The document discusses calculating volumes of objects with non-circular cross-sections. It gives the example of finding the volume of a rectangular pyramid. The volume formula derived is 1/3*base area*height. It also discusses a solid with a circular base where each cross-section is a square perpendicular to the base.

1. Volumes By Non Circular
Cross-Sections

2. Volumes By Non Circular
Cross-Sections
a
b
e.g. Find the volume of this rectangular pyramid
h

3. Volumes By Non Circular
Cross-Sections
y
xa
b
e.g. Find the volume of this rectangular pyramid
h

4. Volumes By Non Circular
Cross-Sections
y
xa
z
b
e.g. Find the volume of this rectangular pyramid
h

5. Volumes By Non Circular
Cross-Sections
y
xa
z
b
e.g. Find the volume of this rectangular pyramid
h

6. Volumes By Non Circular
Cross-Sections
y
xa
z
z
b
e.g. Find the volume of this rectangular pyramid
h

7. Volumes By Non Circular
Cross-Sections
y
xa
z
y2
z
b
e.g. Find the volume of this rectangular pyramid
h

8. Volumes By Non Circular
Cross-Sections
y
xa
z
y2
zx2
b
e.g. Find the volume of this rectangular pyramid
h

9. Volumes By Non Circular
Cross-Sections
y
xa
z
y2
zx2
b
e.g. Find the volume of this rectangular pyramid
h
  xyzA 4

10. Find x in terms of z
x
z
h
b
2
1
b
2
1


11. Find x in terms of z
x
z
h
b
2
1
b
2
1


12. Find x in terms of z
x
z
h
b
2
1
b
2
1

x

13. Find x in terms of z
x
z
h
b
2
1
b
2
1

x (x,z)

14. Find x in terms of z
x
z
h
b
2
1
b
2
1

x (x,z)
ssimilarusing:1Method 

15. Find x in terms of z
x
z
h
b
2
1
b
2
1

x (x,z)
ssimilarusing:1Method 
h
b
2
1

16. Find x in terms of z
x
z
h
b
2
1
b
2
1

x (x,z)
ssimilarusing:1Method 
h
b
2
1
h - z
x

17. Find x in terms of z
x
z
h
b
2
1
b
2
1

x (x,z)
ssimilarusing:1Method 
h
b
2
1
h - z
x
x
zh
b
h 

2
1

18. Find x in terms of z
x
z
h
b
2
1
b
2
1

x (x,z)
ssimilarusing:1Method 
h
b
2
1
h - z
x
x
zh
b
h 

2
1
 
h
zhb
x
x
zh
b
h
2
2





19. Find x in terms of z
x
z
h
b
2
1
b
2
1

x (x,z)
ssimilarusing:1Method 
h
b
2
1
h - z
x
x
zh
b
h 

2
1
 
h
zhb
x
x
zh
b
h
2
2




geometrycoordinateusing:2Method

20. Find x in terms of z
x
z
h
b
2
1
b
2
1

x (x,z)
ssimilarusing:1Method 
h
b
2
1
h - z
x
x
zh
b
h 

2
1
 
h
zhb
x
x
zh
b
h
2
2




geometrycoordinateusing:2Method
(0,h)

21. Find x in terms of z
x
z
h
b
2
1
b
2
1

x (x,z)
ssimilarusing:1Method 
h
b
2
1
h - z
x
x
zh
b
h 

2
1
 
h
zhb
x
x
zh
b
h
2
2




geometrycoordinateusing:2Method
(0,h)





 0,
2
1
b

22. Find x in terms of z
x
z
h
b
2
1
b
2
1

x (x,z)
ssimilarusing:1Method 
h
b
2
1
h - z
x
x
zh
b
h 

2
1
 
h
zhb
x
x
zh
b
h
2
2




geometrycoordinateusing:2Method
b
h
b
h
m
2
2
1
0
0





(0,h)





 0,
2
1
b

23. Find x in terms of z
x
z
h
b
2
1
b
2
1

x (x,z)
ssimilarusing:1Method 
h
b
2
1
h - z
x
x
zh
b
h 

2
1
 
h
zhb
x
x
zh
b
h
2
2




geometrycoordinateusing:2Method
b
h
b
h
m
2
2
1
0
0





 0
2


 x
b
h
hz
(0,h)





 0,
2
1
b

24. Find x in terms of z
x
z
h
b
2
1
b
2
1

x (x,z)
ssimilarusing:1Method 
h
b
2
1
h - z
x
x
zh
b
h 

2
1
 
h
zhb
x
x
zh
b
h
2
2




geometrycoordinateusing:2Method
b
h
b
h
m
2
2
1
0
0





 0
2


 x
b
h
hz
 
 
h
zhb
x
hxhzb
2
2



(0,h)





 0,
2
1
b

25. Find x in terms of z
x
z
h
b
2
1
b
2
1

x (x,z)
ssimilarusing:1Method 
h
b
2
1
h - z
x
x
zh
b
h 

2
1
 
h
zhb
x
x
zh
b
h
2
2




geometrycoordinateusing:2Method
b
h
b
h
m
2
2
1
0
0





 0
2


 x
b
h
hz
 
 
h
zhb
x
hxhzb
2
2



 
h
zha
y
2
Similarly;


(0,h)





 0,
2
1
b

26.      



 



 

h
zha
h
zhb
zA
22
4

27.      



 



 

h
zha
h
zhb
zA
22
4
 
2
2
h
zhab 


28.      



 



 

h
zha
h
zhb
zA
22
4
 
2
2
h
zhab 

 
z
h
zhab
V 

 2
2

29.      



 



 

h
zha
h
zhb
zA
22
4
 
2
2
h
zhab 

 
z
h
zhab
V 

 2
2
 





h
z
z
z
h
zhab
V
0
2
2
0
lim

30.      



 



 

h
zha
h
zhb
zA
22
4
 
2
2
h
zhab 

 
z
h
zhab
V 

 2
2
 





h
z
z
z
h
zhab
V
0
2
2
0
lim
  
h
dzzh
h
ab
0
2
2

31.      



 



 

h
zha
h
zhb
zA
22
4
 
2
2
h
zhab 

 
z
h
zhab
V 

 2
2
 





h
z
z
z
h
zhab
V
0
2
2
0
lim
  
h
dzzh
h
ab
0
2
2
 
h
zh
h
ab
0
3
2
3 









32.      



 



 

h
zha
h
zhb
zA
22
4
 
2
2
h
zhab 

 
z
h
zhab
V 

 2
2
 





h
z
z
z
h
zhab
V
0
2
2
0
lim
  
h
dzzh
h
ab
0
2
2
 
h
zh
h
ab
0
3
2
3 








3
3
2
units
3
3
0
abh
h
h
ab









33. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx

34. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x-2
-2
2
2

35. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2

36. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2

37. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2

38. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2

39. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2

40. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2

41. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2

42. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2

43. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2

44. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2

45. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2

46. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2
x

47. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2
x
y2
y2

48. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2
x
y2
y2
  2
4yxA 

49. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2
x
y2
y2
  2
4yxA 
 2
44 x

50. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2
x
y2
y2
  2
4yxA 
 2
44 x
  xxV  2
44

51. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2
x
y2
y2
  2
4yxA 
 2
44 x
  xxV  2
44
 


2
2
2
0
44lim
x
x
xxV

52. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2
x
y2
y2
  2
4yxA 
 2
44 x
  xxV  2
44
 


2
2
2
0
44lim
x
x
xxV
  
2
0
2
48 dxx

53. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2
x
y2
y2
  2
4yxA 
 2
44 x
  xxV  2
44
 


2
2
2
0
44lim
x
x
xxV
  
2
0
2
48 dxx
2
0
3
3
1
48 


  xx

54. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2
x
y2
y2
  2
4yxA 
 2
44 x
  xxV  2
44
 


2
2
2
0
44lim
x
x
xxV
  
2
0
2
48 dxx
2
0
3
3
1
48 


  xx
3
units
3
128
0
3
8
88






 

55. (ii) A solid has a circular base, , and each cross section is
a square perpendicular to the base.
422
 yx
y
x2
2
-2
-2
x
y2
y2
  2
4yxA 
 2
44 x
  xxV  2
44
 


2
2
2
0
44lim
x
x
xxV
  
2
0
2
48 dxx
2
0
3
3
1
48 


  xx
3
units
3
128
0
3
8
88






 
Exercise 3A; 16ade, 19
Exercise 3C; 2, 7, 8