Can you solve (ii)

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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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|
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.
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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