problemset9s - MAT 137Y 2008-09 Winter Session, Solutions...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
MAT 137Y 2008-09 Winter Session, Solutions to Problem Set 9 1 (SHE 6.3) 12. The volume is V = Z 1 0 2 π x · 2 x dx + Z 2 1 2 π x · 2 2 - x dx = 4 π Z 1 0 x 2 dx - 4 π Z 0 1 ( 2 - u ) u du ( u = 2 - x ) = 4 π ± x 3 3 ² 1 0 - 4 π ± 4 3 u 3 / 2 - 2 5 u 5 / 2 ² 0 1 = 76 15 π . 34. The volume is given by V = Z r 2 - a 2 0 2 π x ( p r 2 - x 2 - a ) dx = 2 π Z r 2 - a 2 0 h x ( r 2 - x 2 ) 1 / 2 - ax i dx = 2 π ± - 1 3 ( r 2 - x 2 ) 3 / 2 - a 2 x 2 ² r 2 - a 2 0 = 1 3 π ( 2 r 3 + a 3 - 3 ar 2 ) . 46. We let u = r 2 - x 2 so du = - 2 x dx . The volume is given by V = 2 Z r q r 2 - h 2 4 2 π x p r 2 - x 2 dx = - 2 π Z 0 h 2 / 4 u 1 / 2 du = 2 π ± 2 3 u 3 / 2 ² h 2 / 4 0 = π h 3 6 . 2 (a) The region and a typical shell is illustrated below. (b) Using shells, the volume is V = Z 1 0 2 π ( y + 1 )( y - y 2 ) dy = 2 π Z 1 0 y 3 / 2 + y 1 / 2 - y 3 - y 2 dy = 2 π ± 2 5 y 5 / 2 + 2 3 y 3 / 2 - 1 4 y 4 - 1 3 y 3 ² 1 0 = 29 π 30 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
(c) If we were to use washers, then the volume would be expressed as V = Z 1 0 π ( x + 1 ) 2 - π ( x 2 + 1 ) 2 dx = π Z 1 0 x + 2 x + 1 - x 4 - 2 x 2 - 1 dx = π Z 1 0 x + 2 x - x 4 - 2 x 2 dx = π ± 1 2 x 2 + 4 3 x 3 / 2 - 1 5 x 5 - 2 3 x 3 ² 1 0 = 29 30 π 3 (SHE 7.1) 40. Since f 0 ( x
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/09/2010 for the course MAT 137 taught by Professor Uppal during the Spring '08 term at University of Toronto- Toronto.

Page1 / 4

problemset9s - MAT 137Y 2008-09 Winter Session, Solutions...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online