4. Calculate: (1,0,2π) (a) | zd. +xdy+ydz (b) (1,0,0) along the curve with the parametrisation x(t) = (cost, sint, t), 0 ≤ t ≤ 2π; (2,3,2) [ x²dx − xzdy + y²dz (1,0,1) along the straight line that connects the starting and endpoints; (1,0,2) dx + (x² + y² + z²)dz (0,1,0) Y along the curve in the first octant given by x² + y² = 1, z = 2x; (0,0,2) (d) \ (2,0,0) ydxy(x 1)dy + y²zdz - along the curve in the first octant given by x² + y² + z² = 4, (x − 1)² + y² = 1.
4. Calculate: (1,0,2π) (a) | zd. +xdy+ydz (b) (1,0,0) along the curve with the parametrisation x(t) = (cost, sint, t), 0 ≤ t ≤ 2π; (2,3,2) [ x²dx − xzdy + y²dz (1,0,1) along the straight line that connects the starting and endpoints; (1,0,2) dx + (x² + y² + z²)dz (0,1,0) Y along the curve in the first octant given by x² + y² = 1, z = 2x; (0,0,2) (d) \ (2,0,0) ydxy(x 1)dy + y²zdz - along the curve in the first octant given by x² + y² + z² = 4, (x − 1)² + y² = 1.
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Usually, when these questions are asked, x=cost and y =sint, I'm confused as to why they are switched in the worked solutions here and how they got the value for z initially.
Transcribed Image Text:4. Calculate:
(1,0,2π)
(a) | zd. +xdy+ydz
(b)
(1,0,0)
along the curve with the parametrisation x(t) = (cost, sint, t), 0 ≤ t ≤ 2π;
(2,3,2)
[ x²dx − xzdy + y²dz
(1,0,1)
along the straight line that connects the starting and endpoints;
(1,0,2)
dx + (x² + y² + z²)dz
(0,1,0)
Y
along the curve in the first octant given by x² + y² = 1, z = 2x;
(0,0,2)
(d) \
(2,0,0)
ydxy(x 1)dy + y²zdz
-
along the curve in the first octant given by x² + y² + z² = 4, (x − 1)² + y² = 1.
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