MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: volumes by slicing, disks and washers (MATH 141)

Q: How do I know which curve gives the outer radius in the washer method?

Measure from the axis, not from the x-axis. The outer radius goes to the boundary of the slice that is farthest from the axis of rotation. When the axis lies below the region, that is the upper curve; when the axis lies above the region, it is the lower curve. Test one point: the outer radius squared minus the inner radius squared must be positive.

Q: When do I integrate in x and when in y with disks and washers?

The slice is always perpendicular to the axis of rotation. A horizontal axis, such as the x-axis or a line y equals a constant, gives vertical slices and an integral in x. A vertical axis gives horizontal slices and an integral in y, and every curve must first be solved for x in terms of y, with the bounds given as y values.

Q: Why is the washer area pi times R squared minus r squared and not pi times R minus r squared?

A washer is a big disk with a hole: its area is the area of the big disk minus the area of the hole, pi R squared minus pi r squared. The other expression is the area of a disk whose radius is the thickness of the region, and it no longer depends on how far the region is from the axis, so it describes a different, smaller solid.

Q: How do I find the volume of a solid with known cross-sections?

Find the area of one cross-section as a function of the position of the cut, then integrate that area along the axis. The side of the cross-section is usually the whole chord of the base at that position, from its lower edge to its upper edge. A square on a chord c has area c squared, a semicircle on it has area pi c squared over eight.

This sheet is not a summary of section 6.2 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on disk and washer volumes in MATH 141 at McGill University, and which precise gesture avoids each loss.

Almost every lost mark in this chapter is lost in the SET-UP, before any integration: the wrong variable, a radius measured from the wrong line, two radii subtracted before being squared. Every number below is exact, as on the exam, and checked by an independent computation.

The thread of the chapter

Slice PERPENDICULAR to the axis, measure every radius FROM THE AXIS, and square each radius before subtracting: a washer is $\pi(R^2 - r^2)$ , never $\pi(R - r)^2$ , and when the axis sits above the region the outer radius goes to the LOWER curve.

This chapter is part of MATH 141, Calculus 2 (McGill)

The essentials

One slice, two radii, one square

• Cut the solid PERPENDICULAR to the axis of rotation. The slice has thickness $dx$ if the axis is horizontal, $dy$ if it is vertical: the variable always runs ALONG the axis.
• If the region touches the axis along the whole slice, the slice is a disk: $dV = \pi R^2\,dx$ . If there is a gap, it is a washer: $dV = \pi(R^2 - r^2)\,dx$ .
• $R$ and $r$ are DISTANCES FROM THE AXIS to the far and near ends of the slice. About $y = k$ : $|k - y|$ . About $x = k$ : $|k - x|$ .
• When the axis lies ABOVE the region (or to its right), the far end is the lower (or left) boundary: the curve on top gives the INNER radius.
• Square each radius, then subtract. $R^2 - r^2 = (R - r)(R + r)$ , and the factor $R + r$ is where the distance to the axis lives.

Region between

y = x^2

and

y = 2

turning about

y = 3

: the outer radius

R = 3 - x^2

reaches the LOWER curve, the inner radius

r = 1

reaches the line on top.

Write the three lines $R = \dots$ , $r = \dots$ , variable and bounds $\dots$ BEFORE the integral. On most MATH 141 marking schemes they carry more marks than the evaluation.

Beyond rotation: the area of the slice is all you need

• For any solid, $V = \int_a^b A(x)\,dx$ where $A(x)$ is the area of the cross-section perpendicular to the $x$ -axis. A disk volume is the case $A = \pi R^2$ .
• Read the side of the cross-section on the BASE: the whole chord, from the lower edge of the base to the upper edge, unless the statement says otherwise.
• Shape constants on a chord $c$ : square $c^2$ , isosceles right triangle with hypotenuse $c$ : $\frac{c^2}{4}$ , with leg $c$ : $\frac{c^2}{2}$ , equilateral: $\frac{\sqrt{3}}{4}c^2$ , semicircle: $\frac{\pi c^2}{8}$ .
• Two solids with the same cross-sectional area at every $x$ have the same volume (Cavalieri): the volume depends on $A(x)$ , not on where the slice sits.
• The same solid sliced in two directions gives the same volume: the second slicing is a check.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Which variable, which radii, by the position of the axis

The region lies between an upper curve $y = f(x)$ and a lower curve $y = g(x)$ for $a \le x \le b$ ; for a vertical axis, between a left curve $x = u(y)$ and a right curve $x = w(y)$ . The examples use the region between $y = x$ and $y = x^2$ on $[0, 1]$ , that is $y \le x \le \sqrt{y}$ for $0 \le y \le 1$ .

Axis	Slices and variable	Radii
$x$ -axis	vertical, $dx$	$R = f$ , $r = g$
Example: $\pi \int_0^1 (x^2 - x^4)\,dx = \frac{2\pi}{15}$
$y = k$ above, $k \ge f$	vertical, $dx$	$R = k - g$ , $r = k - f$
Example: About $y = 1$ : $\pi \int_0^1 \left((1 - x^2)^2 - (1 - x)^2\right)dx = \frac{\pi}{5}$
$y = k$ below, $k \le g$	vertical, $dx$	$R = f - k$ , $r = g - k$
Example: About $y = -1$ : $\pi \int_0^1 \left((x + 1)^2 - (x^2 + 1)^2\right)dx = \frac{7\pi}{15}$
$y$ -axis	horizontal, $dy$	$R = w$ , $r = u$
Example: $\pi \int_0^1 (y - y^2)\,dy = \frac{\pi}{6}$
$x = k$ to the right, $k \ge w$	horizontal, $dy$	$R = k - u$ , $r = k - w$
Example: About $x = 1$ : $\pi \int_0^1 \left((1 - y)^2 - (1 - \sqrt{y})^2\right)dy = \frac{\pi}{6}$
any axis	washer	$\pi(R - r)^2$ no such rule
Example: About the $x$ -axis: $\pi \int_0^1 (x - x^2)^2\,dx = \frac{\pi}{30}$ , not $\frac{2\pi}{15}$ . What to do: Square each radius, then subtract: $\pi(R^2 - r^2)$ . If both share a term, factor $(R - r)(R + r)$ .

The lower curve gives $R$ in the second line and $r$ in the third: the curve on top has no privilege, only the distance to the axis counts.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Subtracting the radii before squaring

the set-up mark and the answer, often 3 to 4 marks out of 5

What not to write

“About the $x$ -axis, between $y = x$ and $y = x^2$ : $V = \pi \int_0^1 (x - x^2)^2\,dx = \frac{\pi}{30}$ .”

What to write

“ $R = x$ , $r = x^2$ , so $V = \pi \int_0^1 (x^2 - x^4)\,dx = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}$ .”

With

R = 3

and

r = 2

the washer has area

5\pi

; the disk of radius

R - r = 1

has area

\pi

. Same thickness, five times less material.

Why: A washer is a big disk minus a hole: $\pi R^2 - \pi r^2$ . The expression $\pi(R - r)^2$ is a disk of radius $R - r$ , the thickness of the region, and it forgets the distance to the axis entirely: it computes a different solid.

2. Taking the height of a curve for its radius about y = k

the whole question: the integral written is the volume about the x-axis

What not to write

“The region under $y = e^x$ , $0 \le x \le 1$ , turns about $y = -2$ : $V = \pi \int_0^1 e^{2x}\,dx$ .”

What to write

“ $R = e^x + 2$ , $r = 2$ : $V = \pi \int_0^1 (e^{2x} + 4e^x)\,dx = \pi\left(\frac{e^2 - 1}{2} + 4(e - 1)\right)$ .”

Why: $e^x$ is the distance from the curve to $y = 0$ . The axis is $y = -2$ , so every distance grows by $2$ , and the region, which sits ON the $x$ -axis, is now separated from the axis by a gap of $2$ : the slices are washers, not disks.

3. Keeping the top curve as the outer radius when the axis is above

2 marks for the radii, even when the absolute value rescues the number

What not to write

“Between $y = x$ and $y = x^2$ about $y = 1$ : $R = 1 - x$ , $r = 1 - x^2$ , so $V = -\frac{\pi}{5}$ , I take $\frac{\pi}{5}$ .”

What to write

“The axis $y = 1$ is above, so the far boundary is the LOWER curve: $R = 1 - x^2$ , $r = 1 - x$ , $V = \pi \int_0^1 (2x - 3x^2 + x^4)\,dx = \frac{\pi}{5}$ .”

Why: A negative volume is the free alarm of the washer method: it means $r > R$ on the interval. The fix is to redo the radii from the axis, not to drop the sign, because the marker reads the radii, not only the final number.

4. Integrating in x with washers about a vertical axis

the whole question: the integral in x is the solid about the x-axis

What not to write

“Between $y = x$ and $y = x^2$ about the $y$ -axis: $V = \pi \int_0^1 (x^2 - x^4)\,dx = \frac{2\pi}{15}$ .”

What to write

“Horizontal slices: $x$ runs from $y$ to $\sqrt{y}$ , so $R = \sqrt{y}$ , $r = y$ and $V = \pi \int_0^1 (y - y^2)\,dy = \frac{\pi}{6}$ .”

Why: With disks and washers the slice is perpendicular to the axis, so a vertical axis means horizontal slices and an integral in $y$ . The curves must be solved for $x$ first. An integral in $x$ about a vertical axis exists, but with cylindrical shells and a different integrand: that is the next chapter.

5. Carrying the x limits into a dy integral

1 to 2 marks, and an answer with a 2 to the power five thirds in it

What not to write

“Region bounded by $y = x^3$ , $y = 8$ and the $y$ -axis, about the $y$ -axis: $V = \pi \int_0^2 y^{2/3}\,dy$ .”

What to write

“ $x = y^{1/3}$ and $y$ runs from $0$ to $8$ : $V = \pi \int_0^8 y^{2/3}\,dy = \frac{3\pi}{5} \cdot 32 = \frac{96\pi}{5}$ .”

Why: The bounds belong to the variable of integration. The region goes from $x = 0$ to $x = 2$ , but from $y = 0$ to $y = 8$ , and the integral is in $y$ . An exact answer containing $2^{5/3}$ on a no-calculator exam is itself a warning.

6. Forgetting to square, and choosing the technique for f instead of its square

3 marks: the set-up is wrong, so nothing after it counts

What not to write

“Under $y = \frac{1}{1 + x}$ on $[0, 1]$ about the $x$ -axis: $V = \pi \int_0^1 \frac{dx}{1 + x} = \pi\ln 2$ .”

What to write

“The slice is a disk of radius $\frac{1}{1 + x}$ : $V = \pi \int_0^1 \frac{dx}{(1 + x)^2} = \pi\left[-\frac{1}{1 + x}\right]_0^1 = \frac{\pi}{2}$ .”

Why: $\pi \int f\,dx$ is $\pi$ times an AREA, in units squared. The integrand of a disk volume is $f^2$ , and it is $f^2$ whose technique you choose: a logarithm for $f$ becomes a power for $f^2$ , an arcsine for $f$ becomes partial fractions.

7. Taking the half-chord as the side of a cross-section

2 marks, and an answer off by a factor of 4

What not to write

“Base $x^2 + y^2 \le 9$ , square cross-sections perpendicular to the $x$ -axis: side $\sqrt{9 - x^2}$ , so $V = \int_{-3}^{3}(9 - x^2)\,dx = 36$ .”

What to write

“The square stands on the whole chord, from $-\sqrt{9 - x^2}$ to $\sqrt{9 - x^2}$ : side $2\sqrt{9 - x^2}$ , $V = \int_{-3}^{3}4(9 - x^2)\,dx = 144$ .”

Why: $\sqrt{9 - x^2}$ is the radius of the rotation problems, the distance from the axis to the circle. Here nothing turns: the slice stands on the full segment the base cuts. A factor $4$ error means a factor $2$ on a length, which squared.

8. Treating a bowl as a cylinder

the whole question

What not to write

“The hemispherical basin of radius $3$ is filled to half its depth, so it holds half its volume, $9\pi$ .”

What to write

“ $V(h) = \pi \int_0^h (6y - y^2)\,dy = \pi\left(3h^2 - \frac{h^3}{3}\right)$ , and $V\left(\frac{3}{2}\right) = \frac{45\pi}{8}$ , that is $\frac{5}{16}$ of $18\pi$ .”

Why: Volume is proportional to depth only when every slice has the same area, which is a cylinder. In a bowl the low slices are small: set up the slices, never scale the depth.

Which method to choose

Which slice, by the POSITION of the axis and the SHAPE of the region

Draw the region and the axis first, then read the branches from the top

The same region under

y = x^2

: about the

x

-axis the slice is vertical and the integral is in

x

; about the

y

-axis it is horizontal and the integral is in

y

If the axis is horizontal ( $x$ -axis or $y = k$ ) → vertical slices, integrate in $x$ , radii are vertical distances $|k - y|$

Example: about $y = -1$ : $R = f + 1$ , $r = g + 1$
If the axis is vertical and every boundary can be solved for $x$ → horizontal slices, integrate in $y$ , rewrite each curve as $x = g(y)$ and find the $y$ bounds

Example: $y = \ln x$ becomes $x = e^y$ , $0 \le y \le 1$
If the axis is vertical and a boundary cannot be solved for $x$ → washers are stuck: this is exactly the case for cylindrical shells, the next chapter

Example: $y = x^3 + x$ about the $y$ -axis
If the region touches the axis along the whole slice → disks, $r = 0$

Example: under $\sin x$ about the $x$ -axis: $\pi \int_0^\pi \sin^2 x\,dx = \frac{\pi^2}{2}$
If there is a gap between the region and the axis → washers, $r > 0$ , and the solid has a tunnel

Example: under $4 - x^2$ about $y = -1$ : $r = 1$
If nothing turns, the cross-sections are described → find $A(x)$ from the chord of the base, then $\int A(x)\,dx$

Example: squares on the chords of $x^2 + y^2 \le 9$ : $A(x) = 4(9 - x^2)$

The sketch is not decoration: on the exam it is where you read which boundary is far from the axis. A set-up written without a sketch is where the swapped radii come from.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Setting up a volume by washers

When to use it: Any question that asks for the volume of the solid obtained by rotating a region about a line

1 Sketch the region and the axis. Find the intersection points of the boundaries.
2 Name the slice: perpendicular to the axis, so vertical ( $dx$ ) or horizontal ( $dy$ ). If $dy$ , rewrite every boundary as $x$ in terms of $y$ and give the $y$ bounds.
3 Write $R$ and $r$ as distances from the axis, in the form axis minus curve or curve minus axis, BEFORE simplifying. Check $R \ge r \ge 0$ at one interior point.
4 Write $V = \pi \int (R^2 - r^2)$ with the bounds of the chosen variable, then simplify $R^2 - r^2$ , factoring when both radii share a term.
5 Evaluate exactly, and check the sign, the units and a bounding cylinder.

Concluding sentence

“The axis $y = 1$ lies above the region, so the slices are vertical washers with $R(x) = 1 - x^2$ and $r(x) = 1 - x$ for $0 \le x \le 1$ , and $V = \pi \int_0^1 \left[(1 - x^2)^2 - (1 - x)^2\right]dx = \frac{\pi}{5}$ .”

The trap: Simplifying $R$ and $r$ in your head before writing them: the sign that tells which is outer disappears, and with it the check.

Marking: Typically 1 mark for the sketch and the bounds, 2 for the two radii, 1 for the integral written correctly, 1 for the evaluation. The set-up is worth more than the arithmetic.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The integrand is never negative

Evaluate $R^2 - r^2$ at one interior point. A negative value means the radii are swapped; a negative final volume means the same thing, too late.

About $y = 1$ between $y = x$ and $y = x^2$ , at $x = \frac{1}{2}$ : $\left(\frac{3}{4}\right)^2 - \left(\frac{1}{2}\right)^2 = \frac{5}{16} > 0$ .
The bounding cylinder

The solid fits in the cylinder of radius $\max R$ over the same length. A disk or washer volume larger than that cylinder is wrong.

Under $\sin x$ on $[0, \pi]$ : $\frac{\pi^2}{2}$ against a cylinder of $\pi \cdot 1^2 \cdot \pi = \pi^2$ .
A farther axis gives a larger volume

Moving the axis away from the region can only widen every washer. If two answers of the same question go the other way, one set-up is wrong.

The arch $y = 4 - x^2$ : $\frac{512\pi}{15}$ about the $x$ -axis, $\frac{832\pi}{15}$ about $y = -1$ .
Units and familiar solids

A volume is in cubic units: pi times an area is not one. When the solid is a cone, a sphere, a cap or a pyramid, the known formula is a free check.

Pyramid with square base $6$ and height $4$ : $\int_0^4 36\left(1 - \frac{y}{4}\right)^2dy = 48 = \frac{1}{3} \cdot 36 \cdot 4$ .

The typical problem, taken apart

Washers about a vertical line to the left of the region

The region $\mathcal{R}$ between $y = x^3$ and $y = x$ in the first quadrant turns about the line $x = -1$ . Find the volume of the solid.

No calculator. Justify every radius as on a MATH 141 final.

The region between

y = x^3

and

y = x

for

0 \le x \le 1

, the vertical axis

x = -1

to its left, and one horizontal slice.

Step 1

Intersections: $x^3 = x$ gives $x = 0$ or $x = 1$ in the first quadrant, so the corners are $(0, 0)$ and $(1, 1)$ . At $x = \frac{1}{2}$ , $x = \frac{1}{2} > \frac{1}{8} = x^3$ : the line is on top.

Why

The bounds and the relative position of the curves come first; everything else depends on them. One test point between the roots is a full justification.

Step 2

The axis $x = -1$ is vertical, so the slices are horizontal, $0 \le y \le 1$ . Rewrite: $y = x$ becomes $x = y$ , and $y = x^3$ becomes $x = y^{1/3}$ . At $y = \frac{1}{8}$ : $y = \frac{1}{8} < \frac{1}{2} = y^{1/3}$ , so the cube root is the RIGHT boundary.

Why

On top in $x$ does not mean on the right in $y$ : the line was the upper curve, it is the LEFT boundary of the horizontal slice. Only the new test point decides.

Step 3

Distances to $x = -1$ , which lies to the left: $R(y) = y^{1/3} - (-1) = 1 + y^{1/3}$ and $r(y) = y - (-1) = 1 + y$ .

Why

Written as curve minus axis before simplifying, each radius shows the $+1$ that the distance to the axis adds. Forgetting it gives the volume about the $y$ -axis, a different solid.

Step 4

$R^2 - r^2 = (1 + 2y^{1/3} + y^{2/3}) - (1 + 2y + y^2) = 2y^{1/3} + y^{2/3} - 2y - y^2$ .

Why

Here the two radii share only the $1$ , so expanding is as quick as factoring. The $1$ 's cancel: that cancellation is the sign that both radii were measured from the same axis.

Step 5

$V = \pi \int_0^1 \left(2y^{1/3} + y^{2/3} - 2y - y^2\right)dy = \pi\left(\frac{3}{2} + \frac{3}{5} - 1 - \frac{1}{3}\right) = \frac{23\pi}{30}$ .

Why

Each power integrates by the table: $\int_0^1 y^{p}\,dy = \frac{1}{p + 1}$ , so $2 \cdot \frac{3}{4}$ , $\frac{3}{5}$ , $2 \cdot \frac{1}{2}$ and $\frac{1}{3}$ . The common denominator $30$ closes the computation.

The conclusion, written out

“The solid obtained by rotating $\mathcal{R}$ about $x = -1$ has volume $V = \pi \int_0^1 \left[(1 + y^{1/3})^2 - (1 + y)^2\right]dy = \frac{23\pi}{30}$ .”

The classic mistake on this problem: Taking the line as outer radius because it is on top, which gives $-\frac{23\pi}{30}$ ; or forgetting the $+1$ , which gives $\pi \int_0^1 (y^{2/3} - y^2)\,dy = \frac{4\pi}{15}$ , the volume about the $y$ -axis.

Learn by heart

• Slice PERPENDICULAR to the axis: horizontal axis $dx$ , vertical axis $dy$ with every curve as $x = g(y)$ .
• Disk: $\pi R^2$ . Washer: $\pi(R^2 - r^2)$ , never $\pi(R - r)^2$ .
• Radii are distances FROM THE AXIS: $|k - y|$ about $y = k$ , $|k - x|$ about $x = k$ . Axis above: $R$ to the lower curve.
• Known cross-sections: $V = \int A(x)\,dx$ , side read on the WHOLE chord of the base.
• The integrand is $f^2$ : choose the technique for $f^2$ , not for $f$ .
• Negative volume: radii swapped. Volume larger than the bounding cylinder: a square or a $\pi$ is wrong.

Frequently asked questions

How do I know which curve gives the outer radius in the washer method?

Measure from the axis, not from the x-axis. The outer radius goes to the boundary of the slice that is farthest from the axis of rotation. When the axis lies below the region, that is the upper curve; when the axis lies above the region, it is the lower curve. Test one point: the outer radius squared minus the inner radius squared must be positive.

When do I integrate in x and when in y with disks and washers?

The slice is always perpendicular to the axis of rotation. A horizontal axis, such as the x-axis or a line y equals a constant, gives vertical slices and an integral in x. A vertical axis gives horizontal slices and an integral in y, and every curve must first be solved for x in terms of y, with the bounds given as y values.

Why is the washer area pi times R squared minus r squared and not pi times R minus r squared?

A washer is a big disk with a hole: its area is the area of the big disk minus the area of the hole, pi R squared minus pi r squared. The other expression is the area of a disk whose radius is the thickness of the region, and it no longer depends on how far the region is from the axis, so it describes a different, smaller solid.

How do I find the volume of a solid with known cross-sections?

Find the area of one cross-section as a function of the position of the cut, then integrate that area along the axis. The side of the cross-section is usually the whole chord of the base at that position, from its lower edge to its upper edge. A square on a chord c has area c squared, a semicircle on it has area pi c squared over eight.

Practise it

Corrected exercises: Volumes: disks and washers, MATH 141 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Areas between curves and average value Next sheet Volumes: cylindrical shells

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Revision sheet: volumes by slicing, disks and washers (MATH 141)

The essentials

One slice, two radii, one square

Beyond rotation: the area of the slice is all you need

The rules in table form

Which variable, which radii, by the position of the axis

The mistakes that cost marks

1. Subtracting the radii before squaring

2. Taking the height of a curve for its radius about y = k

3. Keeping the top curve as the outer radius when the axis is above

4. Integrating in x with washers about a vertical axis

5. Carrying the x limits into a dy integral

6. Forgetting to square, and choosing the technique for f instead of its square

7. Taking the half-chord as the side of a cross-section

8. Treating a bowl as a cylinder

Which method to choose

Which slice, by the POSITION of the axis and the SHAPE of the region

How the answer is expected to be written

Setting up a volume by washers

Check before you hand in

The typical problem, taken apart

Washers about a vertical line to the left of the region

Learn by heart

Frequently asked questions

How do I know which curve gives the outer radius in the washer method?

When do I integrate in x and when in y with disks and washers?

Why is the washer area pi times R squared minus r squared and not pi times R minus r squared?

How do I find the volume of a solid with known cross-sections?

Corrected exercises: Volumes: disks and washers, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?