1

answer

0

watching

188

views

13 Nov 2019

Chapter 15, Section 15.6, Question 019 Find the flux of the vector field Upper F across sigma in the direction of positive orientation. Upper F left-parenthesis x comma y comma z right-parenthesis equals StartRoot x Superscript 2 Baseline plus y Superscript 2 EndRoot â¢ k â¢ where sigma is the portion of the cone r left-parenthesis u comma v right-parenthesis equals u â¢ c o s v â¢ i plus u â¢ s i n v â¢ j plus 10 â¢ u â¢ k with 0 less-than-or-equal-to u less-than-or-equal-to 4 â¢ s i n v comma 0 less-than-or-equal-to v less-than-or-equal-to pi. Enter the exact answer as an improper fraction, if necessary.

Chapter 15, Section 15.6, Question 019 Find the flux of the vector field F across Ï in the direction of positive orientation. F(x, y, z) - vx2+ y2k where ÏÇs the portion of the cone r(u, v)=ucos vi+usin vj+10uk with 0sus4sin v, 0svs Ï. Enter the exact answer as an improper fraction, if necessary. Click here to enter or edit your answer

Chapter 15, Section 15.6, Question 019 Find the flux of the vector field Upper F across sigma in the direction of positive orientation. Upper F left-parenthesis x comma y comma z right-parenthesis equals StartRoot x Superscript 2 Baseline plus y Superscript 2 EndRoot â¢ k â¢ where sigma is the portion of the cone r left-parenthesis u comma v right-parenthesis equals u â¢ c o s v â¢ i plus u â¢ s i n v â¢ j plus 10 â¢ u â¢ k with 0 less-than-or-equal-to u less-than-or-equal-to 4 â¢ s i n v comma 0 less-than-or-equal-to v less-than-or-equal-to pi. Enter the exact answer as an improper fraction, if necessary.

Chapter 15, Section 15.6, Question 019 Find the flux of the vector field F across Ï in the direction of positive orientation. F(x, y, z) - vx2+ y2k where ÏÇs the portion of the cone r(u, v)=ucos vi+usin vj+10uk with 0sus4sin v, 0svs Ï. Enter the exact answer as an improper fraction, if necessary. Click here to enter or edit your answer

Jean KeelingLv2

11 May 2019