Can you solve (ii)
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.3: Lines
Problem 10E
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Question
Can you solve (ii) and show the answer in table as i sent. Thanks for help
![1. (Areas of Plane Regions). For each of the subproblems below:
(a) sketch the curves, shade the region Rof the plane the curves bound;
(b) represent the region Reither as a Type I,
I= a,
R=
- g(z), y = f(z)):
or Type II region,
y=d
(ェ=(y), エ= f(y))
R=
or as the union
R= R1 U...U R.
of several Type I, or Type II regions, if necesary (in the last case please provide description of all the regions R,);
(c) find the area of the region R;
(d) round your result in (c) to five decimal places.
(i) y= 12 - z, y =- 10z|](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa748baf4-8e38-49cd-acc2-c74d4830d926%2F61db0e5d-6bd9-4131-8969-931e3cfaba20%2Fkojtg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. (Areas of Plane Regions). For each of the subproblems below:
(a) sketch the curves, shade the region Rof the plane the curves bound;
(b) represent the region Reither as a Type I,
I= a,
R=
- g(z), y = f(z)):
or Type II region,
y=d
(ェ=(y), エ= f(y))
R=
or as the union
R= R1 U...U R.
of several Type I, or Type II regions, if necesary (in the last case please provide description of all the regions R,);
(c) find the area of the region R;
(d) round your result in (c) to five decimal places.
(i) y= 12 - z, y =- 10z|
![Subproblem () | Answers
R= the region bounded by the curves
y = sin(1), y = cos(z), 1 = 0.
(a)
I= 0,
I= #/4
(y = sin(z), y = coa(r),
(b)
area( R) =
/4
"(con(z) – sin(2) dz = v2- 1;
(c)
(d)
area( R) es 0.41421.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa748baf4-8e38-49cd-acc2-c74d4830d926%2F61db0e5d-6bd9-4131-8969-931e3cfaba20%2Fo64ea5h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Subproblem () | Answers
R= the region bounded by the curves
y = sin(1), y = cos(z), 1 = 0.
(a)
I= 0,
I= #/4
(y = sin(z), y = coa(r),
(b)
area( R) =
/4
"(con(z) – sin(2) dz = v2- 1;
(c)
(d)
area( R) es 0.41421.
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