smc-1039-40-Other Graphs

Calculus, EXT1 C3 2025 HSC 12b

Consider the region bounded by the hyperbola \(y=\dfrac{1}{x}\), the \(y\)-axis and the lines \(y=1\) and \(y=a\) for \(a>1\).

Find the volume of the solid of revolution formed when the region is rotated about the \(y\)-axis. (2 marks)

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\(\pi\left(1-\dfrac{1}{a}\right)\ \text{u}^3\)

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\(y=\dfrac{1}{x} \ \Rightarrow \ x^2=\dfrac{1}{y^2}\)

\(V\)	\(=\pi \displaystyle \int_1^a x^2\, dy\)
	\(=\pi \displaystyle \int_1^a y^{-2}\, d y\)
	\(=-\pi\left[y^{-1}\right]_1^a\)
	\(=-\pi\left(\dfrac{1}{a}-1\right)\)
	\(=\pi\left(1-\dfrac{1}{a}\right)\ \text{u}^3\)

Calculus, EXT1 C3 2023 HSC 12e

The region, \(R\), bounded by the hyperbola \(y=\dfrac{60}{x+5}\), the line \(x=10\) and the coordinate axes is shown.

Find the volume of the solid of revolution formed when the region \(R\) is rotated about the \(y\)-axis. Leave your answer in exact form. (4 marks)

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\(V=1200 \pi-600\pi\ \ln3 \ \ \text{u}^3\)

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\(y=\dfrac{60}{x+5}\ \ \Rightarrow\ \ x=\dfrac{60}{y}-5 \)

♦ Mean mark 49%.

\(V\)	\(=\pi \displaystyle \int_4^{12} x^2\ dy + \pi r^2h\)
	\(=\pi \displaystyle \int_4^{12} \Big{(} \dfrac{60}{y}-5 \Big{)}^2 \ dy + \pi \times 10^2 \times 4 \)
	\(=\ 25\pi \displaystyle \int_4^{12} \Big{(} \dfrac{12}{y}-1 \Big{)}^2 \ dy + 400\pi \)
	\(=\ 25\pi \displaystyle \int_4^{12} \Big{(} \dfrac{144}{y^2}-\dfrac{24}{y} + 1 \Big{)} \ dy + 400\pi \)
	\(=\ 25\pi \Big{[} \dfrac{-144}{y}- 24 \ln y + y\Big{]}_4^{12} + 400\pi \)
	\(=\ 25\pi \Big{[} (-12-24 \ln 12 +12)-(-36-24 \ln 4+4)\Big{]} + 400\pi \)
	\(=\ 25\pi (24 \ln 4-24 \ln 12+32) + 400\pi \)
	\(=600 \pi\ \ln(3^{-1}) + 800 \pi + 400 \pi \)
	\(=1200 \pi-600\pi\ \ln3 \ \ \text{u}^3\)

Calculus, EXT1 C3 SM-Bank 3

Find the volume of the solid of revolution formed when the graph of `y = sqrt((1 + 2x)/(1 + x^2))` is rotated about the `x`-axis over the interval `[0,1]`. (3 marks)

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`pi(pi/4 + ln2)\ \ text(u³)`

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`V`	`= pi int_0^1 (1 + 2x)/(1 + x^2)\ dx`
	`= pi int_0^1 1/(1 + x^2)\ dx + pi int_0^1 (2x)/(1 + x^2)\ dx`
	`= pi [tan^(−1)(x)]_0^1 + pi [ln(1 + x^2)]_0^1`
	`= pi(tan^(−1)1 – tan^(−1)0) + pi(ln2 – ln1)`
	`= pi(pi/4) + pi(ln2)`
	`= pi(pi/4 + ln2)\ \ text(u³)`

Calculus, EXT1* C3 2017 HSC 12b

The diagram shows the region bounded by `y = sqrt (16 - 4x^2)` and the `x`-axis.

The region is rotated about the `x`-axis to form a solid.

Find the exact volume of the solid formed. (3 marks)

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`(128 pi)/3\ text(u³)`

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`y`	`= sqrt (16 – 4x^2)`
`V`	`= pi int_(-2)^2 y^2\ dx`
	`= 2 pi int_0^2 16 – 4x^2\ dx`
	`= 2 pi [16x – 4/3 x^3]_0^2`
	`= 2 pi [(16 ⋅ 2 – 4/3 2^3)-0]`
	`= 2 pi (64/3)`
	`= (128 pi)/3\ text(u³)`

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.`

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 2007 HSC 3a

Find the volume of the solid of revolution formed when the region bounded by the curve `y = 1/(sqrt(9 + x^2))`, the `x`-axis, the `y`-axis and the line `x = 3`, is rotated about the `x`-axis. (3 marks)

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`(pi^2)/(12)\ \ text(u³)`

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

`:.\ text(Volume)`	`= pi int_0^3 y^2\ dx`
	`= pi int_0^3 (1/(sqrt(9 + x^2)))^2\ dx`
	`= pi int_0^3 1/(9 + x^2)\ dx`
	`= pi [1/3 tan^(−1)\ x/3]_0^3`
	`= pi [1/3 tan^(−1)\ 1 − 1/3 tan^(−1)\ 0]`
	`= pi [(1/3 xx pi/4) − 0]`
	`= (pi^2)/(12)\ \ text(u³)`

Calculus, EXT1* C3 2015 HSC 16b

A bowl is formed by rotating the curve `y = 8 log_e (x - 1)` about the `y`-axis for `0 <= y <= 6.`

Find the volume of the bowl. Give your answer correct to 1 decimal place. (3 marks)

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`118.7\ text(u³)`

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♦ Mean mark 38%.

`y`	`=8 log_e (x – 1)`
`y/8`	`=log_e (x – 1)`
`e^(y/8)`	`=x – 1`
`x`	`=e^(y/8) + 1`
`x^2`	`=(e^(y/8) + 1)^2`
	`=(e^(y/8))^2 + 2e^(y/8) + 1`
	`=e^(y/4) + 2e^(y/8) + 1`

`V`	`= pi int_0^6 x^2\ dy`
	`= pi int_0^6 (e^(y/4) + 2e^(y/8) + 1)\ dy`
	`= pi [4e^(y/4) + 16e^(y/8) + y]_0^6`
	`= pi [(4e^(6/4) + 16e^(6/8) + 6) – (4e^0 + 16e^0 + 0)]`
	`= pi [(4e^1.5 + 16e^0.75 + 6) – (4 + 16)]`
	`= pi [4e^1.5 + 16e^0.75 – 14]`
	`= 118.748…`
	`= 118.7\ \ text{(to 1 d.p.)}`

`:.\ text(Volume of the bowl is 118.7 u³.)`

Calculus, EXT1* C3 2008 HSC 6c

The graph of `y = 5/(x - 2)` is shown below.

The shaded region in the diagram is bounded by the curve `y = 5/(x - 2)`, the `x`-axis and the lines `x = 3` and `x = 6`.

Find the volume of the solid of revolution formed when the shaded region is rotated about the `x`-axis. (3 marks)

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`(75pi)/4\ text(u³)`

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`y`	`= 5/(x – 2)`
`V`	`= pi int_3^6 y^2\ dx`
	`= pi int_3^6 (5/(x – 2))^2\ dx`
	`= 25 pi int_3^6 1/((x – 2)^2)\ dx`
	`= 25 pi [(-1)/(x – 2)]_3^6`
	`= 25 pi [-1/4 – (-1)]`
	`=25 pi [3/4]`
	`= (75pi)/4\ text(u³)`

Calculus, EXT1* C3 2014 HSC 14c

The region bounded by the curve `y = 1 + sqrtx` and the `x`-axis between `x = 0` and `x = 4` is rotated about the `x`-axis to form a solid.

Find the volume of the solid. (3 marks)

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`(68 pi)/3\ text(u³)`

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`y = 1 + sqrtx`

`V`	`= pi int_0^4 y^2\ dx`
	`= pi int_0^4 (1 + sqrtx)^2\ dx`
	`= pi int_0^4 (1 + 2 sqrtx + x)\ dx`
	`= pi [x + 4/3 x^(3/2) + 1/2 x^2]_0^4`
	`= pi [(4 + 4/3 xx 4^(3/2) + 1/2 xx 4^2)\ – 0]`
	`= pi (4 + 32/3 + 8)`
	`= (68 pi)/3\ text(u³)`

Calculus, EXT1* C3 2012 HSC 14b

The diagram shows the region bounded by `y = 3/((x+2)^2)`, the `x`-axis, the `y`-axis, and the line `x = 1`.

The region is rotated about the `x`-axis to form a solid.

Find the volume of the solid. (3 marks)

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`(19pi)/72\ text(u³)`

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MARKER’S COMMENT: Most students omitted the `dx` when stating the definite integral in this question!

`V`	`= pi int_0^1 y^2\ dx`
	`= pi int_0^1 (3/((x+2)^2))^2\ dx`
	`= pi int_0^1 9/((x+2)^4)\ dx`
	`= 9pi int_0^1 (x + 2)^-4\ dx`
	`= 9pi [-1/3 (x + 2)^-3]_0^1`
	`= 9pi [(-1/3 xx 1/(3^3))\ – (-1/3 xx 1/(2^3))]`
	`= 9 pi [-1/81 + 1/24]`
	`= 9 pi (19/648)`
	`= (19pi)/72\ text(u³)`

`:.\ text(Volume of the solid is)\ (19pi)/72\ text(u³)`.