When should you use the washer method ?

When there is a gap between the region being rotated and the line it is rotated around. The solid then has a hole through it, so each cross section is a washer rather than a disc.

The two curves swap places partway across the interval. What must you do?

Split the calculation into separate integrals, one for each stretch. Within each piece the same curve stays further from the axis, so the labels stay fixed.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 8: Applications of IntegrationVolumes of Revolution

Volumes of Revolution (College Board AP® Calculus BC): Flashcards

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FrontDisc Method Around the x-Axis
Define a solid of revolution.

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Cards in this collection (37)

Define a solid of revolution.
The solid formed when an area bounded by $y = f (x)$ and other boundaries is rotated $2 π$ radians about a line.
Its volume is called the volume of revolution.
The region bounded by $y = f (x)$ , the $x$ -axis and the lines $x = a$ and $x = b$ is rotated $2 π$ radians about the $x$ -axis. Its volume is
$V = π \int_{a}^{b}______d x$
The completed formula is $V = π \int_{a}^{b} y^{2} d x$ .
Each disc has radius $|y|$ , so its circular cross section has area $π y^{2}$ .
True or False?
A volume of revolution is a special case of finding a volume from areas of known cross sections.
True.
Each slice is a disc, so the cross-sectional area is $A (x) = π y^{2}$ .
Substituting that into $\int_{a}^{b} A (x) d x$ gives the volume of revolution formula.
A region bounded by the vertical lines $x = a$ and $x = b$ is rotated about the $x$ -axis. What shape are the ends of the solid?
Flat.
The ends come from rotating straight vertical lines, which sweep out flat discs.
Three-dimensional sketches can make them look rounded.
A question gives no limits for a volume of revolution. Where might they come from?
From the $y$ -axis, where $x = 0$ , or from an $x$ -intercept of the curve, where $f (x) = 0$ .
Why is a square root in $f (x)$ less of a problem than it looks?
The formula squares the function, so the square root is removed.
${(\sqrt{3 x^{2} + 2})}^{2}$ is simply $3 x^{2} + 2$ .
Region $R$ is bounded by $f (x) = \sqrt{3 x^{2} + 2}$ , the axes and $x = 3$ . Find the exact volume when $R$ is rotated about the $x$ -axis.
Square the function first.
$π \int_{0}^{3} (3 x^{2} + 2) d x = π {[x^{3} + 2 x]}_{0}^{3} = π (27 + 6) = 33 π$
The region bounded by $y = f (x)$ , the $y$ -axis and the lines $y = a$ and $y = b$ is rotated $2 π$ radians about the $y$ -axis. Its volume is
$V = π \int_{a}^{b}______d y$
The completed formula is $V = π \int_{a}^{b} x^{2} d y$ .
The discs now have radius $|x|$ , so $x$ must be written as a function of $y$ before integrating.
Which radius and which thickness go together for each of the two orientations?
Rotating about the $x$ -axis pairs radius $|y|$ with thickness $d x$ .
Rotating about the $y$ -axis pairs radius $|x|$ with thickness $d y$ .
True or False?
Rotating a region that lies to the left of the $y$ -axis gives a negative volume.
False.
The radius is squared, so $x^{2}$ is positive wherever the region lies.
A volume of revolution is never negative, unlike an area integral.
A rotation about the $y$ -axis uses $y = \arcsin (2 x + 1)$ . What is $x$ in terms of $y$ ?
It is $x = \frac{\sin y - 1}{2}$ .
Taking the sine of both sides gives $\sin y = 2 x + 1$ , which rearranges for $x$ .
A region is bounded by $y = \arcsin (2 x + 1)$ , the negative $x$ -axis and the positive $y$ -axis. What are the $y$ -limits?
They are $y = 0$ and $y = \frac{π}{2}$ .
The lower limit is the $x$ -axis itself, and the upper limit is the $y$ -intercept, where $x = 0$ gives $\arcsin 1$ .
Evaluate $π \int_{0}^{\frac{π}{2}} {(\frac{\sin y - 1}{2})}^{2} d y$ to 3 decimal places.
It is $0.280$ units cubed.
Expanding gives $\frac{π}{4} \int_{0}^{\frac{π}{2}} (\sin^{2} y - 2 \sin y + 1) d y$ , which a calculator evaluates directly.
The region between $y = f (x)$ and the line $y = k$ , from $x = a$ to $x = b$ , is rotated about the line $y = k$ . Its volume is
$V = π \int_{a}^{b} {(______)}^{2} d x$
The completed radius is $y - k$ .
Each disc reaches from the line $y = k$ out to the curve, so its radius is $|y - k|$ .
True or False?
When a region is rotated about the line $y = k$ , each disc has radius $|y|$ .
False.
The radius is measured from the axis of revolution, so it is $|y - k|$ .
Using $|y|$ would measure from the $x$ -axis instead.
How does the formula change for a rotation about the vertical line $x = k$ ?
It becomes $V = π \int_{a}^{b} {(x - k)}^{2} d y$ , with $x$ written as a function of $y$ and the limits taken as $y$ -values.
Why does rotating about $y = - 2$ make each radius larger, not smaller?
Because $y - k$ becomes $y - (- 2) = y + 2$ .
Subtracting a negative adds, which is right, since the axis has moved further away from the curve.
Region $R$ is bounded by $f (x) = 1 + e^{- x}$ and the lines $x = 1$ , $x = 3$ and $y = - 2$ . Set up the integral for a rotation about $y = - 2$ .
The radius is $(1 + e^{- x}) - (- 2) = 3 + e^{- x}$ , so the integral is $π \int_{1}^{3} {(3 + e^{- x})}^{2} d x$ , which comes to $62.753$ units cubed.
The curve $y = \ln (x - 3)$ is rotated about $x = 1$ between $y = - 2$ and $y = 1$ . Set up the integral.
Rearranging gives $x = 3 + e^{y}$ , so the radius is $(3 + e^{y}) - 1 = 2 + e^{y}$ .
The integral is $π \int_{- 2}^{1} {(2 + e^{y})}^{2} d y$ , giving $81.735$ units cubed.
When should you use the washer method?
When there is a gap between the region being rotated and the line it is rotated around.
The solid then has a hole through it, so each cross section is a washer rather than a disc.
The region between the curves $y_{1}$ and $y_{2}$ , from $x = a$ to $x = b$ , is rotated about the $x$ -axis. Its volume is
$V = π \int_{a}^{b} ({(y_{2})}^{2} -______) d x$
The completed formula is $V = π \int_{a}^{b} ({(y_{2})}^{2} - {(y_{1})}^{2}) d x$ .
A washer's area is the outer circle minus the inner circle, so the two squares are subtracted.
True or False?
${(y_{2})}^{2} - {(y_{1})}^{2}$ is the same as ${(y_{2} - y_{1})}^{2}$ .
False.
Expanding the second gives ${(y_{2})}^{2} - 2 y_{1} y_{2} + {(y_{1})}^{2}$ , which has an extra term and the wrong sign.
Square each radius separately, then subtract.
Which curve is $y_{2}$ in the washer formula?
The one further from the $x$ -axis, giving the outer radius.
The curve closer to the axis is $y_{1}$ and gives the inner radius.
What are the inner and outer radii of one washer?
The inner radius is $|y_{1}|$ and the outer radius is $|y_{2}|$ .
A washer of thickness $Δ x$ then has volume $π ({(y_{2})}^{2} - {(y_{1})}^{2}) \cdot Δ x$ .
The two curves swap places partway across the interval. What must you do?
Split the calculation into separate integrals, one for each stretch.
Within each piece the same curve stays further from the axis, so the labels stay fixed.
The region enclosed by $f (x) = \frac{1}{4} x^{2}$ and $g (x) = x$ is rotated about the $x$ -axis. Find the volume.
The line is further from the axis, so $y_{2} = x$ and $y_{1} = \frac{1}{4} x^{2}$ .
Between the intersections at $0$ and $4$ , $π \int_{0}^{4} (x^{2} - \frac{1}{16} x^{4}) d x = \frac{128 π}{15}$ .
The region between the curves $x_{1}$ and $x_{2}$ , from $y = a$ to $y = b$ , is rotated about the $y$ -axis. Its volume is
$V = π \int_{a}^{b} ({(x_{2})}^{2} - {(x_{1})}^{2})______$
The completed formula is $V = π \int_{a}^{b} ({(x_{2})}^{2} - {(x_{1})}^{2}) d y$ .
The thickness of each washer is a small change in $y$ , so the integration runs between $y$ -limits.
Rearranging $y = \frac{1}{4} x^{2}$ gives $x = \pm 2 \sqrt{y}$ . Which sign do you use?
Take $x = 2 \sqrt{y}$ .
The graph shows the relevant part of the curve lying where $x$ is positive.
True or False?
When a curve is rewritten as $x$ in terms of $y$ , both square roots must be kept.
False.
Only the branch matching the region being rotated is used.
Keeping both would describe a curve on either side of the $y$ -axis, which is not the region in the question.
The curves $y = \frac{1}{4} x^{2}$ and $y = x$ are to be rotated about the $y$ -axis. Write both as functions of $y$ and say which is $x_{2}$ .
They become $x = 2 \sqrt{y}$ and $x = y$ .
The line is closer to the $y$ -axis, so $x_{1} = y$ and $x_{2} = 2 \sqrt{y}$ .
Evaluate $π \int_{0}^{4} (4 y - y^{2}) d y$ .
It is $\frac{32 π}{3}$ .
Integrating gives $π {[2 y^{2} - \frac{1}{3} y^{3}]}_{0}^{4} = π (32 - \frac{64}{3})$ , which is about $33.510$ units cubed.
The region between the curves $y_{1}$ and $y_{2}$ , from $x = a$ to $x = b$ , is rotated about the horizontal line $y = k$ . Its volume is
$V = π \int_{a}^{b} ({(y_{2} - k)}^{2} - {(______)}^{2}) d x$
The completed radius is $y_{1} - k$ .
Both radii are measured from the line $y = k$ , so $k$ is subtracted from each curve before squaring.
True or False?
The curve labelled $y_{1}$ is always the lower of the two on the graph.
False.
$y_{1}$ is whichever curve is closer to the axis of revolution.
Rotating about a line above the region makes the upper curve the closer one, so the labels swap.
The region between $y = \frac{1}{4} x^{2}$ and $y = x$ on $[0, 4]$ is rotated about $y = 5$ . Which curve is $y_{1}$ ?
The line $y = x$ , because it lies above the parabola here and so is closer to $y = 5$ .
Rotating the same region about the $x$ -axis would make the parabola $y_{1}$ instead.
Set up and evaluate the integral for the region between $y = \frac{1}{4} x^{2}$ and $y = x$ rotated about $y = 5$ .
Expanding both squares gives $π \int_{0}^{4} ({(\frac{1}{4} x^{2} - 5)}^{2} - {(x - 5)}^{2}) d x = π \int_{0}^{4} (\frac{1}{16} x^{4} - \frac{7}{2} x^{2} + 10 x) d x = \frac{272 π}{15}$
How does the washer formula change for a rotation about the vertical line $x = k$ ?
It becomes $V = π \int_{a}^{b} ({(x_{2} - k)}^{2} - {(x_{1} - k)}^{2}) d y$ , with both curves written as functions of $y$ and $y$ -limits used.
The region between $x = 2 \sqrt{y}$ and $x = y$ is rotated about $x = - 2$ . Set up the integral.
The line $x = y$ is closer to $x = - 2$ , so it is $x_{1}$ .
The integral $π \int_{0}^{4} ({(2 \sqrt{y} + 2)}^{2} - {(y + 2)}^{2}) d y$ simplifies to $π \int_{0}^{4} (8 \sqrt{y} - y^{2}) d y = \frac{64 π}{3}$ .

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