Consider the region D C R² defined by the inequalities æ² + y² > 1, x² + y² < 4, y > 0, and y
Consider the region D C R² defined by the inequalities æ² + y² > 1, x² + y² < 4, y > 0, and y
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 12T
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![Consider the region D C R? defined by the inequalities x2 + y?> 1, x² + y? < 4, y > 0, and y < x. Let w be the two-form w = (x + 8) dx ^ dy. In this
equation we evaluate the integral of w over the region D with canonical orientation using polar coordinates.
Let ø : D2 → D be the change of coordinates from Cartesian to polar coordinates:
$(r, t) = (r cos(t), r sin(t)).
The map o is bijective if D2 is the following rectangular region in the (r, t)-plane:
D2 = {(r, t) E R² | r E[|
,t E[
]}.
The transformation o is orientation-preserving (the determinant of its Jacobian is positive), and so we know that the integral of w over D is equal to the integral
of the pullback o*w over D2. We calculate the pullback two-form:
6*w =
dr A dt.
Finally we can evaluate the integral of ø*w over D2 using double integrals. We get:
=
O*w =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F001eda53-19f3-49af-8959-1aa5b0b6fb9b%2F5d6d7171-680a-4645-a06a-95b07168d56c%2Fx2nngmy_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the region D C R? defined by the inequalities x2 + y?> 1, x² + y? < 4, y > 0, and y < x. Let w be the two-form w = (x + 8) dx ^ dy. In this
equation we evaluate the integral of w over the region D with canonical orientation using polar coordinates.
Let ø : D2 → D be the change of coordinates from Cartesian to polar coordinates:
$(r, t) = (r cos(t), r sin(t)).
The map o is bijective if D2 is the following rectangular region in the (r, t)-plane:
D2 = {(r, t) E R² | r E[|
,t E[
]}.
The transformation o is orientation-preserving (the determinant of its Jacobian is positive), and so we know that the integral of w over D is equal to the integral
of the pullback o*w over D2. We calculate the pullback two-form:
6*w =
dr A dt.
Finally we can evaluate the integral of ø*w over D2 using double integrals. We get:
=
O*w =
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