Unit 8 Beta

Applications of Integration: the one-page sheet

8.1 Average Value of a Function

The average value of f on [a, b] is 1/(b − a) ∫ from a to b of f(x) dx , the height of the equal-area rectangle, in the units of f . A continuous f reaches it somewhere on the interval. Do not confuse it with the average rate of change (f(b) − f(a))/(b − a) , which is the average value of f′ .

Average value of $f$ on $[a, b]$ : $\frac{1}{b - a}\int_a^b f(x)\,dx$ . The integrand is $f$ itself.
Average value is a height, in the units of $f$ . Average rate of change, $\frac{f(b) - f(a)}{b - a}$ , is a slope, in units of $f$ per unit of $x$ .
Areas below the axis count as negative, exactly as in the integral.
If $f$ is continuous, some $c$ in $[a, b]$ has $f(c) = f_{\text{avg}}$ .
From a graph use geometry for the integral; from a table use a Riemann or trapezoidal sum; from a formula use the Fundamental Theorem or a calculator.

spread it evenly over the interval to get one steady height f_avg = 1/(b − a) ∫ from a to b of f(x) dx , in the units of f f takes its average value at some c in [a, b] (between its minimum and maximum, by the Intermediate Value Theorem) 1/(b − a) ∫ from a to b of f′(x) dx = (f(b) − f(a))/(b − a) , the average rate of change of f

average value of a function: The average value of a function <math class="calc-m"><semantics><mrow><mi>f</mi></mrow><annotation encoding="application/x-tex">f</annotation></semantics></math> on <math class="calc-m"><semantics><mrow><mo stretchy="false">[</mo><mi>a</mi><mo>,</mo><mi>b</mi><mo stretchy="false">]</mo></mrow><annotation encoding="application/x-tex">[a, b]</annotation></semantics></math> is <math class="calc-m"><semantics><mrow><mfrac><mn>1</mn><mrow><mi>b</mi><mo>−</mo><mi>a</mi></mrow></mfrac><msubsup><mo largeop="true" movablelimits="false">∫</mo><mi>a</mi><mi>b</mi></msubsup><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mspace width="0.1667em"></mspace><mi>d</mi><mi>x</mi></mrow><annotation encoding="application/x-tex">\frac{1}{b - a}\int_a^b f(x)\,dx</annotation></semantics></math>: the height of the rectangle on <math class="calc-m"><semantics><mrow><mo stretchy="false">[</mo><mi>a</mi><mo>,</mo><mi>b</mi><mo stretchy="false">]</mo></mrow><annotation encoding="application/x-tex">[a, b]</annotation></semantics></math> with the same signed area as the region under <math class="calc-m"><semantics><mrow><mi>f</mi></mrow><annotation encoding="application/x-tex">f</annotation></semantics></math>. It has the units of <math class="calc-m"><semantics><mrow><mi>f</mi></mrow><annotation encoding="application/x-tex">f</annotation></semantics></math>.

8.2 Motion with Integrals

Integrate velocity for displacement, x(b) − x(a) ; integrate speed |v| for total distance. To find a position, start from a known position and add the integral of velocity. Split at the zeros of v when working by hand, and always state units.

Displacement on $[a, b]$ : $\int_a^b v(t)\,dt$ $= x(b) - x(a)$ . Its sign gives the direction of the net change.
Total distance traveled: $\int_a^b |v(t)|\,dt$ . Without a calculator, split at the zeros of $v$ and add the sizes of the pieces.
Position: $x(t)$ $= x(t_0)$ $+ \int_{t_0}^{t} v(s)\,ds$ . The initial position is not optional.
Velocity from acceleration works the same way: $v(t)$ $= v(t_0)$ $+ \int_{t_0}^{t} a(s)\,ds$ .
Units: integrating meters per second over seconds gives meters.

∫ from a to b of v(t) dt = x(b) − x(a) , the displacement motion to the left cancels motion to the right ∫ from a to b of |v(t)| dt counts every stretch as positive: the total distance x(t) = x(a) +∫ from a to t of v(s) ds gives the position at any time

displacement: The change in position of a particle over a time interval: <math class="calc-m"><semantics><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>b</mi><mo stretchy="false">)</mo><mo>−</mo><mi>x</mi><mo stretchy="false">(</mo><mi>a</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">x(b) - x(a)</annotation></semantics></math><wbr><math class="calc-m"><semantics><mrow><mo form="infix">=</mo><msubsup><mo largeop="true" movablelimits="false">∫</mo><mi>a</mi><mi>b</mi></msubsup><mi>v</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mspace width="0.1667em"></mspace><mi>d</mi><mi>t</mi></mrow><annotation encoding="application/x-tex">= \int_a^b v(t)\,dt</annotation></semantics></math>. It can be positive, negative or zero.
total distance: The length of the path a particle travels over a time interval, counting motion in both directions as positive: <math class="calc-m"><semantics><mrow><msubsup><mo largeop="true" movablelimits="false">∫</mo><mi>a</mi><mi>b</mi></msubsup><mo stretchy="false">|</mo><mi>v</mi><mo stretchy="false">(</mo><mi>t</mi><mo stretchy="false">)</mo><mo stretchy="false">|</mo><mspace width="0.1667em"></mspace><mi>d</mi><mi>t</mi></mrow><annotation encoding="application/x-tex">\int_a^b |v(t)|\,dt</annotation></semantics></math>, the integral of speed.

8.3 Accumulation in Context

With a rate in R and a rate out D , the amount satisfies A′(t) = R(t) − D(t) and A(t) = A(a) +∫ from a to t of (R − D) . Integrals of rates carry accumulated units and need an interpretation with quantity, units and interval. The amount peaks or bottoms out where R − D changes sign or at an endpoint.

Amount now = amount at the start + $\int$ (rate in − rate out). The initial amount is part of the answer.
$A'(t) = R(t) - D(t)$ : the amount increases when the rate in is larger and decreases when the rate out is larger.
An integral of a rate has the accumulated unit: gallons per hour integrated over hours gives gallons.
An interpretation names the quantity, its units and the time interval.
For the greatest or least amount on a closed interval: find where R − D changes sign, then compare with the endpoints (candidates test).

its net rate of change is A′(t) = R(t) − D(t) the net change in the amount: ∫ from a to b of (R(t) − D(t)) dt A(t) = A(a) +∫ from a to t of (R(s) − D(s)) ds the amount has a relative extremum there; compare it with the endpoints to find the absolute one

rate in: The rate in (inflow rate) is how fast a quantity is added; the rate out (outflow rate) is how fast it is removed. The amount changes at rate in minus rate out, so amount at time <math class="calc-m"><semantics><mrow><mi>t</mi></mrow><annotation encoding="application/x-tex">t</annotation></semantics></math> = amount at the start + <math class="calc-m"><semantics><mrow><mo largeop="true" movablelimits="false">∫</mo></mrow><annotation encoding="application/x-tex">\int</annotation></semantics></math> (rate in − rate out).

8.4 Area Between Curves (Functions of x)

The area between two curves is a sum of thin vertical rectangles whose heights are top minus bottom. Find where the curves meet for the limits, check which function is on top, put the bottom function in parentheses, and integrate: A = ∫ from a to b of (f(x) − g(x)) dx .

Area between curves: if $f(x) \ge g(x)$ on $[a, b]$ , the area is $\int_a^b \left(f(x) - g(x)\right)\,dx$ , top minus bottom.
Find the limits by solving $f(x) = g(x)$ . They are $x$ -values, because the integral is in $dx$ .
Put the bottom function in parentheses before you subtract: $(x + 1) - (x^2 - 1)$ $= x + 2 - x^2$ .
To tell which curve is on top, test one $x$ between the crossings in both functions.
The formula works wherever the curves are, above or below the $x$ -axis: only the difference matters.

each one reaches from the bottom curve g up to the top curve f , so its height is f(x) − g(x) adding them is a Riemann sum for f − g the sum becomes ∫ from a to b of (f(x) − g(x)) dx the limits a and b are the solutions of f(x) = g(x) the integral is positive, so a negative answer means the subtraction is backward

area between curves: The area of the region bounded by two graphs: on an interval where <math class="calc-m"><semantics><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>≥</mo><mi>g</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">f(x) \ge g(x)</annotation></semantics></math>, it is <math class="calc-m"><semantics><mrow><msubsup><mo largeop="true" movablelimits="false">∫</mo><mi>a</mi><mi>b</mi></msubsup><mrow><mo fence="true" stretchy="true">(</mo><mrow><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>−</mo><mi>g</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow><mo fence="true" stretchy="true">)</mo></mrow><mspace width="0.1667em"></mspace><mi>d</mi><mi>x</mi></mrow><annotation encoding="application/x-tex">\int_a^b \left(f(x) - g(x)\right)\,dx</annotation></semantics></math>, top minus bottom, with limits where the region starts and ends (often where the curves meet).

8.5 Area Between Curves (Functions of y)

When a region is easier to cut sideways, write its edges as x in terms of y and integrate right minus left from the lowest y to the highest: A = ∫ from c to d of (x_right − x_left) dy . The limits are y -values, found by setting the two edges equal.

Integrating with respect to $y$ : area $= \int_c^d \left(x_{\text{right}}(y) - x_{\text{left}}(y)\right)\,dy$ , right minus left.
Rewrite each curve as $x =$ (something in $y$ ) first: $y = \sqrt{x}$ becomes $x = y^2$ , $y = \ln x$ becomes $x = e^y$ .
Limits are $y$ -values. Find them by setting the two $x$ -expressions equal and solving for $y$ .
Choose $y$ when each side edge is one formula but the top or bottom edge changes partway across; slicing in $x$ would take two integrals.
Both ways give the same area when done correctly.

a horizontal strip at height y runs from x_left(y) to x_right(y) its area is (x_right − x_left)Δy the sum becomes ∫ from c to d of (x_right − x_left) dy the limits c and d are the y -values where the region starts and ends

integrating with respect to y: Finding an area with horizontal strips: write each edge as <math class="calc-m"><semantics><mrow><mi>x</mi></mrow><annotation encoding="application/x-tex">x</annotation></semantics></math> in terms of <math class="calc-m"><semantics><mrow><mi>y</mi></mrow><annotation encoding="application/x-tex">y</annotation></semantics></math>, and integrate right minus left, <math class="calc-m"><semantics><mrow><msubsup><mo largeop="true" movablelimits="false">∫</mo><mi>c</mi><mi>d</mi></msubsup><mrow><mo fence="true" stretchy="true">(</mo><mrow><msub><mi>x</mi><mtext>right</mtext></msub><mo>−</mo><msub><mi>x</mi><mtext>left</mtext></msub></mrow><mo fence="true" stretchy="true">)</mo></mrow><mspace width="0.1667em"></mspace><mi>d</mi><mi>y</mi></mrow><annotation encoding="application/x-tex">\int_c^d \left(x_{\text{right}} - x_{\text{left}}\right)\,dy</annotation></semantics></math>, between <math class="calc-m"><semantics><mrow><mi>y</mi></mrow><annotation encoding="application/x-tex">y</annotation></semantics></math>-values. Best when each side edge is one formula.

8.6 Area Between Curves That Cross More Than Twice

When curves cross more than twice, top and bottom swap at each crossing. Find the crossings, split the integral there, use top minus bottom on each piece, and add: the area is ∫ from a to b of |f(x) − g(x)| dx . A single integral of f − g gives a net amount, not the area.

Splitting the integral: find every crossing inside the interval by solving $f(x) = g(x)$ , then integrate top minus bottom on each piece and add.
One integral $\int_a^b (f - g)\,dx$ is a net value: areas where $g$ is on top are subtracted.
Area $= \int_a^b |f(x) - g(x)|\,dx$ . On a calculator, this is the fastest setup.
Crossings outside the interval do not create pieces, and the interval's ends are not crossings.
The same rule holds in $y$ : split where the right and left edges swap.

the curve on top changes at each crossing one integral of f − g counts that piece as negative and cancels area every piece is positive, and their sum is the area the area is also ∫ from a to b of |f(x) − g(x)| dx , which a calculator can evaluate in one step

splitting the integral: Breaking an area integral at every point where the curves cross, so each piece uses top minus bottom (or right minus left) and every piece adds a positive area. Equivalent to integrating <math class="calc-m"><semantics><mrow><mo stretchy="false">|</mo><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>−</mo><mi>g</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo stretchy="false">|</mo></mrow><annotation encoding="application/x-tex">|f(x) - g(x)|</annotation></semantics></math>.

8.7 Volumes with Square and Rectangle Cross Sections

A solid with known cross sections has volume ∫ from a to b of A(x) dx . When the slices are squares standing on a base, each side is the base's width across the slice, f(x) − g(x) , and A(x) = (f(x) − g(x))² . Square the difference, not each curve, and leave out π .

Volume by cross sections: $V$ $= \int_a^b A(x)\,dx$ , where $A(x)$ is the area of the slice at $x$ .
For squares standing on the base, the side is top minus bottom of the base: $s = f(x) - g(x)$ , and $A(x) = s^2$ .
Square the whole difference: $(f - g)^2$ is not $f^2 - g^2$ .
Rectangles work the same way: $A = s \cdot h$ , with $h$ given (a number, or a multiple of $s$ ).
Slices perpendicular to the $y$ -axis use right minus left and $dy$ .

each slice is a thin slab: a cross section of area A(x) with thickness Δx adding the slabs is a Riemann sum for A the volume is V = ∫ from a to b of A(x) dx A(x) = s(x)² = (f(x) − g(x))²

cross section: The flat shape you get by slicing a solid with a plane, here perpendicular to an axis. If the slice at <math class="calc-m"><semantics><mrow><mi>x</mi></mrow><annotation encoding="application/x-tex">x</annotation></semantics></math> has area <math class="calc-m"><semantics><mrow><mi>A</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">A(x)</annotation></semantics></math>, the solid's volume is <math class="calc-m"><semantics><mrow><msubsup><mo largeop="true" movablelimits="false">∫</mo><mi>a</mi><mi>b</mi></msubsup><mi>A</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mspace width="0.1667em"></mspace><mi>d</mi><mi>x</mi></mrow><annotation encoding="application/x-tex">\int_a^b A(x)\,dx</annotation></semantics></math>. For squares with side <math class="calc-m"><semantics><mrow><mi>s</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">s(x)</annotation></semantics></math>, <math class="calc-m"><semantics><mrow><mi>A</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>=</mo><mi>s</mi><mo stretchy="false">(</mo><mi>x</mi><msup><mo stretchy="false">)</mo><mn>2</mn></msup></mrow><annotation encoding="application/x-tex">A(x) = s(x)^2</annotation></semantics></math>.

8.8 Volumes with Triangle and Semicircle Cross Sections

Triangle and semicircle cross sections use the same integral as squares, V = ∫ from a to b of A(x) dx , with s = f(x) − g(x) across the base. Learn the areas: equilateral (√3)/4 s² , isosceles right 1/2 s² (leg in base) or 1/4 s² (hypotenuse in base), semicircle π/8 s² .

Triangular and semicircular cross sections: same method as squares, $V$ $= \int_a^b A(x)\,dx$ , with a new area formula.
Equilateral triangle with side $s$ : $A = \frac{\sqrt{3}}{4}s^2$ .
Isosceles right triangle: leg $s$ in the base gives $\frac{1}{2}s^2$ ; hypotenuse $s$ in the base gives $\frac{1}{4}s^2$ .
Semicircle with diameter $s$ in the base: radius $\frac{s}{2}$ , so $A = \frac{\pi}{8}s^2$ .
On one base, every volume is a constant times $\int s^2\,dx$ : squares $1$ , equilateral $0.433$ , semicircles $0.393$ , right isosceles $0.5$ or $0.25$ .

the segment is a side of the triangle, or the diameter of the semicircle its area is a fixed number times s² A = 1/2 ⋅s⋅(√3)/2 s = (√3)/4 s² A = 1/2 π(s/2)² = π/8 s² each volume is its constant times ∫ from a to b of s² dx

triangular cross section: A slice of a solid shaped like a triangle or a semicircle, standing on a segment of length <math class="calc-m"><semantics><mrow><mi>s</mi></mrow><annotation encoding="application/x-tex">s</annotation></semantics></math> across the base. Areas: equilateral triangle <math class="calc-m"><semantics><mrow><mfrac><msqrt><mn>3</mn></msqrt><mn>4</mn></mfrac><msup><mi>s</mi><mn>2</mn></msup></mrow><annotation encoding="application/x-tex">\frac{\sqrt{3}}{4}s^2</annotation></semantics></math>; isosceles right triangle with a leg in the base <math class="calc-m"><semantics><mrow><mfrac><mn>1</mn><mn>2</mn></mfrac><msup><mi>s</mi><mn>2</mn></msup></mrow><annotation encoding="application/x-tex">\frac{1}{2}s^2</annotation></semantics></math>, with the hypotenuse in the base <math class="calc-m"><semantics><mrow><mfrac><mn>1</mn><mn>4</mn></mfrac><msup><mi>s</mi><mn>2</mn></msup></mrow><annotation encoding="application/x-tex">\frac{1}{4}s^2</annotation></semantics></math>; semicircle with diameter <math class="calc-m"><semantics><mrow><mi>s</mi></mrow><annotation encoding="application/x-tex">s</annotation></semantics></math>, <math class="calc-m"><semantics><mrow><mfrac><mi>π</mi><mn>8</mn></mfrac><msup><mi>s</mi><mn>2</mn></msup></mrow><annotation encoding="application/x-tex">\frac{\pi}{8}s^2</annotation></semantics></math>.

8.9 Disc Method About the Axes

A region that touches the axis of rotation along its whole length makes a solid whose slices are discs. With radius r measured from the axis to the curve, V = π∫ from a to b of r(x)² dx about the x -axis, or π∫ from c to d of r(y)² dy about the y -axis. Keep the π and square the radius.

Disc method: $V$ $= \pi\int_a^b r(x)^2\,dx$ , the cross-section method with $A(x) = \pi r^2$ .
The radius runs from the axis of rotation to the curve. About the $x$ -axis, $r = f(x)$ .
About the $y$ -axis, slices are horizontal: $r$ is the curve's $x$ written in $y$ , and the limits are $y$ -values.
Do not forget $\pi$ , and square the radius inside the integral.
Checks: a cylinder gives $\pi r^2 h$ , a cone $\frac{1}{3}\pi r^2 h$ , a ball $\frac{4}{3}\pi r^3$ .

every slice perpendicular to the axis is a solid disc it sweeps out a disc whose radius is that distance, r(x) has volume πr² Δx V = π∫ from a to b of r(x)² dx write the radius as x in terms of y and integrate in dy

disc method: A way to find the volume of a solid of revolution whose slices are solid discs: <math class="calc-m"><semantics><mrow><mi>V</mi></mrow><annotation encoding="application/x-tex">V</annotation></semantics></math><wbr><math class="calc-m"><semantics><mrow><mo form="infix">=</mo><mi>π</mi><msubsup><mo largeop="true" movablelimits="false">∫</mo><mi>a</mi><mi>b</mi></msubsup><mi>r</mi><mo stretchy="false">(</mo><mi>x</mi><msup><mo stretchy="false">)</mo><mn>2</mn></msup><mspace width="0.1667em"></mspace><mi>d</mi><mi>x</mi></mrow><annotation encoding="application/x-tex">= \pi\int_a^b r(x)^2\,dx</annotation></semantics></math>, where <math class="calc-m"><semantics><mrow><mi>r</mi></mrow><annotation encoding="application/x-tex">r</annotation></semantics></math> is the distance from the axis of rotation to the curve. About the <math class="calc-m"><semantics><mrow><mi>y</mi></mrow><annotation encoding="application/x-tex">y</annotation></semantics></math>-axis, the discs are horizontal and <math class="calc-m"><semantics><mrow><mi>V</mi></mrow><annotation encoding="application/x-tex">V</annotation></semantics></math><wbr><math class="calc-m"><semantics><mrow><mo form="infix">=</mo><mi>π</mi><msubsup><mo largeop="true" movablelimits="false">∫</mo><mi>c</mi><mi>d</mi></msubsup><mi>r</mi><mo stretchy="false">(</mo><mi>y</mi><msup><mo stretchy="false">)</mo><mn>2</mn></msup><mspace width="0.1667em"></mspace><mi>d</mi><mi>y</mi></mrow><annotation encoding="application/x-tex">= \pi\int_c^d r(y)^2\,dy</annotation></semantics></math>.

8.10 Disc Method About Other Lines

To revolve a region about a line y = k or x = h that it touches all the way across, use discs whose radius is the distance from that line to the curve: V = π∫ from a to b of (k − f(x))² dx (or the mirror image). Measure from the axis, subtract in the order that makes the radius positive, and square the whole radius.

Revolving about a line: the disc method still applies; only the radius changes.
The radius is a distance from the axis: larger value minus smaller value. With the line $y = k$ above the curve, $r = k - f(x)$ ; below it, $r = f(x) - k$ .
With a negative $k$ , subtract carefully: about $y = -2$ , $r$ $= f(x) - (-2)$ $= f(x) + 2$ .
A vertical line $x = h$ gives horizontal discs: $r = h - x(y)$ or $x(y) - h$ , integrated in $dy$ .
Square $(k - f(x))$ as a whole; $k^2 - f(x)^2$ is a different number.

every slice is a disc centered on that line each rectangle reaches from the line to the curve and sweeps out a solid disc r = k − f(x) if the line is above the curve, f(x) − k if it is below V = π∫ from a to b of (f(x) − k)² dx

revolving about a line: Revolving a region about a line other than an axis, such as <math class="calc-m"><semantics><mrow><mi>y</mi><mo>=</mo><mi>k</mi></mrow><annotation encoding="application/x-tex">y = k</annotation></semantics></math> or <math class="calc-m"><semantics><mrow><mi>x</mi><mo>=</mo><mi>h</mi></mrow><annotation encoding="application/x-tex">x = h</annotation></semantics></math>. The radius is the distance from that line to the curve: <math class="calc-m"><semantics><mrow><mi>r</mi><mo>=</mo><mo stretchy="false">|</mo><mi>f</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo><mo>−</mo><mi>k</mi><mo stretchy="false">|</mo></mrow><annotation encoding="application/x-tex">r = |f(x) - k|</annotation></semantics></math> for a horizontal line, <math class="calc-m"><semantics><mrow><mi>r</mi><mo>=</mo><mo stretchy="false">|</mo><mi>h</mi><mo>−</mo><mi>x</mi><mo stretchy="false">(</mo><mi>y</mi><mo stretchy="false">)</mo><mo stretchy="false">|</mo></mrow><annotation encoding="application/x-tex">r = |h - x(y)|</annotation></semantics></math> for a vertical one.

8.11 Washer Method About the Axes

When the region does not touch the axis, each slice is a washer. Measure both radii from the axis: the outer radius R to the far curve and the inner radius r to the near one. Then V = π∫ from a to b of (R² − r²) dx : the difference of squares, never the square of the difference.

Washer method: $V$ $= \pi\int_a^b \left(R^2 - r^2\right)\,dx$ , outer radius squared minus inner radius squared.
Both radii are measured from the axis: $R$ to the far curve, $r$ to the near one. About the $x$ -axis, $R = f(x)$ (top) and $r = g(x)$ (bottom).
Square each radius, then subtract: $R^2 - r^2$ , not $(R - r)^2$ .
About the $y$ -axis, the washers are horizontal: write each curve as $x$ in terms of $y$ , and the farther curve from the $y$ -axis is the outer radius.
Discs are washers with $r = 0$ : use washers whenever the region leaves a gap at the axis.

each rectangle starts above the axis and sweeps out a ring, a washer outer radius R = axis to the far curve; inner radius r = axis to the near curve its area is πR² − πr² = π(R² − r²) V = π∫ from a to b of (R(x)² − r(x)²) dx

washer method: The volume of a solid of revolution whose slices have a hole: each slice is a washer with outer radius <math class="calc-m"><semantics><mrow><mi>R</mi></mrow><annotation encoding="application/x-tex">R</annotation></semantics></math> (axis to the far edge of the region) and inner radius <math class="calc-m"><semantics><mrow><mi>r</mi></mrow><annotation encoding="application/x-tex">r</annotation></semantics></math> (axis to the near edge), so <math class="calc-m"><semantics><mrow><mi>V</mi></mrow><annotation encoding="application/x-tex">V</annotation></semantics></math><wbr><math class="calc-m"><semantics><mrow><mo form="infix">=</mo><mi>π</mi><msubsup><mo largeop="true" movablelimits="false">∫</mo><mi>a</mi><mi>b</mi></msubsup><mrow><mo fence="true" stretchy="true">(</mo><mrow><msup><mi>R</mi><mn>2</mn></msup><mo>−</mo><msup><mi>r</mi><mn>2</mn></msup></mrow><mo fence="true" stretchy="true">)</mo></mrow><mspace width="0.1667em"></mspace><mi>d</mi><mi>x</mi></mrow><annotation encoding="application/x-tex">= \pi\int_a^b \left(R^2 - r^2\right)\,dx</annotation></semantics></math>.

8.12 Washer Method About Other Lines

Revolving about a line other than an axis uses the same washer integral, V = π∫ (R² − r²) . Draw one rectangle perpendicular to the axis; R is the distance from the axis to its far end and r to its near end. When the axis is above (or to the right of) the region, the far end is the bottom (or left) curve.

Washers about a line: $V = \pi\int \left(R^2 - r^2\right)$ with both radii measured from the line $y = k$ or $x = h$ .
Axis below the region ( $y = k$ , $k$ small): $R = f(x) - k$ , $r = g(x) - k$ . With $k = -1$ , add 1 to each height.
Axis above the region: the far edge is the bottom curve, so $R = k - g(x)$ and $r = k - f(x)$ .
Vertical axis $x = h$ : horizontal washers in $dy$ ; the far edge may be the left curve if the axis is on the right.
Square each radius separately; $(R - r)^2$ is the region's height squared, not a washer's area.

each slice is a washer centered on that line its far end gives R and its near end gives r , both measured from y = k R = (top) − k and r = (bottom) − k the bottom curve is now the far edge: R = k − (bottom) and r = k − (top) V = π∫ from a to b of (R² − r²) dx

washer about a line: The washer method with the axis of rotation on a line <math class="calc-m"><semantics><mrow><mi>y</mi><mo>=</mo><mi>k</mi></mrow><annotation encoding="application/x-tex">y = k</annotation></semantics></math> or <math class="calc-m"><semantics><mrow><mi>x</mi><mo>=</mo><mi>h</mi></mrow><annotation encoding="application/x-tex">x = h</annotation></semantics></math>. Both radii are distances from that line: <math class="calc-m"><semantics><mrow><mi>R</mi></mrow><annotation encoding="application/x-tex">R</annotation></semantics></math> to the edge of the region farther from the line, <math class="calc-m"><semantics><mrow><mi>r</mi></mrow><annotation encoding="application/x-tex">r</annotation></semantics></math> to the nearer edge, and <math class="calc-m"><semantics><mrow><mi>V</mi><mo>=</mo><mi>π</mi><mo largeop="true" movablelimits="false">∫</mo><mrow><mo fence="true" stretchy="true">(</mo><mrow><msup><mi>R</mi><mn>2</mn></msup><mo>−</mo><msup><mi>r</mi><mn>2</mn></msup></mrow><mo fence="true" stretchy="true">)</mo></mrow></mrow><annotation encoding="application/x-tex">V = \pi\int \left(R^2 - r^2\right)</annotation></semantics></math>.