# 15_5_11 - -1 a R Thus the area of 1 is S 1 = R 2 Z 2 π d...

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11. Let the sphere be x 2 + y 2 + z 2 = R 2 , and the cylinder be x 2 + y 2 = R 2 . Let 1 and 2 be the parts of the sphere and the cylinder, respectively, lying between the planes z = a and z = b , where - R a b R . Evidently, the area of 2 is S 2 = 2 π R ( b - a ) square units. An area element on the sphere is given in terms of spherical coordinates by dS = R 2 sin φ d φ d θ. On 1 we have z = R cos φ , so 1 lies between φ = cos - 1 ( b / R ) and φ = cos
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Unformatted text preview: -1 ( a / R ) . Thus the area of 1 is S 1 = R 2 Z 2 π d θ Z cos-1 ( a / R ) cos-1 ( b / R ) sin φ d φ = 2 π R 2 (-cos φ) ± ± ± cos-1 ( a / R ) cos-1 ( b / R ) = 2 π R ( b-a ) sq. units. Observe that 1 and 2 have the same area. x y z z = b z = a z = R z =-R Fig. 5-11...
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