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Calculus, EXT1 C3 SM-Bank 1

The region enclosed by the semicircle  `y = sqrt(1 - x^2)`  and the `x`-axis is to be divided into two pieces by the line  `x = h`, when  `0 <= h <1`.
 


 

The two pieces are rotated about the `x`-axis to form solids of revolution. The value of `h` is chosen so that the volumes of the solids are in the ratio `2 : 1`.

Show that `h` satisfies the equation  `3h^3 - 9h + 2 = 0`.  (3 marks)

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`text(Show Worked Solution)`

Show Worked Solution

(i)   `text(Volume of smaller solid)`

`= pi int_h^1 (sqrt(1 – x^2))^2\ dx`

`= pi int_h^1 1 – x^2\ dx`

`= pi[x – (x^3)/3]_h^1`

`= pi[(1 – 1/3) – (h – (h^3)/3)]`

`= pi(2/3 – h + (h^3)/3)`

 
`text(S)text(ince smaller solid is)\ 1/3\ text(volume of sphere,)`

`pi(2/3 – h + (h^3)/3)` `= 1/3 xx 4/3 · pi · 1^3`
`(h^3)/3 – h + 2/3` `= 4/9`
`3h^3 – 9h + 6` `= 4`
`:. 3h^3 – 9h + 2` `= 0\ \ text(… as required)`

Calculus, EXT1* C3 2016 HSC 15a

The diagram shows two curves  `C_1` and `C_2.` The curve `C_1` is the semicircle  `x^2 + y^2 = 4, \ -2 <= x <= 0.` The curve `C_2` has equation  `x^2/9 + y^2/4 = 1, \ 0 <= x <= 3.`
 

hsc-2016-15a
 

An egg is modelled by rotating the curves about the `x`-axis to form a solid of revolution.

Find the exact value of the volume of the solid of revolution.  (4 marks)

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`(40 pi)/3\ text(u³)`

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`text(Consider)\ \ C_1,`

`V_1` `= pi int_-2^0 y^2\ dx`
  `= pi int_-2^0 4 – x^2\ dx`
  `= pi [4x – x^3/3]_-2^0`
  `= pi [0 – (-8 + 8/3)]`
  `= (16 pi)/3\ u³`

 

`text(Consider)\ \ C_2`

`x^2/9 + y^2/4` `= 1`
`y^2` `= 4 – (4x^2)/9`

 

`V_2` `= pi int_0^3 4 – (4x^2)/9\ dx`
  `= pi [4x – (4x^3)/27]_0^3`
  `= pi [(12 – (4 · 3^3)/27) – 0]`
  `= 8 pi\ text(u³)`

 

`text(Volume)` `= V_1 + V_2`
  `= (16 pi)/3 + 8 pi`
  `= (40 pi)/3\ text(u³)`

Calculus, EXT1* C3 2010 HSC 10b

The circle  `x^2 + y^2 = r^2`  has radius `r` and centre `O`. The circle meets the positive `x`-axis at `B`. The point `A` is on the interval `OB`. A vertical line through `A` meets the circle at `P`. Let  `theta = /_OPA`.
  

2010 10b1

  1. The shaded region bounded by the arc `PB` and the intervals `AB` and `AP` is rotated about the `x`-axis. Show that the volume, `V`, formed is given by
  2. `V = (pi r^3)/3 (2-3 sin theta + sin^3 theta)`   (3 marks)

  3. A container is in the shape of a hemisphere of radius `r` metres. The container is initially horizontal and full of water. The container is then tilted at an angle of `theta` to the horizontal so that some water spills out. 

  1. (1) Find `theta` so that the depth of water remaining is one half of the original depth.   (1 mark)

  2. (2) What fraction of the original volume is left in the container?   (2 marks)

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  1. `text(Proof)  text{(See Worked Solutions)}`
  2. (1) `theta = pi/6\ text(radians)`
  3. (2) `5/16`
Show Worked Solution

i.    `text(Show that)\ V = (pir^3)/3 (2-3sin theta + sin^3 theta)`

♦♦♦ Mean mark (i) 16%.
MARKER’S COMMENT: A common error was to integrate `r^2` to `1/3 r^3` instead of `r^2 x` (note that `r` is a constant).  
`sin theta` `= (OA)/r`
`:.\ OA` `= r sin theta`
`=> A` `= (r sin theta, 0),\ \ \ \ B = (r,0)`

 

`:.V` `= pi int_(rsintheta)^r y^2\ dx`
  `= pi int_(rsintheta)^r (r^2-x^2)\ dx\ \ \ \ text{(using}\ x^2+y^2=r^2text{)}`
  `= pi [r^2 x-(x^3)/3]_(rsintheta)^r`
  `= pi [(r^3-r^3/3)-(r^3 sin theta-(r^3 sin^3 theta)/3)]`
  `= pi ((2r^3)/3-r^3 sin theta + (r^3 sin^3 theta)/3)`
  `= (pir^3)/3 (2-3 sin theta + sin^3 theta)\ \ \ text(… as required)`

 

ii. (1) `text(Depth of water remaining) = 1/2 xx text(original depth:)`

♦♦♦ Part (ii) mean marks 3% and 2% for (ii)(1) and (ii)(2) respectively.
`r-r sin theta` `=1/2 r`
`r (1-sin theta)` `= 1/2 r`
`1-sin theta` `= 1/2`
`sin theta` `= 1/2`
`:.\ theta` `= pi/6\ text(radians)`

 

 MARKER’S COMMENT: Previous parts of a question should always be at the front and centre of a student’s mind and direct their strategy.
ii. (2)    `text(Original Volume)` `= 1/2 xx 4/3 pi r^3`
    `= 2/3 pi r^3`

 

`text(New Volume)` `= (pi r^3)/3 [2-3 sin(pi/6) + sin^3(pi/6)]`
  `= (pir^3)/3 [2-(3 xx 1/2) + (1/2)^3]`
  `= (pi r^3)/3 [2-3/2 + 1/8]`
  `= (pir^3)/3 (5/8)`
  `= (5 pi r^3)/24`

 

`:.\ text(Fraction of original volume left)`

`= ((5pir^3)/24)/(2/3 pi r^3)`

`= 5/24 xx 3/2`

`= 5/16`