Mixing
Finding the area of the unit circle using an integral in polar coordinates
$begingroup$
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 the circle using an integral, I would get
$$int_0^{2pi}left(cos^2theta+sin^2theta-1right)dtheta$$
But this equals $0$; therefore, you can't find the area using polar coordinates. Or am I wrong?
polar-coordinates
edited Jan 1 at 11:09
Blue
47.7k870151
asked Mar 10 '17 at 18:54
Heavenly96Heavenly96
279217
$endgroup$
|
show 1 more comment
$begingroup$
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 the circle using an integral, I would get
$$int_0^{2pi}left(cos^2theta+sin^2theta-1right)dtheta$$
But this equals $0$; therefore, you can't find the area using polar coordinates. Or am I wrong?
polar-coordinates
edited Jan 1 at 11:09
Blue
47.7k870151
asked Mar 10 '17 at 18:54
Heavenly96Heavenly96
279217
$endgroup$
3
$begingroup$
The polar form for the unit circle is $r=1$. Thus the area is $int_{theta = 0}^{2pi}int_{r=0}^1 rdr dtheta=int_{theta = 0}^{2pi} frac 12 dtheta = pi$.
$endgroup$
– lulu
Mar 10 '17 at 18:58
$begingroup$
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 $0 le theta le 2pi$.
$endgroup$
– Decaf-Math
Mar 10 '17 at 19:01
$begingroup$
I can't use a Jacobian here, the person is in Calc 2 that I'm trying to show this to. @pyrazolam
$endgroup$
– Heavenly96
Mar 10 '17 at 19:01
$begingroup$
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).
$endgroup$
– lulu
Mar 10 '17 at 19:03
$begingroup$
Well, I know that the area is just $pi r^2$ but the professor requires polar for this question @lulu
$endgroup$
– Heavenly96
Mar 10 '17 at 19:04
|
show 1 more comment
$begingroup$
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 the circle using an integral, I would get
$$int_0^{2pi}left(cos^2theta+sin^2theta-1right)dtheta$$
But this equals $0$; therefore, you can't find the area using polar coordinates. Or am I wrong?
polar-coordinates
edited Jan 1 at 11:09
Blue
47.7k870151
asked Mar 10 '17 at 18:54
Heavenly96Heavenly96
279217
$endgroup$
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 the circle using an integral, I would get
$$int_0^{2pi}left(cos^2theta+sin^2theta-1right)dtheta$$
But this equals $0$; therefore, you can't find the area using polar coordinates. Or am I wrong?
polar-coordinates
polar-coordinates
edited Jan 1 at 11:09
Blue
47.7k870151
asked Mar 10 '17 at 18:54
Heavenly96Heavenly96
279217
edited Jan 1 at 11:09
Blue
47.7k870151
asked Mar 10 '17 at 18:54
Heavenly96Heavenly96
279217
edited Jan 1 at 11:09
Blue
47.7k870151
edited Jan 1 at 11:09
Blue
47.7k870151
edited Jan 1 at 11:09
Blue
47.7k870151
47.7k870151
asked Mar 10 '17 at 18:54
Heavenly96Heavenly96
279217
asked Mar 10 '17 at 18:54
Heavenly96Heavenly96
279217
asked Mar 10 '17 at 18:54
Heavenly96Heavenly96
279217
279217
3
$begingroup$
The polar form for the unit circle is $r=1$. Thus the area is $int_{theta = 0}^{2pi}int_{r=0}^1 rdr dtheta=int_{theta = 0}^{2pi} frac 12 dtheta = pi$.
$endgroup$
– lulu
Mar 10 '17 at 18:58
$begingroup$
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 $0 le theta le 2pi$.
$endgroup$
– Decaf-Math
Mar 10 '17 at 19:01
$begingroup$
I can't use a Jacobian here, the person is in Calc 2 that I'm trying to show this to. @pyrazolam
$endgroup$
– Heavenly96
Mar 10 '17 at 19:01
$begingroup$
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).
$endgroup$
– lulu
Mar 10 '17 at 19:03
$begingroup$
Well, I know that the area is just $pi r^2$ but the professor requires polar for this question @lulu
$endgroup$
– Heavenly96
Mar 10 '17 at 19:04
|
show 1 more comment
3
$begingroup$
The polar form for the unit circle is $r=1$. Thus the area is $int_{theta = 0}^{2pi}int_{r=0}^1 rdr dtheta=int_{theta = 0}^{2pi} frac 12 dtheta = pi$.
$endgroup$
– lulu
Mar 10 '17 at 18:58
$begingroup$
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 $0 le theta le 2pi$.
$endgroup$
– Decaf-Math
Mar 10 '17 at 19:01
$begingroup$
I can't use a Jacobian here, the person is in Calc 2 that I'm trying to show this to. @pyrazolam
$endgroup$
– Heavenly96
Mar 10 '17 at 19:01
$begingroup$
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).
$endgroup$
– lulu
Mar 10 '17 at 19:03
$begingroup$
Well, I know that the area is just $pi r^2$ but the professor requires polar for this question @lulu
$endgroup$
– Heavenly96
Mar 10 '17 at 19:04
3
3
$begingroup$
The polar form for the unit circle is $r=1$. Thus the area is $int_{theta = 0}^{2pi}int_{r=0}^1 rdr dtheta=int_{theta = 0}^{2pi} frac 12 dtheta = pi$.
$endgroup$
– lulu
Mar 10 '17 at 18:58
$begingroup$
The polar form for the unit circle is $r=1$. Thus the area is $int_{theta = 0}^{2pi}int_{r=0}^1 rdr dtheta=int_{theta = 0}^{2pi} frac 12 dtheta = pi$.
$endgroup$
– lulu
Mar 10 '17 at 18:58
$begingroup$
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 $0 le theta le 2pi$.
$endgroup$
– Decaf-Math
Mar 10 '17 at 19:01
$begingroup$
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 $0 le theta le 2pi$.
$endgroup$
– Decaf-Math
Mar 10 '17 at 19:01
$begingroup$
I can't use a Jacobian here, the person is in Calc 2 that I'm trying to show this to. @pyrazolam
$endgroup$
– Heavenly96
Mar 10 '17 at 19:01
$begingroup$
I can't use a Jacobian here, the person is in Calc 2 that I'm trying to show this to. @pyrazolam
$endgroup$
– Heavenly96
Mar 10 '17 at 19:01
$begingroup$
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).
$endgroup$
– lulu
Mar 10 '17 at 19:03
$begingroup$
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).
$endgroup$
– lulu
Mar 10 '17 at 19:03
$begingroup$
Well, I know that the area is just $pi r^2$ but the professor requires polar for this question @lulu
$endgroup$
– Heavenly96
Mar 10 '17 at 19:04
$begingroup$
Well, I know that the area is just $pi r^2$ but the professor requires polar for this question @lulu
$endgroup$
– Heavenly96
Mar 10 '17 at 19:04
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
@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 $pi r^2, {piover2}r^2, {piover4}r^2$, and ${piover8}r^2$. Notice something interesting here? The area of each division ends up yielding half the angle required to find such area: $$A_{Deltatheta}approx{Deltathetaover2}r^2.$$ As $Delta theta to 0$, summing up each of the pieces to gain the full area gives us the formula $$A_{r(theta)} = int_alpha^beta{dthetaover2}r^2.$$
This is how the ${1over 2}$ comes about.
answered Mar 10 '17 at 19:11
Decaf-MathDecaf-Math
3,224825
$endgroup$
add a comment |
$begingroup$
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 MDoug M
44.4k31854
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
@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 $pi r^2, {piover2}r^2, {piover4}r^2$, and ${piover8}r^2$. Notice something interesting here? The area of each division ends up yielding half the angle required to find such area: $$A_{Deltatheta}approx{Deltathetaover2}r^2.$$ As $Delta theta to 0$, summing up each of the pieces to gain the full area gives us the formula $$A_{r(theta)} = int_alpha^beta{dthetaover2}r^2.$$
This is how the ${1over 2}$ comes about.
answered Mar 10 '17 at 19:11
Decaf-MathDecaf-Math
3,224825
$endgroup$
add a comment |
$begingroup$
@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 $pi r^2, {piover2}r^2, {piover4}r^2$, and ${piover8}r^2$. Notice something interesting here? The area of each division ends up yielding half the angle required to find such area: $$A_{Deltatheta}approx{Deltathetaover2}r^2.$$ As $Delta theta to 0$, summing up each of the pieces to gain the full area gives us the formula $$A_{r(theta)} = int_alpha^beta{dthetaover2}r^2.$$
This is how the ${1over 2}$ comes about.
answered Mar 10 '17 at 19:11
Decaf-MathDecaf-Math
3,224825
$endgroup$
add a comment |
$begingroup$
@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 $pi r^2, {piover2}r^2, {piover4}r^2$, and ${piover8}r^2$. Notice something interesting here? The area of each division ends up yielding half the angle required to find such area: $$A_{Deltatheta}approx{Deltathetaover2}r^2.$$ As $Delta theta to 0$, summing up each of the pieces to gain the full area gives us the formula $$A_{r(theta)} = int_alpha^beta{dthetaover2}r^2.$$
This is how the ${1over 2}$ comes about.
answered Mar 10 '17 at 19:11
Decaf-MathDecaf-Math
3,224825
$endgroup$
@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 $pi r^2, {piover2}r^2, {piover4}r^2$, and ${piover8}r^2$. Notice something interesting here? The area of each division ends up yielding half the angle required to find such area: $$A_{Deltatheta}approx{Deltathetaover2}r^2.$$ As $Delta theta to 0$, summing up each of the pieces to gain the full area gives us the formula $$A_{r(theta)} = int_alpha^beta{dthetaover2}r^2.$$
This is how the ${1over 2}$ comes about.
answered Mar 10 '17 at 19:11
Decaf-MathDecaf-Math
3,224825
answered Mar 10 '17 at 19:11
Decaf-MathDecaf-Math
3,224825
answered Mar 10 '17 at 19:11
Decaf-MathDecaf-Math
3,224825
answered Mar 10 '17 at 19:11
Decaf-MathDecaf-Math
3,224825
3,224825
add a comment |
add a comment |
$begingroup$
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 MDoug M
44.4k31854
$endgroup$
add a comment |
$begingroup$
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 MDoug M
44.4k31854
$endgroup$
add a comment |
$begingroup$
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 MDoug M
44.4k31854
$endgroup$
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 MDoug M
44.4k31854
answered Mar 10 '17 at 19:20
Doug MDoug M
44.4k31854
answered Mar 10 '17 at 19:20
Doug MDoug M
44.4k31854
answered Mar 10 '17 at 19:20
Doug MDoug M
44.4k31854
44.4k31854
add a comment |
add a comment |
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$begingroup$
The polar form for the unit circle is $r=1$. Thus the area is $int_{theta = 0}^{2pi}int_{r=0}^1 rdr dtheta=int_{theta = 0}^{2pi} frac 12 dtheta = pi$.
$endgroup$
– lulu
Mar 10 '17 at 18:58
$begingroup$
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 $0 le theta le 2pi$.
$endgroup$
– Decaf-Math
Mar 10 '17 at 19:01
$begingroup$
I can't use a Jacobian here, the person is in Calc 2 that I'm trying to show this to. @pyrazolam
$endgroup$
– Heavenly96
Mar 10 '17 at 19:01
$begingroup$
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).
$endgroup$
– lulu
Mar 10 '17 at 19:03
$begingroup$
Well, I know that the area is just $pi r^2$ but the professor requires polar for this question @lulu
$endgroup$
– Heavenly96
Mar 10 '17 at 19:04