Explanation Torus is obtained by rotating a circle about the axis. Given: x 2 + ( y − b ) = r 2 -------------is the circle with centre ( 0 , b ) and radius r Calculation: By rotating the circle (i) about the x -axis we obtain torus. From (i) ( y − b ) = r 2 − x 2 Taking square root ( y − b ) = ± r 2 − x 2 y = b ± r 2 − x 2 The inner part of the circle is y = b − r 2 − x 2 and the outer part of the circle is y = b + r 2 − x 2 Then the surface of the torus is the surface generated by the upper of the circle. i.e., y = b + r 2 − x 2 plus the area generated by the lower half of the circle y = b − r 2 − x 2 Surface area, s = ∫ − r r 2 π y d s S = ∫ − r r 2 π y d s d x d x = 2 ∫ 0 r 2 π y d s d x d x -------- (ii), [By symmetry] Where d s d x = 1 + ( d y d x ) 2 y = b ± r 2 − x 2 Differentiate with respect to x d d x x n = n x n − 1 ( A B ) n = A n B n ( A m ) n = A m n d y d x = 0 ± ( 1 2 r 2 − x 2 ) ( − 2 x ) d y d x = ∓ ( x r 2 − x 2 ) d s d x = 1 + ( d y d x ) 2 = 1 + ( ∓ x r 2 − x 2 ) 2 = 1 + ( ∓ x r 2 − x 2 ) 2 = 1 + x 2 ( r 2 − x 2 ) 2 = 1 + x 2 r 2 − x 2 = r 2 − x 2 + x 2 r 2 − x 2 = r 2 r 2 − x 2 = [ r 2 r 2 − x 2 ] 1 2 = [ r 2 ] 1 2 [ r 2 − x 2 ] 1 2 = r r 2 − x 2 ∴ d s

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 9.1, Problem 49E

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The surface area of the torus

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Chapter 9.1, Problem 48E

Chapter 9.1, Problem 50E

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The surface area of the torus

Solution Summary: The author calculates the surface area of the torus by rotating a circle about the x-axis.

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The surface area of the torus

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