Surface Integrals – On a cone – Exercise 4120 Post category:Surface Integrals Post comments:0 Comments Exercise Calculate the integral ∫∫SPdydz+Qdxdz+Rdxdy\int\int_S Pdydz + Qdxdz +Rdxdy∫∫SPdydz+Qdxdz+Rdxdy Where P=y−z,Q=z−x,R=x−yP=y-z, Q=z-x, R=x-yP=y−z,Q=z−x,R=x−y And the surface S is the outside of some of the cone z2=x2+y2,0≤z≤4z^2=x^2+y^2, 0\leq z\leq 4z2=x2+y2,0≤z≤4 Final Answer Show final answer ∫∫SF⃗n^ds=0\int\int_S \vec{F}\hat{n}ds=0∫∫SFn^ds=0 Solution Coming soon… Share with Friends Read more articles Previous PostSurface Integrals – On a cylinder – Exercise 4048 Next PostSurface Integrals – On a plane – Exercise 4109 You Might Also Like Surface Integrals – On a closed domain – Exercise 4782 May 8, 2019 Surface Integrals – On a hemisphere – Exercise 4089 March 17, 2019 Surface Integrals – On a cone – Exercise 4103 March 17, 2019 Surface Integrals – On a plane – Exercise 4109 March 17, 2019 Surface Integrals – On a cylinder – Exercise 4048 March 15, 2019 Surface Integrals – On a paraboloid – Exercise 4055 March 15, 2019 Leave a Reply Cancel replyCommentEnter your name or username to comment Enter your email address to comment Enter your website URL (optional) Δ
