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Consider the Steady-State Temperature Distribution in a Circular Disc of Radius

Question 17

Multiple Choice

Consider the steady-state temperature distribution in a circular disc of radius CC centere at the origin, with temperature given as a function, f(θ) f ( \theta ) on the boundary r=cr = c and zero on the boundaries θ=0\theta = 0 and θ=π\theta = \pi The mathematical model of this situation is


A) ∂2u∂r2−1r∂u∂r+1r2∂2u∂θ2=0,u(c,θ) =f(θ) ,u(r,0) =0,u(r,π) =0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta ) , u ( r , 0 ) = 0 , u ( r , \pi ) = 0
B) ∂2u∂r2+1r∂u∂r−1r2∂2u∂θ2=0,u(c,θ) =f(θ) ,u(r,0) =0,u(r,π) =0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta ) , u ( r , 0 ) = 0 , u ( r , \pi ) = 0
C) ∂2u∂r2−1r∂u∂r−1r2∂2u∂θ2=0,u(c,θ) =f(θ) ,u(r,0) =0,u(r,π) =0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } - \frac { 1 } { r } \frac { \partial u } { \partial r } - \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta ) , u ( r , 0 ) = 0 , u ( r , \pi ) = 0
D) ∂2u∂r2+1r2∂u∂r+1r∂2u∂θ2=0,u(c,θ) =f(θ) ,u(r,0) =0,u(r,π) =0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r ^ { 2 } } \frac { \partial u } { \partial r } + \frac { 1 } { r } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta ) , u ( r , 0 ) = 0 , u ( r , \pi ) = 0
E) ∂2u∂r2+1r∂u∂r+1r2∂2u∂θ2=0,u(c,θ) =f(θ) ,u(r,0) =0,u(r,π) =0\frac { \partial ^ { 2 } u } { \partial r ^ { 2 } } + \frac { 1 } { r } \frac { \partial u } { \partial r } + \frac { 1 } { r ^ { 2 } } \frac { \partial ^ { 2 } u } { \partial \theta ^ { 2 } } = 0 , u ( c , \theta ) = f ( \theta ) , u ( r , 0 ) = 0 , u ( r , \pi ) = 0

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