1. Volumes By Non Circular
Cross-Sections

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

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

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

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

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

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

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

9. Volumes By Non Circular
Cross-Sections
e.g. Find the volume of this rectangular pyramid
y
h
z
a x
2y
b
2x z
A z   4 xy

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

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

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

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

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

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

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

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

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

19. Find x in terms of z Method 2 : using coordinate geometry
z
h
x (x,z)
1 1 x
 b b
2 2
Method 1 : using similar s
h hz
h 
1 x
b
1 2
b 2h h  z
2 
b x
h-z b h  z 
x
x 2h

20. Find x in terms of z Method 2 : using coordinate geometry
z
h (0,h)
x (x,z)
1 1 x
 b b
2 2
Method 1 : using similar s
h hz
h 
1 x
b
1 2
b 2h h  z
2 
b x
h-z b h  z 
x
x 2h

21. Find x in terms of z Method 2 : using coordinate geometry
z
h (0,h)
x (x,z)
 1 b,0 
1 1 x 2 
 b b  
2 2
Method 1 : using similar s
h hz
h 
1 x
b
1 2
b 2h h  z
2 
b x
h-z b h  z 
x
x 2h

22. Find x in terms of z Method 2 : using coordinate geometry
z
h0
m
h (0,h) 1
0 b
x (x,z) 2
 2h

 1 b,0  b
1 1 x 2 
 b b  
2 2
Method 1 : using similar s
h hz
h 
1 x
b
1 2
b 2h h  z
2 
b x
h-z b h  z 
x
x 2h

23. Find x in terms of z Method 2 : using coordinate geometry
z
h0
m
h (0,h) 1
0 b
x (x,z) 2
 2h

 1 b,0  b
1 1 x 2   2h
 b b   zh   x  0
2 2 b
Method 1 : using similar s
h hz
h 
1 x
b
1 2
b 2h h  z
2 
b x
h-z b h  z 
x
x 2h

24. Find x in terms of z Method 2 : using coordinate geometry
z
h0
m
h (0,h) 1
0 b
x (x,z) 2
 2h

 1 b,0  b
1 1 x 2   2h
 b b   zh   x  0
2 2 b
Method 1 : using similar s b z  h   2hx
h hz
 b h  z 
h
1 x
b x 2h
1 2
b 2h h  z
2 
b x
h-z b h  z 
x
x 2h

25. Find x in terms of z Method 2 : using coordinate geometry
z
h0
m
h (0,h) 1
0 b
x (x,z) 2
 2h

 1 b,0  b
1 1 x 2   2h
 b b   zh   x  0
2 2 b
Method 1 : using similar s b z  h   2hx
h hz
 b h  z 
h
1 x
b x 2h
1 2
b 2h h  z Similarly;
2 
b x a h  z 
h-z b h  z  y
x 2h
x 2h

26.  bh  z   ah  z 
A z   4 
 2h   2h 
 

27.  bh  z   ah  z 
A z   4 
 2h   2h 
 
abh  z 
2

h2

28.  bh  z   ah  z 
A z   4 
 2h   2h 
 
abh  z 
2

h2
abh  z 
2
V  2
 z
h

29.  b h  z    a  h  z  
A z   4 
 2h   2h  abh  z 
h 2
  V  lim  2
 z
z  0 h
abh  z 
2 z 0

h2
abh  z 
2
V  2
 z
h

30.  bh  z   ah  z 
A z   4 
 2h   2h  abh  z 
h 2
  V  lim  2
 z
z  0 h
abh  z 
2 z 0
 ab
h
h2  2  h  z  dz
2
abh  z  h 0
2
V  2
 z
h

31.  bh  z   ah  z 
A z   4 
 2h   2h  abh  z 
h 2
  V  lim  2
 z
z  0 h
abh  z 
2 z 0
 ab
h
h2  2  h  z  dz
2
abh  z  h 0
2
V   z
ab  h  z  
3 h
2
h  2
h   3 0 

32.  b h  z    a  h  z  
A z   4 
 2h   2h  abh  z 
h 2
  V  lim  2
 z
z  0 h
abh  z 
2 z 0
 ab
h
h2  2  h  z  dz
2
abh  z  h 0
2
V   z
ab  h  z  
3 h
2
h  2
h   3 0 
ab  h 3 
 2 0  
h  3
abh
 units 3
3

33. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.

34. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
y
2
-2 2 x
-2

35. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
y
2
-2 2 x
-2

36. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
y
2
-2 2 x
-2

37. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
y
2
-2 2 x
-2

38. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
y
2
-2 2 x
-2

39. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
y
2
-2 2 x
-2

40. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
y
2
-2 2 x
-2

41. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
y
2
-2 2 x
-2

42. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
y
2
-2 2 x
-2

43. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
y
2
-2 2 x
-2

44. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
y
2
-2 2 x
-2

45. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
y
2
-2 2 x
-2

46. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
y
2
-2 2 x
-2
x

47. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
y
2
-2 2 x
-2
2y
x
2y

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

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

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

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

52. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
44  x 2  x
2
y V  lim
x  0

x  2
2
2  8 4  x 2 dx
0
-2 2 x
-2
A x   4 y 2
 44  x 2 
2y
V  44  x 2  x
x
2y

53. (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is
a square perpendicular to the base.
44  x 2  x
2
y V  lim
x  0

x  2
2
2  8 4  x 2 dx
0
2
4 x  1 x 3 
 8
-2 2 x  3 0 
-2
A x   4 y 2
 44  x 2 
2y
V  44  x 2  x
x
2y

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

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