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Thomas Calculus 13th [Solutions]

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Section 6.1 Volumes Using Cross-Sections 433<br />

11. The slices perpendicular to the edge labeled 5 are triangles, and by similar triangles we have b 4 h 3 b<br />

h 3 4 .<br />

The equation of the line through (5, 0) and (0, 4) is y 4 x 4, thus the length of the base 4 x 4 and<br />

5<br />

5<br />

the height 3 4 x 4 3 x 3. Thus A( x) 1 (base) (height) 1 4 x 4 3 x 3<br />

4 5 5<br />

2 2 5 5<br />

6 2 12<br />

b<br />

5 2 3 2<br />

5<br />

x x 6 and V A<br />

6 12 2 6<br />

25 5<br />

( x ) dx x x<br />

0 25 5 6 dx x x<br />

25 5<br />

6 x (10 30 30) 0 10<br />

a<br />

0<br />

12. The slices parallel to the base are squares. The cross section of the pyramid is a triangle, and by similar<br />

triangles we have<br />

b 3<br />

b<br />

3<br />

h Thus<br />

3<br />

5<br />

3<br />

y<br />

25 0<br />

h<br />

15 0 15<br />

5 5 .<br />

2 3<br />

2<br />

9 2 d<br />

5<br />

9 2<br />

5 25 c<br />

0 25<br />

A( y) (base) y y V A( y)<br />

dy y dy<br />

13. (a) It follows from Cavalieris Principle that the volume of a column is the same as the volume of a right<br />

prism with a square base of side length s and altitude h. Thus,<br />

2 2<br />

STEP 1) A( x) (sidelength) s ;<br />

STEP 2) a 0, b h;<br />

STEP 3)<br />

b<br />

h 2 2<br />

V A( x)<br />

dx s dx s h<br />

a<br />

0<br />

(b) From Cavalieris Principle we conclude that the volume of the column is the same as the volume of the<br />

2<br />

prism described above, regardless of the number of turns V s h<br />

14. 1) The solid and the cone have the same altitude<br />

of 12.<br />

2) The cross sections of the solid are disks of<br />

diameter x x x<br />

If we place the vertex of<br />

2 2 .<br />

the cone at the origin of the coordinate system<br />

and make its axis or symmetry coincide with<br />

the x -axis then the cones cross sections will<br />

be circular disks of diameter x x x<br />

4 4 2<br />

(see accompanying figure).<br />

3) The solid and the cone have equal altitudes and<br />

identical parallel cross sections. From<br />

Cavaliers Principle we conclude that the solid<br />

and the cone have the same volume.<br />

15.<br />

16.<br />

17.<br />

2 2 2 2 2<br />

R ( x ) y 1 x V R<br />

2 0 ( x ) dx<br />

0 1 x dx<br />

2 0 1 x x dx x x x<br />

4 2 12<br />

0<br />

2 4 8 2<br />

2 12 3<br />

2 2 3 2<br />

( ) 3y<br />

2 2 2 3<br />

2 2 2 3<br />

2<br />

9 3 3<br />

2 0 ( ) y<br />

R y x V R y dy dy y dy y<br />

0 2 0 4 4 0 4<br />

8 6<br />

R( y) tan y<br />

4<br />

; u y<br />

4<br />

du dy<br />

4<br />

4 du dy;<br />

2<br />

y 0 u 0, y 1 u ;<br />

4<br />

1 2 1 /4 2 /4 2<br />

/4<br />

V R<br />

0 ( y ) dy<br />

0 tan y dy<br />

4<br />

4 0 tan u du 4 0<br />

1 sec u du 4 u tan u<br />

0<br />

4 1 0 4<br />

4<br />

Copyright<br />

2014 Pearson Education, Inc.

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