Surface Integrals – On a cone – Exercise 4068 Post category:Surface Integrals Post comments:0 Comments Exercise Calculate the integral ∫∫S(x2+y2+z2)ds\int\int_S (x^2+y^2+z^2) ds∫∫S(x2+y2+z2)ds Where S is part of the cone z=x2+y2z=\sqrt{x^2+y^2}z=x2+y2 That is on the circle x2+y2≤2xx^2+y^2\leq 2xx2+y2≤2x Final Answer Show final answer ∫∫S(x2+y2+z2)ds=32π\int\int_S (x^2+y^2+z^2) ds=3\sqrt{2}\pi∫∫S(x2+y2+z2)ds=32π Solution Coming soon… Share with Friends Read more articles Previous PostSurface Integrals – Surface area of a plane – Exercise 4074 Next PostSurface Integrals – On a paraboloid – Exercise 4055 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 cone – Exercise 4120 March 19, 2019 Surface Integrals – On a cylinder – Exercise 4048 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
