2theta-1)dtheta$ which equals $0$ and therefore you can't find the area using polar coordinates? Or am I wrong
polar-coordinates
asked Mar 10 '17 at 18:54
Heavenly96
279215
I just want to ask if the polar form of the unit circle is
$cos^2theta+sin^2theta=1$ so if I were to try and find the area of it using an integral I would get
$int_0^2pi(cos^2theta+sin^2theta-1)dtheta$ which equals $0$ and therefore you can't find the area using polar coordinates? Or am I wrong
polar-coordinates
asked Mar 10 '17 at 18:54
Heavenly96
279215
asked Mar 10 '17 at 18:54
Heavenly96
279215
asked Mar 10 '17 at 18:54
Heavenly96
279215
asked Mar 10 '17 at 18:54
Heavenly96
279215
279215
3
The polar form for the unit circle is $r=1$. Thus the area is $int_theta = 0^2piint_r=0^1 rdr dtheta=int_theta = 0^2pi frac 12 dtheta = pi$.
– lulu
Mar 10 '17 at 18:58
There is a slight subtlety when you create the change of coordinates from rectangular to polar. You must involve a concept called the Jacobian matrix of the change of variables: en.wikipedia.org/wiki/Jacobian_matrix_and_determinant Moreover, the unit disk which you can use to describe the region you're trying to find the area of is the region such that $0le r le 1$ and $0 le theta le 2pi$.
– Decaf-Math
Mar 10 '17 at 19:01
I can't use a Jacobian here, the person is in Calc 2 that I'm trying to show this to. @pyrazolam
– Heavenly96
Mar 10 '17 at 19:01
Why use polar coordinates then? Just write the upper part of the circle as the graph of $y=sqrt 1-x^2$ and integrate that from $-1$ to $1$ (and then multiply by $2$ to get the lower half).
– lulu
Mar 10 '17 at 19:03
Well, I know that the area is just $pi r^2$ but the professor requires polar for this question @lulu
– Heavenly96
Mar 10 '17 at 19:04
 |Â
show 1 more comment
3
The polar form for the unit circle is $r=1$. Thus the area is $int_theta = 0^2piint_r=0^1 rdr dtheta=int_theta = 0^2pi frac 12 dtheta = pi$.
– lulu
Mar 10 '17 at 18:58
There is a slight subtlety when you create the change of coordinates from rectangular to polar. You must involve a concept called the Jacobian matrix of the change of variables: en.wikipedia.org/wiki/Jacobian_matrix_and_determinant Moreover, the unit disk which you can use to describe the region you're trying to find the area of is the region such that $0le r le 1$ and $0 le theta le 2pi$.
– Decaf-Math
Mar 10 '17 at 19:01
I can't use a Jacobian here, the person is in Calc 2 that I'm trying to show this to. @pyrazolam
– Heavenly96
Mar 10 '17 at 19:01
Why use polar coordinates then? Just write the upper part of the circle as the graph of $y=sqrt 1-x^2$ and integrate that from $-1$ to $1$ (and then multiply by $2$ to get the lower half).
– lulu
Mar 10 '17 at 19:03
Well, I know that the area is just $pi r^2$ but the professor requires polar for this question @lulu
– Heavenly96
Mar 10 '17 at 19:04
3
3
The polar form for the unit circle is $r=1$. Thus the area is $int_theta = 0^2piint_r=0^1 rdr dtheta=int_theta = 0^2pi frac 12 dtheta = pi$.
– lulu
Mar 10 '17 at 18:58
The polar form for the unit circle is $r=1$. Thus the area is $int_theta = 0^2piint_r=0^1 rdr dtheta=int_theta = 0^2pi frac 12 dtheta = pi$.
– lulu
Mar 10 '17 at 18:58
There is a slight subtlety when you create the change of coordinates from rectangular to polar. You must involve a concept called the Jacobian matrix of the change of variables: en.wikipedia.org/wiki/Jacobian_matrix_and_determinant Moreover, the unit disk which you can use to describe the region you're trying to find the area of is the region such that $0le r le 1$ and $0 le theta le 2pi$.
– Decaf-Math
Mar 10 '17 at 19:01
There is a slight subtlety when you create the change of coordinates from rectangular to polar. You must involve a concept called the Jacobian matrix of the change of variables: en.wikipedia.org/wiki/Jacobian_matrix_and_determinant Moreover, the unit disk which you can use to describe the region you're trying to find the area of is the region such that $0le r le 1$ and $0 le theta le 2pi$.
– Decaf-Math
Mar 10 '17 at 19:01
I can't use a Jacobian here, the person is in Calc 2 that I'm trying to show this to. @pyrazolam
– Heavenly96
Mar 10 '17 at 19:01
I can't use a Jacobian here, the person is in Calc 2 that I'm trying to show this to. @pyrazolam
– Heavenly96
Mar 10 '17 at 19:01
Why use polar coordinates then? Just write the upper part of the circle as the graph of $y=sqrt 1-x^2$ and integrate that from $-1$ to $1$ (and then multiply by $2$ to get the lower half).
– lulu
Mar 10 '17 at 19:03
Why use polar coordinates then? Just write the upper part of the circle as the graph of $y=sqrt 1-x^2$ and integrate that from $-1$ to $1$ (and then multiply by $2$ to get the lower half).
– lulu
Mar 10 '17 at 19:03
Well, I know that the area is just $pi r^2$ but the professor requires polar for this question @lulu
– Heavenly96
Mar 10 '17 at 19:04
Well, I know that the area is just $pi r^2$ but the professor requires polar for this question @lulu
– Heavenly96
Mar 10 '17 at 19:04
 |Â
show 1 more comment
2 Answers
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@Heavenly96, see the very first formula in teal here:
The derivation for this is mainly due to an analogy with the area of a circle. What are the areas of a full, semi, quarter, and eighth circle respectively? They are $pi r^2, piover2r^2, piover4r^2$, and $piover8r^2$. Notice something interesting here? The area of each division ends up yielding half the angle required to find such area: $$A_DeltathetaapproxDeltathetaover2r^2.$$ As $Delta theta to 0$, summing up each of the pieces to gain the full area gives us the formula $$A_r(theta) = int_alpha^betadthetaover2r^2.$$
This is how the $1over 2$ comes about.
answered Mar 10 '17 at 19:11
Decaf-Math
2,742721
add a comment |Â
up vote
0
down vote
The equation of the unit circle is polar coorinates is $r =1$
Integration in polar coordinates... If you are tutoring someone who is not in multivariate calculus and you cannot formally introduce the Jacobian, you need to introduce it in a less formal way.
When you learned the Riemann integral in Cartesian space, you divided the area under the curve into a sequence of rectangles. For a fine enough partition, the sum of the rectangles would equal your area. The base of each rectangle is $dx,$ and the height is $y(x),$ the area of each is $y dx$ and the area of them all is $int_a^b y dx$
Moving to polar, rather than rectangles you sections of circles. The area of each is $frac 12 r^2 dtheta.$ and the area inside a polar curve is $int_0^2pi frac 12 r^2 dr$ (usually, the limits are $0$ to $2pi$, but not always, and so,must be checked).
answered Mar 10 '17 at 19:20
Doug M
39.3k31749
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
@Heavenly96, see the very first formula in teal here:
The derivation for this is mainly due to an analogy with the area of a circle. What are the areas of a full, semi, quarter, and eighth circle respectively? They are $pi r^2, piover2r^2, piover4r^2$, and $piover8r^2$. Notice something interesting here? The area of each division ends up yielding half the angle required to find such area: $$A_DeltathetaapproxDeltathetaover2r^2.$$ As $Delta theta to 0$, summing up each of the pieces to gain the full area gives us the formula $$A_r(theta) = int_alpha^betadthetaover2r^2.$$
This is how the $1over 2$ comes about.
answered Mar 10 '17 at 19:11
Decaf-Math
2,742721
add a comment |Â
up vote
0
down vote
@Heavenly96, see the very first formula in teal here:
The derivation for this is mainly due to an analogy with the area of a circle. What are the areas of a full, semi, quarter, and eighth circle respectively? They are $pi r^2, piover2r^2, piover4r^2$, and $piover8r^2$. Notice something interesting here? The area of each division ends up yielding half the angle required to find such area: $$A_DeltathetaapproxDeltathetaover2r^2.$$ As $Delta theta to 0$, summing up each of the pieces to gain the full area gives us the formula $$A_r(theta) = int_alpha^betadthetaover2r^2.$$
This is how the $1over 2$ comes about.
answered Mar 10 '17 at 19:11
Decaf-Math
2,742721
add a comment |Â
up vote
0
down vote
up vote
0
down vote
@Heavenly96, see the very first formula in teal here:
The derivation for this is mainly due to an analogy with the area of a circle. What are the areas of a full, semi, quarter, and eighth circle respectively? They are $pi r^2, piover2r^2, piover4r^2$, and $piover8r^2$. Notice something interesting here? The area of each division ends up yielding half the angle required to find such area: $$A_DeltathetaapproxDeltathetaover2r^2.$$ As $Delta theta to 0$, summing up each of the pieces to gain the full area gives us the formula $$A_r(theta) = int_alpha^betadthetaover2r^2.$$
This is how the $1over 2$ comes about.
answered Mar 10 '17 at 19:11
Decaf-Math
2,742721
@Heavenly96, see the very first formula in teal here:
The derivation for this is mainly due to an analogy with the area of a circle. What are the areas of a full, semi, quarter, and eighth circle respectively? They are $pi r^2, piover2r^2, piover4r^2$, and $piover8r^2$. Notice something interesting here? The area of each division ends up yielding half the angle required to find such area: $$A_DeltathetaapproxDeltathetaover2r^2.$$ As $Delta theta to 0$, summing up each of the pieces to gain the full area gives us the formula $$A_r(theta) = int_alpha^betadthetaover2r^2.$$
This is how the $1over 2$ comes about.
answered Mar 10 '17 at 19:11
Decaf-Math
2,742721
answered Mar 10 '17 at 19:11
Decaf-Math
2,742721
answered Mar 10 '17 at 19:11
Decaf-Math
2,742721
answered Mar 10 '17 at 19:11
Decaf-Math
2,742721
2,742721
add a comment |Â
add a comment |Â
up vote
0
down vote
The equation of the unit circle is polar coorinates is $r =1$
Integration in polar coordinates... If you are tutoring someone who is not in multivariate calculus and you cannot formally introduce the Jacobian, you need to introduce it in a less formal way.
When you learned the Riemann integral in Cartesian space, you divided the area under the curve into a sequence of rectangles. For a fine enough partition, the sum of the rectangles would equal your area. The base of each rectangle is $dx,$ and the height is $y(x),$ the area of each is $y dx$ and the area of them all is $int_a^b y dx$
Moving to polar, rather than rectangles you sections of circles. The area of each is $frac 12 r^2 dtheta.$ and the area inside a polar curve is $int_0^2pi frac 12 r^2 dr$ (usually, the limits are $0$ to $2pi$, but not always, and so,must be checked).
answered Mar 10 '17 at 19:20
Doug M
39.3k31749
add a comment |Â
up vote
0
down vote
The equation of the unit circle is polar coorinates is $r =1$
Integration in polar coordinates... If you are tutoring someone who is not in multivariate calculus and you cannot formally introduce the Jacobian, you need to introduce it in a less formal way.
When you learned the Riemann integral in Cartesian space, you divided the area under the curve into a sequence of rectangles. For a fine enough partition, the sum of the rectangles would equal your area. The base of each rectangle is $dx,$ and the height is $y(x),$ the area of each is $y dx$ and the area of them all is $int_a^b y dx$
Moving to polar, rather than rectangles you sections of circles. The area of each is $frac 12 r^2 dtheta.$ and the area inside a polar curve is $int_0^2pi frac 12 r^2 dr$ (usually, the limits are $0$ to $2pi$, but not always, and so,must be checked).
answered Mar 10 '17 at 19:20
Doug M
39.3k31749
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The equation of the unit circle is polar coorinates is $r =1$
Integration in polar coordinates... If you are tutoring someone who is not in multivariate calculus and you cannot formally introduce the Jacobian, you need to introduce it in a less formal way.
When you learned the Riemann integral in Cartesian space, you divided the area under the curve into a sequence of rectangles. For a fine enough partition, the sum of the rectangles would equal your area. The base of each rectangle is $dx,$ and the height is $y(x),$ the area of each is $y dx$ and the area of them all is $int_a^b y dx$
Moving to polar, rather than rectangles you sections of circles. The area of each is $frac 12 r^2 dtheta.$ and the area inside a polar curve is $int_0^2pi frac 12 r^2 dr$ (usually, the limits are $0$ to $2pi$, but not always, and so,must be checked).
answered Mar 10 '17 at 19:20
Doug M
39.3k31749
The equation of the unit circle is polar coorinates is $r =1$
Integration in polar coordinates... If you are tutoring someone who is not in multivariate calculus and you cannot formally introduce the Jacobian, you need to introduce it in a less formal way.
When you learned the Riemann integral in Cartesian space, you divided the area under the curve into a sequence of rectangles. For a fine enough partition, the sum of the rectangles would equal your area. The base of each rectangle is $dx,$ and the height is $y(x),$ the area of each is $y dx$ and the area of them all is $int_a^b y dx$
Moving to polar, rather than rectangles you sections of circles. The area of each is $frac 12 r^2 dtheta.$ and the area inside a polar curve is $int_0^2pi frac 12 r^2 dr$ (usually, the limits are $0$ to $2pi$, but not always, and so,must be checked).
answered Mar 10 '17 at 19:20
Doug M
39.3k31749
answered Mar 10 '17 at 19:20
Doug M
39.3k31749
answered Mar 10 '17 at 19:20
Doug M
39.3k31749
answered Mar 10 '17 at 19:20
Doug M
39.3k31749
39.3k31749
add a comment |Â
add a comment |Â
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The polar form for the unit circle is $r=1$. Thus the area is $int_theta = 0^2piint_r=0^1 rdr dtheta=int_theta = 0^2pi frac 12 dtheta = pi$.
– lulu
Mar 10 '17 at 18:58
There is a slight subtlety when you create the change of coordinates from rectangular to polar. You must involve a concept called the Jacobian matrix of the change of variables: en.wikipedia.org/wiki/Jacobian_matrix_and_determinant Moreover, the unit disk which you can use to describe the region you're trying to find the area of is the region such that $0le r le 1$ and $0 le theta le 2pi$.
– Decaf-Math
Mar 10 '17 at 19:01
I can't use a Jacobian here, the person is in Calc 2 that I'm trying to show this to. @pyrazolam
– Heavenly96
Mar 10 '17 at 19:01
Why use polar coordinates then? Just write the upper part of the circle as the graph of $y=sqrt 1-x^2$ and integrate that from $-1$ to $1$ (and then multiply by $2$ to get the lower half).
– lulu
Mar 10 '17 at 19:03
Well, I know that the area is just $pi r^2$ but the professor requires polar for this question @lulu
– Heavenly96
Mar 10 '17 at 19:04