11.28 - Section 13.6 wrap-up Stewart computes the surface...

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Unformatted text preview: Section 13.6 wrap-up: Stewart computes the surface area of a sphere of radius a using a spherical coordinate parametrization. We’re doing it a different way, with polar coordinates instead. Last time we parametrized the top half of the sphere of radius a by r = 〈 r cos θ , r sin θ , sqrt( a 2 – r 2 ) 〉 with ( r , θ ) varying over D = [0, a ] × [0,2 π ]. We computed that | r r × r θ | = ar / sqrt( a 2 – r 2 ), so (picking up from where we left off) we get A ( S ) = ∫ ∫ D | r r × r θ | dA = … ..?.. ∫ 2 π ∫ a ( ar / sqrt( a 2 – r 2 )) r dr d θ ? (take a vote) No! In our formula for surface area, dA refers to area in the u , v plane (which here is the r , θ plane), NOT the x , y plane! So the area element is just dr d θ . A ( S ) = ∫ 2 π ∫ a ( ar / sqrt( a 2 – r 2 )) dr d θ = ( ∫ 2 π d θ ) ( ∫ a ( ar / sqrt( a 2 – r 2 )) dr ) = (2 π ) (– a sqrt( a 2 – r 2 ) | a ) = (2 π ) ( a 2 ) which agrees with the fact that the surface area of the sphere of radius a is 4 π a 2 ....
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11.28 - Section 13.6 wrap-up Stewart computes the surface...

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