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# X2 T05 03 parallel crosssections (2011)

## by Nigel Simmons, Teacher at Baulkham Hills High School on May 16, 2011

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## X2 T05 03 parallel crosssections (2011)Webinar Transcript

• Volumes By Non Circular Cross-Sections
• Volumes By Non Circular Cross-Sectionse.g. Find the volume of this rectangular pyramid h a b
• Volumes By Non Circular Cross-Sectionse.g. Find the volume of this rectangular pyramid y h a x b
• Volumes By Non Circular Cross-Sectionse.g. Find the volume of this rectangular pyramid y h z a x b
• Volumes By Non Circular Cross-Sectionse.g. Find the volume of this rectangular pyramid y h z a x b
• Volumes By Non Circular Cross-Sectionse.g. Find the volume of this rectangular pyramid y h z a x b z
• Volumes By Non Circular Cross-Sectionse.g. Find the volume of this rectangular pyramid y h z a x 2y b z
• Volumes By Non Circular Cross-Sectionse.g. Find the volume of this rectangular pyramid y h z a x 2y b 2x z
• Volumes By Non Circular Cross-Sectionse.g. Find the volume of this rectangular pyramid y h z a x 2y b 2x z A z   4 xy
• Find x in terms of z z h 1 1 x b b 2 2
• Find x in terms of z z h 1 1 x b b 2 2
• Find x in terms of z z h x 1 1 x b b 2 2
• Find x in terms of z z h x (x,z) 1 1 x b b 2 2
• Find x in terms of z z h x (x,z) 1 1 x b b 2 2Method 1 : using similar s
• Find x in terms of z z h x (x,z) 1 1 x b b 2 2Method 1 : using similar s h 1 b 2
• Find x in terms of z z h x (x,z) 1 1 x b b 2 2Method 1 : using similar s h 1 b 2h-z x
• Find x in terms of z z h x (x,z) 1 1 x b b 2 2Method 1 : using similar s h hz h  1 x b 1 2 b 2h-z x
• Find x in terms of z z h x (x,z) 1 1 x b b 2 2Method 1 : using similar s h hz h  1 x b 1 2 b 2h h  z 2  b xh-z b h  z  x x 2h
• Find x in terms of z Method 2 : using coordinate geometry z h x (x,z) 1 1 x b b 2 2Method 1 : using similar s h hz h  1 x b 1 2 b 2h h  z 2  b xh-z b h  z  x x 2h
• Find x in terms of z Method 2 : using coordinate geometry z h (0,h) x (x,z) 1 1 x b b 2 2Method 1 : using similar s h hz h  1 x b 1 2 b 2h h  z 2  b xh-z b h  z  x x 2h
• 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 2Method 1 : using similar s h hz h  1 x b 1 2 b 2h h  z 2  b xh-z b h  z  x x 2h
• 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 2Method 1 : using similar s h hz h  1 x b 1 2 b 2h h  z 2  b xh-z b h  z  x x 2h
• 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 bMethod 1 : using similar s h hz h  1 x b 1 2 b 2h h  z 2  b xh-z b h  z  x x 2h
• 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 bMethod 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 xh-z b h  z  x x 2h
• 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 bMethod 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
•  bh  z   ah  z A z   4   2h   2h   
•  bh  z   ah  z A z   4   2h   2h    abh  z  2  h2
•  bh  z   ah  z A z   4   2h   2h    abh  z  2  h2 abh  z  2 V  2  z h
•  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
•  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
•  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 
•  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
• (ii) A solid has a circular base, x 2  y 2  4 , and each cross section is a square perpendicular to the base.
• (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
• (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
• (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
• (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
• (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
• (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
• (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
• (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
• (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
• (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
• (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
• (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
• (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
• (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 -22y x 2y
• (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 22y x 2y
• (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
• (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
• (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
• (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
• (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
• (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
• (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