solution to Area Volume and Length problems

# Thomas' Calculus: Early Transcendentals

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MAT104—Problems on Area, Volume and Length From Old Exams. Answers. 1. (a) 2 π ; (b) 9 π . 2. ( e 2 - 1) 2. 3. Volume = π ( e - 2 - e - 4 ) / 2; Surface Area = π e 2 + 1 e 2 - e 4 + 1 e 4 + ln 1 + e 2 + 1 1 + e 4 + 1 e ! . (or add π/e 2 + π/e 4 .) 4. Area( R ) = 1 - π/ 4; Volume( S ) = π (5 / 3 - π/ 2); Surface Area( S ) = π 2 - 2 π ( π 2 - π ). 5. (a) 5 / 6; (b) 17 π/ 15; (c) 11 π/ 6. 6. (a) y = 3; (b) 2 π 2 . 7. 16 π/ 3. 8. 5 / 2 + ln 3 2 / 4. 9. (a) 2 π R 2 1 x ( x + x 2 ) d x = 73 π/ 6; (b) 2 π R 2 1 (3 - x )( x + x 2 ) d x = 65 π/ 6; (c) π R 2 1 ( x + x 2 ) 2 d x = 481 π/ 30. 10. d y d x = sin t + t cos t cos t - t sin t . ; y = - 2 x/π + π/ 2. 11. π ( e - 1); π R 1 0 e 2 x 2 d x . 12. π/ 6 ( (1 + 4 e 2 ) 3 / 2 - 5 5 ) . 13. (a) 3 π/ 5; (b) 6
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• Fall '07
• Nelson

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