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...


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:<br>(a) sketch the curves, shade the region Rof the plane the curves bound;<br>(b) represent the region Reither as a Type I,<br>I= a,<br>R=<br>- g(z), y = f(z)):<br>or Type II region,<br>y=d<br>(ェ=(y), エ= f(y))<br>R=<br>or as the union<br>R= R1 U...U R.<br>of several Type I, or Type II regions, if necesary (in the last case please provide description of all the regions R,);<br>(c) find the area of the region R;<br>(d) round your result in (c) to five decimal places.<br>(i) y= 12 - z, y =- 10z|<br>

Extracted 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<br>R= the region bounded by the curves<br>y = sin(1), y = cos(z), 1 = 0.<br>(a)<br>I= 0,<br>I= #/4<br>(y = sin(z), y = coa(r),<br>(b)<br>area( R) =<br>/4<br>

Extracted 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.

Jun 04, 2022
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