S 1 If we were to use the washer method we would first have to locate the local

# S 1 if we were to use the washer method we would

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1. If we were to use the “washer” method, we would first have to locate the local maximum point ( a, b ) of y = x ( x 1) 2 using the methods of Chapter 4. Then we would have to solve the equation y = x ( x 1) 2 for x in terms of y to obtain the functions x = g 1 ( y ) and x = g 2 ( y ) shown in the first figure. This step would be difficult because it involves the cubic formula. Finally we would find the volume using V = π R b 0 © [ g 1 ( y )] 2 [ g 2 ( y )] 2 ª dy . Using shells, we find that a typical approximating shell has radius x , so its circumference is 2 π x . Its height is y , that is, x ( x 1) 2 . So the total volume is V = R 1 0 2 π x £ x ( x 1) 2 ¤ dx = 2 π R 1 0 ¡ x 4 2 x 3 + x 2 ¢ dx = 2 π · x 5 5 2 x 4 4 + x 3 3 ¸ 1 0 = π 15 3. V = Z 2 1 2 π x · 1 x dx = 2 π Z 2 1 1 dx = 2 π [ x ] 2 1 = 2 π (2 1) = 2 π 5. V = R 1 0 2 π xe x 2 dx . Let u = x 2 . Thus, du = 2 x dx , so V = π R 1 0 e u du = π £ e u ¤ 1 0 = π (1 1 /e ) VOLUMES BY CYLINDRICAL SHELLS 7 Solutions: Volumes by Cylindrical Shells
7. The curves intersect when 4( x 2) 2 = x 2 4 x + 7 4 x 2 16 x + 16 = x 2 4 x + 7 3 x 2 12 x + 9 = 0 3( x 2 4 x + 3) = 0 3( x 1)( x 3) = 0 , so x = 1 or 3 . V = 2 π R 3 1 © x £¡ x 2 4 x + 7 ¢ 4( x 2) 2 ¤ª dx = 2 π R 3 1 £ x ( x 2 4 x + 7 4 x 2 + 16 x 16) ¤ dx = 2 π R 3 1 £ x ( 3 x 2 + 12 x 9) ¤ dx = 2 π ( 3) R 3 1 ( x 3 4 x 2 + 3 x ) dx = 6 π £ 1 4 x 4 4 3 x 3 + 3 2 x 2 ¤ 3 1 = 6 π £¡ 81 4 36 + 27 2 ¢ ¡ 1 4 4 3 + 3 2 ¢¤ = 6 π ¡ 20 36 + 12 + 4 3 ¢ = 6 π ¡ 8 3 ¢ = 16 π 9. V = R 2 1 2 π y ¡ 1 + y 2 ¢ dy = 2 π R 2 1 ¡ y + y 3 ¢ dy = 2 π £ 1 2 y 2 + 1 4 y 4 ¤ 2 1 = 2 π £ (2 + 4) ¡ 1 2 + 1 4 ¢¤ = 2 π ¡ 21 4 ¢ = 21 π 2 11. V = 2 π Z 8 0 [ y ( 3 y 0)] dy = 2 π Z 8 0 y 4 / 3 dy = 2 π h 3 7 y 7 / 3 i 8 0 = 6 π 7 (8 7 / 3 ) = 6 π 7 (2 7 ) = 768 π 7 8 VOLUMES BY CYLINDRICAL SHELLS
13. The curves intersect when 4 x 2 = 6 2 x 2 x 2 + x 3 = 0 (2 x + 3)( x 1) = 0 x = 3 2 or 1 . Solving the equations for x gives us y = 4 x 2 x = ± 1 2 y and 2 x + y = 6 x = 1 2 y + 3 . V = 2 π Z 4 0 © y £¡ 1 2 y ¢ ¡ 1 2 y ¢¤ª dy + 2 π Z 9 4 © y £¡ 1 2 y + 3 ¢ ¡ 1 2 y ¢¤ª dy = 2 π Z 4 0 ( y y ) dy + 2 π Z 9 4 ³ 1 2 y 2 + 3 y + 1 2 y 3 / 2 ´ dy = 2 π h 2 5 y 5 / 2 i 4 0 + 2 π h 1 6 y 3 + 3 2 y 2 + 1 5 y 5 / 2 i 9 4 = 2 π ¡ 2 5 · 32 ¢ + 2 π £¡ 243 2 + 243 2 + 243 5 ¢ ¡ 32 3 + 24 + 32 5 ¢¤ = 128 5 π + 2 π ¡ 433 15 ¢ = 1250 15 π = 250 3 π 15. V = R 2 1 2 π ( x 1) x 2 dx = 2 π £ 1 4 x 4 1 3 x 3 ¤ 2 1 = 2 π £¡ 4 8 3 ¢ ¡ 1 4 1 3 ¢¤ = 17 6 π 17. V = R 2 1 2 π (4 x ) x 2 dx = 2 π £ 4 3 x 3 1 4 x 4 ¤ 2 1 = 2 π £¡ 32 3 4 ¢ ¡ 4 3 1 4 ¢¤ = 67 6 π 19. V = R 2 0 2 π (3 y )(5 x ) dy = R 2 0 2 π (3 y ) ¡ 5 y 2 1 ¢ dy = R 2 0 2 π ¡ 12 4 y 3 y 2 + y 3 ¢ dy = 2 π £ 12 y 2 y 2 y 3 + 1 4 y 4 ¤ 2 0 = 2 π (24 8 8 + 4) = 24 π VOLUMES BY CYLINDRICAL SHELLS 9
21. V = R 2 1 2 π x ln x dx 23. V = R 1 0 2 π [ x ( 1)] ¡ sin π 2 x x 4 ¢ dx 25. V = R π 0 2 π (4 y ) sin y dy 27. x = π / 4 0 4 = π 16 .