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CHAPTER 1 - 2 - 10 .1 10 cm. radius A =pR =p(3.8 10 cm...

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10. We assume an uncertainty of 0.1 × 10 4  cm.  We compare the area for the specified radius to the area for the extreme  radius. A 1  = p R 1 2  = p(3.8 × 10 4  cm) 2  = 4.54 × 10 9  cm 2  ;       A 2  = p R 2 2  = p[(3.8 + 0.1) × 10 4   cm] 2  = 4.78 × 10 9  cm 2  , so the uncertainty in the area is ? A  = A 2  –  A 1  = 0.24 × 10 9  cm 2  = 0.2 × 10 9  cm 2 . We write the area as         A  = (4.5  ±  0.2) × 10 9  cm 2 . We could also treat the change as a differential: d A  =2p R  d R  = 2p(3.8 × 10 4  cm)(0.1 × 10 4  cm) = 2 × 10 8  cm 2 . 11. We compare the volume with the specified radius to the volume for the extreme radius. V 1  =  ) p R 1 3  =  ) p(2.86 m) 3  = 98.0 m 3 ;       V 2  =  ) p R 2 3  =  ) p(2.86 m + 0.08 m) 3  = 106.45 m 3 , so the uncertainty in the volume is ? V  = V 2  – V 1  = 8.5 m 3 ; and the % uncertainty is % uncertainty = [(8.5 m 3 )/(98.0 m 3 )] 100 =        9% .
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