x-y-z-e-cos-e-sin-0-0-2pi-R-r-x-i-y-j-z-k-a-r-a-r-a-r-a- Tinku Tara June 3, 2023 Operation Research 0 Comments FacebookTweetPin Question Number 1858 by 123456 last updated on 16/Oct/15 [x(ρ,θ,ξ)y(ρ,θ,ξ)z(ρ,θ,ξ)]=[ρeξcosθρeξsinθξ]{ρ∈[0,+∞)θ∈[0,2π)ξ∈Rr(ρ,θ,ξ)=x(ρ,θ,ξ)i+y(ρ,θ,ξ)j+z(ρ,θ,ξ)kaρ=∂r∂ρaθ=∂r∂θaξ=∂r∂ξaρ⋅aρ+aθ⋅aθ+aξ⋅aξ=???aρ⋅aθ+aρ⋅aξ+aθ⋅aξ=???aρ×aθ+aρ×aξ+aθ×aξ=??? Answered by 112358 last updated on 16/Oct/15 r(ρ,θ,ξ)=(ρeξcosθ)i+(ρeξsinθ)j+ξkaρ=∂r∂ρ=∂∂ρ(ρeξcosθ)i+∂∂ρ(ρeξsinθ)j+∂∂ρ(ξ)kaρ=(eξcosθ)i+(eξsinθ)j+0kBythesimilarprocessinvolvingpartialdifferentiationwegetaθ=(−ρeξsinθ)i+(ρeξcosθ)j+0kaξ=(ρeξcosθ)i+(ρeξsinθ)j+kForanyvectorwehavex.x=∣x∣2.∴S1=Σa.a=∣aθ∣2+∣aρ∣2+∣aξ∣2S1=e2ξ(sin2θ+cos2θ)+2ρ2e2ξ(sin2θ+cos2θ)+1S1=e2ξ(2ρ2+1)+1LetS2=aθ.aρ+aξ.aρ+aθ.aξaρ.aθ=−ρe2ξcosθsinθ+ρe2ξsinθcosθ=0aξ.aρ=ρe2ξcos2θ+ρe2ξsin2θ=ρe2ξaθ.aξ=−ρ2e2ξsinθcosθ+ρ2e2ξsinθcosθ=0∴S2=0+ρe2ξ+0=ρe2ξaρ×aθ=i(0−0)−j(0−0)+k(eξcosθ×ρeξcosθ−eξsinθ(−ρeξsinθ))aρ×aθ=0i+0j+ρe2ξkaρ×aξ=i(eξsinθ×1−0)−j(1×eξcosθ−0)+k(ρe2ξcosθsinθ−ρe2ξcosθsinθ)aρ×aξ=(eξsinθ)i−(eξcosθ)j+0kaθ×aξ=i(ρeξcosθ−0)−j(−ρeξsinθ−0)+k(−ρ2e2ξsin2θ−ρ2e2ξcos2θ)aθ×aξ=(ρeξcosθ)i+(ρeξsinθ)j−(ρ2e2ξ)k∴aρ×aθ+aρ×aξ+aθ×aξ=eξ(ρcosθ+sinθ)i+eξ(ρsinθ−cosθ)j+ρe2ξ(1−ρ)k Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: let-y-x-3-2x-1-find-dy-dx-when-x-2-Next Next post: V-pi-2-e-2z-dz-S-2pi-e-z-1-2-e-2z-dz-V-S- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.