What tells you that two volumes of revolution must be subtracted rather than added?

Subtract when the region being rotated does not touch the axis of rotation, because the solid then has a hole running through it and that hole has to be taken out of a larger solid. A region that does have the axis of rotation as one of its boundaries produces a solid piece, and needs no subtraction.

What assumption is normally made about the thickness of a container modelled as a solid of revolution, and when does it stop being safe?

The thickness of the material is normally ignored, because it is small compared with the size of the object. It stops being safe when the thickness is significant, and the object then has to be modelled as the difference between two solids of revolution, an outer one and an inner one.

True or False? When a vase is modelled as a solid of revolution, its handles are part of the volume found.

False. Only the main body of the object is modelled, so handles, and features such as the lip around the rim of a bucket, are left out. They are not produced by rotating the profile curve, so nothing in the integral corresponds to them.

A LevelFurther MathsEdexcelCore PureFlashcardsFurther CalculusVolumes of Revolution

Volumes of Revolution (Edexcel A Level Further Maths: Core Pure): Flashcards

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FrontVolumes of Revolution
Define a solid of revolution.

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Cards in this collection (18)

Define a solid of revolution.
A solid of revolution is the three-dimensional shape formed when a region bounded by a curve, and usually by lines or an axis as well, is rotated through $360 °$ about a straight line.
Its volume of revolution is the volume of that solid.
The region under $y = f (x)$ between $x = a$ and $x = b$ is rotated through $360 °$ about the $x$ -axis. Fill in the missing parts of the volume formula:
$V =______\int_{a}^{b}______d x$
The completed formula is:
$V = π \int_{a}^{b} y^{2} d x$
Note that it is $y^{2}$ and not $y^{3}$ , even though a volume is three-dimensional.
How is a volume of revolution built up from a sum of simpler solids?
Split the solid into thin discs perpendicular to the axis of rotation, each one a cylinder whose radius is the $y$ -value there and whose thickness is a small step $δ x$ .
Adding the discs only gives an approximation, so the exact volume is the limit as $δ x \to 0$ , and that limit is what the integral means.
The region bounded by $y = \sqrt{3 x^{2} + 2}$ , the coordinate axes and the line $x = 3$ is rotated through $360 °$ about the $x$ -axis. Find the exact volume.
Square $y$ first, giving $y^{2} = 3 x^{2} + 2$ , and read the limits $a = 0$ and $b = 3$ from the $y$ -axis and the line.
The volume is then:
$V = π \int_{0}^{3} (3 x^{2} + 2) d x = π {[x^{3} + 2 x]}_{0}^{3} = 33 π cubic units$
Before finding a volume of revolution about the $y$ -axis, what must you do to an equation given as $y = f (x)$ , and in what order?
Rearrange it into the form $x = g (y)$ first, which is finding the inverse function, and only then square to get $x^{2}$ .
Doing the two steps together is what leads to squaring $x$ twice, and the formula $V = π \int_{c}^{d} x^{2} d y$ needs everything written in terms of $y$ .
True or False?
For a volume of revolution about the $y$ -axis, the limits of the integral are $x$ -values.
False.
The integration is with respect to $y$ , so the limits $c$ and $d$ are $y$ -values read from the boundaries of the region, not the $x$ -values of that same region.
The region bounded by $y = \arcsin (2 x + 1)$ and the coordinate axes is rotated about the $y$ -axis. What is the integrand, and what are the limits?
Rearranging gives $\sin y = 2 x + 1$ , so $x = \frac{1}{2} (\sin y - 1)$ and the integrand is $x^{2} = \frac{1}{4} {(\sin y - 1)}^{2}$ .
The region reaches from the $x$ -axis, where $y = 0$ , up to the $y$ -axis, where $y = \arcsin 1 = \frac{π}{2}$ , so those are the limits.
A curve has parametric equations $x = f (t)$ and $y = g (t)$ . Fill in the missing factor in the volume of revolution about the $x$ -axis:
$V = π \int_{t_{1}}^{t_{2}} {[g (t)]}^{2}______d t$
The completed formula is:
$V = π \int_{t_{1}}^{t_{2}} {[g (t)]}^{2} f' (t) d t$
The extra factor appears because $d x = f' (t) d t$ , which is how $d x$ is converted into $d t$ .
True or False?
A volume of revolution can be found from parametric equations without ever writing $y$ in terms of $x$ .
True.
Every part of the integral is rewritten in terms of the parameter instead, including the limits, which become $t$ -values.
That freedom matters because for many pairs of parametric equations eliminating $t$ is inconvenient or impossible.
A curve has $x = \sec t$ and $y = \sqrt{cosec t}$ . When this curve is rotated about the $x$ -axis, what does the integrand simplify to?
Multiplying $y^{2} = cosec t$ by $\frac{d x}{d t} = \sec t \tan t$ gives $\frac{1}{\sin t} \times \frac{1}{\cos t} \times \frac{\sin t}{\cos t}$ , in which the sines cancel.
The integrand is therefore just $\sec^{2} t$ , which integrates to $\tan t$ .
What tells you that two volumes of revolution must be subtracted rather than added?
Subtract when the region being rotated does not touch the axis of rotation, because the solid then has a hole running through it and that hole has to be taken out of a larger solid.
A region that does have the axis of rotation as one of its boundaries produces a solid piece, and needs no subtraction.
A rectangle bounded by $x = a$ , $x = b$ , $y = c$ and $y = d$ , with $0 < c < d$ , is rotated about the $x$ -axis to form a ring-shaped prism. Fill in its volume:
$V = π \int_{a}^{b}______d x - π \int_{a}^{b}______d x$
The completed formula is:
$V = π \int_{a}^{b} d^{2} d x - π \int_{a}^{b} c^{2} d x$
The outer edge $y = d$ sweeps out a cylinder, and the inner edge $y = c$ sweeps out the cylinder that is removed from it.
When two volumes of revolution are subtracted and both integrals have the same limits, how can they be combined into one?
They can be written as the single integral $V = π \int_{a}^{b} ({(y_{1})}^{2} - {(y_{2})}^{2}) d x$ , since the limits and the factor $π$ are shared.
This does not work for an addition problem, where the two parts cover different intervals and forcing the same limits would count part of the volume twice.
True or False?
A rotation about the $x$ -axis can need two volumes to be added together, not just subtracted.
True.
When different curves form the upper boundary of the region over different stretches of the $x$ -axis, each stretch generates its own solid and the two solids are added.
For the region under $y = 2^{x}$ from $x = 0$ to $x = 1$ and under $y = 4 - 2^{x}$ from $x = 1$ to $x = 2$ , the volume is the sum of two separate integrals.
What assumption is normally made about the thickness of a container modelled as a solid of revolution, and when does it stop being safe?
The thickness of the material is normally ignored, because it is small compared with the size of the object.
It stops being safe when the thickness is significant, and the object then has to be modelled as the difference between two solids of revolution, an outer one and an inner one.
An object is described as a bowl standing upright on its base. Which axis is it modelled as a rotation about, and why?
About the $y$ -axis, because the axis of symmetry of a bowl standing upright runs vertically through it.
The curve giving the bowl's profile is rotated about that vertical axis, so the integration is with respect to $y$ and the limits are heights.
True or False?
When a vase is modelled as a solid of revolution, its handles are part of the volume found.
False.
Only the main body of the object is modelled, so handles, and features such as the lip around the rim of a bucket, are left out.
They are not produced by rotating the profile curve, so nothing in the integral corresponds to them.
A container is modelled as a solid of revolution, and its volume is $40 π {cm}^{3}$ . What is its capacity in litres?
Divide by $1000$ , since $1000 {cm}^{3}$ is equal to $1$ litre.
Here that gives $\frac{40 π}{1000} = 0.126$ litres to three significant figures.

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