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Area in Polar Coordinates
Ó=  <= /span>= Fra= nk Ma 2= 006
------=_NextPart_01C97CE7.92FACF50 Content-Location: file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0003.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Area in Polar Coordinates
Area in Polar Coordinates
Ó=  <= /span>= Fra= nk Ma 2= 006
Objective:
Integral formula of the area swept
out by a polar function between
two angles
------=_NextPart_01C97CE7.92FACF50 Content-Location: file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0026.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Area in Polar Coordinates
Area in Polar Coordinates
Given a polar function r =3D = f(q<= span style=3D'font-size:178%'>) with A < q < B, it "sweeps&q= uot; out an area between the= m;
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Area in Polar Coordinates
Given= a polar function r =3D = f(q<= span style=3D'font-size:178%'>) with A < q < B, it "sweeps&q= uot; out an area between the= m;
r=3Df(q)
q=3DA
q=3DB
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Area in Polar Coordinates
Ó <= /span>Frank Ma 2006
Given a polar function r =3D = f(q<= span style=3D'font-size:178%'>) with A < q < B, it "sweeps&q= uot; out an area between the= m;
r=3Df(q)
q=3DA
The Polar Area Formula:
The area swept out by r =3D f(q<= span style=3D'font-size:200%'>) from = ;
q<= span style=3D'font-size:200%'>=3DA to q<= span style=3D'font-size:200%'>=3DB is:
q=3DB
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Area in Polar Coordinates
Ó <= /span>Frank Ma 2006
Example:
Find the area enc= losed = by r =3D f(q) =3D k with
0 < q < 2π.
= ;
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Area in Polar Coordinates
Ó <= /span>Frank Ma 2006
Example:
Find the area enc= losed = by r =3D f(q) =3D k with
0 < q < 2π.
k
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Area in Polar Coordinates
Ó <= /span>Frank Ma 2006
Example:
Find the area enc= losed = by r =3D f(q) =3D k with
0 < q < 2π.
= ;
We have:
k
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Area in Polar Coordinates
Ó <= /span>Frank Ma 2006
Example:
Find the area enc= losed = by r =3D f(q) =3D k with
0 < q < 2π.
= ;
We have:
k
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Area in Polar Coordinates
Ó <= /span>Frank Ma 2006
Example:
Find the area enc= losed = by r =3D f(q) =3D k with
0 < q < 2π.
= ;
We have:
k
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Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Area in Polar Coordinates
Area in Polar Coordinates
Ó=  <= /span>Frank Ma 2006
cos2(A) =3D
From cosine double-= angle formulas, we get the following:
 1 + cos(2A) 
2
sin2(A) =3D
 1 – cos(2A) 
2
------=_NextPart_01C97CE7.92FACF50 Content-Location: file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0007_image034.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhYwACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAAAAABj AAEAgAAAAAAAAAIIjI+py+0PIysAOy== ------=_NextPart_01C97CE7.92FACF50 Content-Location: file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0007_image035.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhYwACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAAAAABj AAEAgAAAAAAAAAIIjI+py+0PIysAOy== ------=_NextPart_01C97CE7.92FACF50 Content-Location: file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0031.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Area in Polar Coordinates
Area in Polar Coordinates
Ó=  <= /span>Frank Ma 2006
cos2(A) =3D
From cosine double-= angle formulas, we get the following:
 1 + cos(2A) 
2
sin2(A) =3D
 1 – cos(2A) 
2
We use these formul= as for the integral:
------=_NextPart_01C97CE7.92FACF50 Content-Location: file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0031_image036.wmz Content-Transfer-Encoding: base64 Content-Type: image/x-wmz H4sIAAAAAAACC7t+9tgsBjA4wNHAzMgJYgWEMjIAGcxpjAwMTAwyYFlWIOZkgrGYGBmhLEam//// g1l6jBJQMW64Oh6mBuYDHEJAlhobP4MUw3+QYgYBkG1A1logbuBgYNBlBuqBquFh8E0syQipLEhl YNgAtvs304G/EBdOAFnKxMLEIBBcmZuUn8PAcJqbq0znqRDDjiWfyw8CsYLB1/IunrRLLECFukDV QMMZjLhWMTkwgLR+gprX8A9mHiOaeZv4uR7iNo+RAUR/gJurwFTJDjK3EM6fxgLhY7GHAWRPSGZu arGCX2q5QlB+bmIeTndD7GGEm7uTDWRCCoa5DCSZi+n+SawgIzThfAOw+zVAsQ/mNzECo4YhOb8Y arPCf/JsRvdRCeNUZpARRvlc0HjnAqcNsLFQuwQY2MG8PeDUxMjEpBRcWVySmstwCWQuA1CHIkMX WAOI8HguxIDuS2awfgBdTfVE4gIAAA== ------=_NextPart_01C97CE7.92FACF50 Content-Location: file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0031_image037.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhcwAvAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAMACQBs AB0AhAAAAAAAAAUAAA0ICAAFCw0LCwAABQgIDQ0IBQUICwAFCAsLDQsIBQsFAAUIDQgFAAUABQUF CwUFCAgFBQsICAgICwsFBQUFBQECAwECAwECAwECAwECAwECAwECAwECAwX/ICCKgxAQY6qubOu+ cCzPQGEGxkHvfO/ziERJsfgZj8gdI5HcNXC6pvTVQE1hCAeAQbx6R4WH9fsqQIrka7ibbhUirIEE PXoGiIUJHRg4qexsImt7bSkUhFsGezZaJUwlWj6QYAKNAkxgD4qFK0GdgTWVdV2MPgiboZEADYFh qJwiCCgDUal/Y4OxrzSsaFWZrQI5MjZ9wX3DgjcJcAx9OHRce6e1NrgPmDFPBhWXutUCY6HJL9QA Q0WTYQFaeVG/bmIpuSTeo4gqNkzF7Q+BjvOE1YqXrYGiMLiELShVA06LSSNKPHtGLtHAFfrqbJI4 EUrAinEgXDxnT8QTLdvw6qVAwC4FA3I2QLFUxaKXIHkWKYGyAbJTSxUss2356cxji5kpbMb6OQLp Q1FLMSkF4JRSTxUlxC0V93Kgs10rmSIERqjqipdoeOoYexPUuJH5TGRjcIAnnV5vml4lmY1tIkRd XUgTYeHCJr9oMQqMwXJiF5ZdBogMhemUSr9/c66YenZTAwdVLBRJHFjxXqDP/t1AVaCCHbAqBt/s ozWTUIwP+jhYF2mdn3iwYcWlyTi4Gk0qhetNPhyWK+bKqZ6mRLyNQejRC8wpMwFumuvRw0uRLb78 D1nm0zth50m9+zKrsb9P/9X7/PchAAA7 ------=_NextPart_01C97CE7.92FACF50 Content-Location: file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0031_image038.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhYwACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAAAAABj AAEAgAAAAAAAAAIIjI+py+0PIysAOy== ------=_NextPart_01C97CE7.92FACF50 Content-Location: file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0031_image039.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhYwACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAAAAABj AAEAgAAAAAAAAAIIjI+py+0PIysAOy== ------=_NextPart_01C97CE7.92FACF50 Content-Location: file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0032.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Area in Polar Coordinates
Area in Polar Coordinates
Ó=  <= /span>Frank Ma 2006
cos2(A) =3D
From cosine double-= angle formulas, we get the following:
 1 + cos(2A) 
2
sin2(A) =3D
 1 – cos(2A) 
2
We use these formul= as for the integral:
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Area in Polar Coordinates
Ó=  <= /span>Frank Ma 2006
cos2(A) =3D
From cosine double-= angle formulas, we get the following:
 1 + cos(2A) 
2
sin2(A) =3D
 1 – cos(2A) 
2
We use these formul= as for the integral:
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Area in Polar Coordinates
Ó=  <= /span>Frank Ma 2006
cos2(A) =3D
From cosine double-= angle formulas, we get the following:
 1 + cos(2A) 
2
sin2(A) =3D
 1 – cos(2A) 
2
We use these formul= as for the integral:
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file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0034_image051.gif Content-Transfer-Encoding: base64 Content-Type: image/gif R0lGODlhYwACAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAAAAABj AAEAgAAAAAAAAAIIjI+py+0PIysAOy== ------=_NextPart_01C97CE7.92FACF50 Content-Location: file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0035.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Area in Polar Coordinates
Area in Polar Coordinates
Ó=  <= /span>Frank Ma 2006
cos2(A) =3D
From cosine double-= angle formulas, we get the following:
 1 + cos(2A) 
2
sin2(A) =3D
 1 – cos(2A) 
2
We use these formul= as for the integral:
We integrate sin2(x) similarly.  
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Area in Polar Coordinates
Ó=  <= /span>Frank Ma 2006
cos2(A) =3D
From cosine double-= angle formulas, we get the following:
 1 + cos(2A) 
2
sin2(A) =3D
 1 – cos(2A) 
2
We use these formul= as for the integral:
= We integrate sin2(x) similarly.  These are useful in polar = area calculations d= ue to f2(q) in the integrand.=  
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Area in Polar Coordinates
Example:
Find the area enc= losed by r =3D f(q) =3D 2sin(q). =
------=_NextPart_01C97CE7.92FACF50 Content-Location: file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0038.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Area in Polar Coordinates
Area in Polar Coordinates
Example:
Find the area enc= losed by r =3D f(q) =3D 2sin(q). =
1
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Area in Polar Coordinates
Example:
Find the area enc= losed = by r =3D f(q) =3D 2sin(q). The = circle is traced = once round with q from 0 to π.
q=3Dπ
q=3D0
1
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Area in Polar Coordinates
Example:
Find the area enc= losed = by r =3D f(q) =3D 2sin(q). The = circle is traced = once round with q from 0 to π.
1
q=3Dπ
q=3D0
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Area in Polar Coordinates
Ó <= /span>Frank Ma 2006
Example:
Find the area enc= losed = by r =3D f(q) =3D 2sin(q). The = circle is traced = once round with q from 0 to π.
So We have:
1
q=3Dπ
q=3D0
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------=_NextPart_01C97CE7.92FACF50 Content-Location: file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0041.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Area in Polar Coordinates
Area in Polar Coordinates
Ó <= /span>Frank Ma 2006
Example:
Find the area enc= losed = by r =3D f(q) =3D 2sin(q). The = circle is traced = once round with q from 0 to π.
So We have:
1
q=3Dπ
q=3D0
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file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0042.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Area in Polar Coordinates
Area in Polar Coordinates
Ó <= /span>Frank Ma 2006
Example:
Find the area enc= losed = by r =3D f(q) =3D 2sin(q). The = circle is traced = once round with q from 0 to π.
So We have:
1
q=3Dπ
q=3D0
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Area in Polar Coordinates
Ó <= /span>Frank Ma 2006
Example:
Find the area enc= losed = by r =3D f(q) =3D 2sin(q). The = circle is traced = once round with q from 0 to π.
So We have:
1
q=3Dπ
q=3D0
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Area in Polar Coordinates
Ó <= /span>Frank Ma 2006
Example:
Find the area enc= losed = by r =3D f(q) =3D 2sin(q). The = circle is traced = once round with q from 0 to π.
So We have:
1
q=3Dπ
q=3D0
Area of a circle with r=3D1
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R0lGODlhFAAUAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAYAAQAO AA4AgAAAAAAAAAIchI8WyRDfzlvRwTqrpTHjm0AeM4HkeJYdSmpHAQA7 ------=_NextPart_01C97CE7.92FACF50 Content-Location: file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0049.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Area in Polar Coordinates
Area in Polar Coordinates
Ó <= /span>Frank Ma 2006
Example:
Find the area enc= losed = by r =3D f(q) =3D 2sin(q). The = circle is traced = once round with q from 0 to π.
So We have:
1
q=3Dπ
q=3D0
If we integrate q from 0 to 2π, we get 2= p for the area.
Area of a circle with r=3D1
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file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0050.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Area in Polar Coordinates
Area in Polar Coordinates
Ó <= /span>Frank Ma 2006
Example:
Find the area enc= losed = by r =3D f(q) =3D 2sin(q). The = circle is traced = once round with q from 0 to π.
So We have:
1
q=3Dπ
q=3D0
If we integrate q from 0 to 2π, we get 2= p for the area. This because the graph traced over the ci= rcle twice hence the answer for the area of two circles.
Area of a circle with r=3D1
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file:///C:/4EC892E5/43AreainPolarCoordinate_files/slide0006.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Area in Polar Coordinates
Area in Polar Coordinates
Ó <= /span>Frank Ma 2006
Example:
Find the area enclosed by
r =3D f(q) =3D 1 – cos(q) with
0 < q <
`= 0;
2 =
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Area in Polar Coordinates
Ó <= /span>Frank Ma 2006
Example:
Find the area enclosed by
r =3D f(q) =3D 1 – cos(q) with
0 < q <
`= 0;
2 =