Spin a curve once around the x-axis and you get a solid.
Slice that solid thinly at right angles to the x-axis and each slice is a disk of radius f(x), so integrating the disk area π[f(x)]² with respect to x gives the volume.
The length of a curve uses the same idea: cut the curve into tiny segments ds = √(dx² + dy²), measure each with the Pythagorean theorem, and add them all up.
Turn the rotation angle and the curve sweeps around the x-axis drawing cross-section circles; move the upper limit t and the curve length and its exact value change together.
Volume of a Solid of Revolution
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The curve turned around the x-axis by the rotation angle; the blue disk has radius f(x)
270°
√x
🍩 Stacking disks
①Spinning the curve once around the x-axis makes a solid. The dashed curve is the curve turned by the rotation angle, and the orange arcs are the cross-section circles traced at each x
②The slice at x is a disk of radius f(x), with area π[f(x)]²
③Integrating this area with respect to x gives the volume; the exact volume for the chosen function is written under the picture
Three Ways to Find a Volume
Disk method
V = π ∫ab [f(x)]² dx
For rotation about the x-axis, integrate the area of a disk of radius f(x)
Washer method
V = π ∫ab ([R(x)]² − [r(x)]²) dx
For rotating the region between two curves, integrate the area of a ring: the circle of outer radius R minus the circle of inner radius r
Shell method
V = 2π ∫ab x · f(x) dx
For rotation about the y-axis, integrate the lateral area 2πx · f(x) of a thin cylindrical shell of radius x and height f(x)
💡 Which method to use
①Rotating about the x-axis and integrating in x is the disk method
②For a solid with a hole in the middle, the washer method subtracts the inner circle from the outer one
③When rotating about the y-axis but integrating in x, use the shell method. If both work, choose whichever makes the calculation simpler
Length of a Curve
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The length of y = x√x, found by cutting it into short segments ds and joining them
3
📏 Measuring along the curve
①Cut the curve into very short pieces and each one is close to a straight segment ds
②ds = √(dx² + dy²) is the Pythagorean theorem; since dy = f′(x) dx, ds = √(1 + [f′(x)]²) dx
③For y = x√x, f′(x) = (3/2)√x, so 1 + [f′(x)]² = 1 + 9x/4, and integrating it from 0 to t gives the L under the picture
④A straight segment joining the same two endpoints is shorter than L
Length of a curve
L = ∫ab √(1 + [f′(x)]²) dx
Integrate ds = √(1 + (dy/dx)²) dx
Area and Distance Traveled
Area between two curves
S = ∫ab |f(x) − g(x)| dx
Integrate the upper function minus the lower one; if they cross, split the interval at the crossing points
Integrating the absolute value of velocity gives the distance actually traveled; integrating velocity itself gives the change in position (displacement)
🚗 Quantities built up by integrals
①Integrating velocity v(t) gives the change in position (displacement); integrating |v(t)| gives the distance traveled
②Integrating acceleration a(t) gives the change in velocity
③An integral adds up small amounts: stacking disk areas gives a volume, and joining short segments gives the length of a curve
Worked Examples
Example 1
Find the volume of the solid formed by rotating y = √x (0 ≤ x ≤ 4) about the x-axis.
1
Use the disk method V = π∫[f(x)]² dx, with [f(x)]² = x.
V = π ∫04 (√x)2 dx = π ∫04 x dx
2
Evaluate the definite integral.
π · 422 = π · 8 = 8π
▸ 8π
The volume of a solid of revolution is the integral of the cross-section area π[f(x)]². In the picture y = √x runs over 0 ≤ x ≤ 2, giving 2π; over 0 ≤ x ≤ 4 it becomes 8π.
Example 2
Find the area enclosed by the curves y = x2 and y = 2x.
1
Find the intersections to set the interval. For 0 < x < 2, 2x is the upper curve and x2 the lower one.
x2 = 2x ⇒ x = 0, 2
2
Integrate (upper − lower).
∫02 (2x − x2) dx = 4 − 83 = 43
▸ 43
An area between curves is the integral of (upper − lower) between their intersections. Check which curve is on top first.
Summary
Three formulas for applications
V = π∫[f(x)]²dx, L = ∫√(1 + [f′(x)]²)dx, S = ∫|f − g|dx
Volume of a solid of revolution, length of a curve, area between two curves
CSAT-style problem, adapted
A point P moves on a number line with velocity v(t) = 3t2 − 6t. How far does P travel from t = 0 to t = 3?
①4
②6
③8
④10
⑤12
▸ ③ 8
1
Look at the sign of v(t) = 3t(t − 2): negative for 0 < t < 2, positive for 2 < t < 3. Let the position be s(t) = t3 − 3t2.
v(t) = 3t(t − 2), s(t) = t3 − 3t2
2
The distance traveled is ∫|v| dt, so add the absolute values of the displacement on each interval.
|s(2) − s(0)| + |s(3) − s(2)| = 4 + 4 = 8
🎯 Exam Points
①The disk method V = π∫[f(x)]² dx is the basic tool for rotation about the x-axis
②The washer method uses a difference of squares, [R(x)]² − [r(x)]², not (R − r)²
③The shell method 2π∫x · f(x) dx is convenient for rotation about the y-axis
④For the length of a curve, find f′(x) first and simplify √(1 + [f′(x)]²)