Once you can evaluate a definite integral, AP Calculus AB puts that skill to work measuring shapes: the area enclosed between two curves, and the volume of a three-dimensional solid formed by rotating a two-dimensional region around a line. These problems reward careful setup far more than they reward computational speed — the actual integration is often routine once the bounds, the integrand, and (for volumes) the right method have all been identified correctly.
For area between curves, the recurring challenge is finding where the curves intersect (to get the bounds) and figuring out which curve is on top over each piece of the interval, since the top-minus-bottom setup flips whenever the curves cross. Some regions are naturally described as functions of y rather than x — a sideways parabola or an S-shaped cubic is often far easier to integrate with respect to y, integrating right-minus-left instead of top-minus-bottom. For volumes of revolution, the central decision is disk/washer versus shell: disks and washers integrate perpendicular slices (circles or rings) and work naturally when the axis of rotation is one of the boundaries, while shells integrate cylindrical "wrapping paper" layers and often avoid messy algebra when the axis of rotation is parallel to the strips being summed but not directly touching the region.
Below are 20 AP Calculus AB-style practice problems covering area between curves (in both x and y), the ratio of areas bounded by a power function, and volumes of revolution using disks, washers, and shells around horizontal, vertical, and off-axis lines. Work through each problem, select your answer, and check whether you got it right — every problem includes a complete, step-by-step explanation. Once you've worked through these, keep building your skills with thousands more official-style questions in our free AP Calculus AB QBank.
Find the area between the curve y = 3/x and the line x+y = 4.
Find intersections by substituting y=3/x into x+y=4: x + 3/x = 4 → x2−4x+3 = 0 → (x−1)(x−3)=0, giving x=1 and x=3.
On (1,3), the line lies above the curve (e.g., at x=2: line gives 2, curve gives 1.5). Area = ∫13[(4−x)−3/x]dx = [4x−x2/2−3ln x]13.
At x=3: 12−4.5−3ln3 = 7.5−3ln3. At x=1: 4−0.5−0 = 3.5.
Subtract: (7.5−3ln3)−3.5 = 4−3ln3. The correct answer is A.
In the first quadrant, find the area bounded below by the x-axis and above by the curves y = 2sin x and y = 2cos x.
The two curves intersect where 2sin x = 2cos x, i.e. at x=π/4. The region bounded above by both curves is bounded by whichever curve is lower at each x.
On [0,π/4], 2sin x ≤ 2cos x, so the boundary is 2sin x. On [π/4,π/2], 2cos x ≤ 2sin x, so the boundary is 2cos x.
Area = ∫0π/42sin x dx + ∫π/4π/22cos x dx = [−2cos x]0π/4 + [2sin x]π/4π/2.
= (−√2+2) + (2−√2) = 4−2√2. The correct answer is A.
Keep the momentum going.
Explore 1,500+ Free AP Calculus AB ProblemsFind the area bounded above by y = 2cos x and below by y = 2sin x, from x = π/4 to x = 5π/4.
Area = ∫π/45π/4(2cos x−2sin x)dx = [2sin x+2cos x]π/45π/4.
At x=5π/4: 2(−√2/2)+2(−√2/2) = −√2−√2 = −2√2. At x=π/4: 2(√2/2)+2(√2/2) = √2+√2 = 2√2.
Subtract: −2√2 − 2√2 = −4√2. Taking the absolute value (since area must be positive, and this reflects the direction of subtraction): 4√2.
The correct answer is A.
Find the area bounded by the curve y = tan x, the line x = π/3, and the x-axis.
This region runs from x=0 (where tan x meets the x-axis) to x=π/3. Its area is ∫0π/3tan x dx = [−ln|cos x|]0π/3.
Evaluate: −ln(cos(π/3)) + ln(cos(0)) = −ln(1/2) + ln 1 = −ln(1/2).
Since −ln(1/2) = ln 2 (using the property that negating a log flips the fraction inside), the area is ln 2.
The correct answer is A.
Find the total area bounded by the curve y = x3−3x2−4x and the x-axis.
Factor to find the x-intercepts: y = x(x2−3x−4) = x(x−4)(x+1), giving roots at x=−1, 0, 4.
Since the curve changes sign at each root, compute the two pieces separately and take absolute values. The antiderivative is F(x) = x4/4−x3−2x2.
From −1 to 0: F(0)−F(−1) = 0−(1/4+1−2) = 0−(−3/4) = 3/4. From 0 to 4: F(4)−F(0) = (64−64−32)−0 = −32, with absolute value 32.
Total area: 3/4 + 32 = 3/4+128/4 = 131/4. The correct answer is B.
5 down, 15 to go — keep practicing.
Explore 1,500+ Free AP Calculus AB ProblemsThe area bounded by y = ex, y = 3, and the y-axis is equal to
Find where the curves meet: ex=3 → x=ln3. The region runs from x=0 (the y-axis) to x=ln3, with y=3 on top and y=ex on the bottom.
Area = ∫0ln3(3−ex)dx = [3x−ex]0ln3.
At x=ln3: 3ln3−3. At x=0: 0−1 = −1.
Subtract: (3ln3−3)−(−1) = 3ln3−2. The correct answer is A.
The area enclosed by the curve y2 = x(2−x) is given by
Since x(2−x) ≥ 0 exactly for x between 0 and 2, this curve is defined there, and y = ±√(x(2−x)) = ±√(2x−x2), symmetric above and below the x-axis.
The enclosed area is twice the area of just the upper half (y ≥ 0): Area = 2∫02√(2x−x2)dx.
The correct answer is B.
7 down, 13 to go.
Explore 1,500+ Free AP Calculus AB ProblemsThe figure below shows part of the curve y = x2 and a rectangle with two vertices at (0,0) and (c,0). What is the ratio of the area of the rectangle to the shaded part of it above the curve?
The rectangle has width c and height c2 (matching the curve's height at x=c), so its area is c · c2 = c3.
The area under the curve from 0 to c is ∫0cx2dx = c3/3. The shaded area (above the curve, inside the rectangle) is the rest: c3 − c3/3 = (2/3)c3.
Ratio of rectangle to shaded area: c3 : (2/3)c3. Dividing both sides by (1/3)c3: 3 : 2.
The correct answer is A.
The region whose boundaries are y = x2, x = 3, and y = 0 is rotated about the x-axis. Choose the alternative that gives the volume of the solid generated.
Since the region is rotated about the x-axis (a boundary of the region), use the disk method: V = π∫03(x2)2dx = π∫03x4dx.
= π[x5/5]03 = π(243/5).
The correct answer is C.
The region whose boundaries are y = x2, x = 3, and y = 0 is rotated about the y-axis. Choose the alternative that gives the volume of the solid generated.
Since the region is rotated about the y-axis, and the region is naturally described in terms of x (from x=0 to x=3), the shell method avoids needing to solve for x in terms of y: V = 2π∫03 x · x2 dx = 2π∫03x3dx.
= 2π[x4/4]03 = 2π(81/4) = 81π/2.
The correct answer is C.
10 down, 10 to go — over halfway there.
Explore 1,500+ Free AP Calculus AB ProblemsThe first-quadrant region bounded by y = x2, the y-axis, and y = 16 is rotated about the y-axis. Choose the alternative that gives the volume of the solid generated.
Since the axis of rotation is the y-axis and the region's other boundary (y=16) is horizontal, it's cleanest to integrate with respect to y. Solve for x: x = √y.
Use the disk method: V = π∫016(√y)2dy = π∫016y dy.
= π[y2/2]016 = π(256/2) = 128π.
The correct answer is A.
y = x2 and y = 9; about the x-axis. Choose the alternative that gives the volume of the solid generated.
The curves meet where x2=9, i.e. x=±3. Rotating about the x-axis, each cross-section is a washer: outer radius 9 (from the line), inner radius x2 (from the curve).
V = π∫−33[92−(x2)2]dx = 2π∫03(81−x4)dx (using symmetry to double the 0-to-3 piece).
= 2π[81x−x5/5]03 = 2π[243−243/5] = 2π(972/5) = 1944π/5.
The correct answer is B.
y = x2 and y = 9; about the line y = 9. Choose the alternative that gives the volume of the solid generated.
Rotating about y=9 (which coincides with one boundary of the region), each cross-section is a full disk with radius equal to the vertical distance from the curve up to y=9: radius = 9−x2.
V = π∫−33(9−x2)2dx = 2π∫03(81−18x2+x4)dx.
= 2π[81x−6x3+x5/5]03 = 2π[243−162+243/5] = 2π[81+48.6] = 2π(648/5).
= 1296π/5. The correct answer is B.
An arch of y = 2sin x and the x-axis is rotated about the x-axis. Choose the alternative that gives the volume of the solid generated.
One arch of 2sin x runs from x=0 to x=π. Using the disk method: V = π∫0π(2sin x)2dx = 4π∫0πsin2x dx.
The standard result ∫0πsin2x dx = π/2 (using the half-angle identity and integrating) applies here.
V = 4π · (π/2) = 2π2. The correct answer is C.
14 down, 6 to go.
Explore 1,500+ Free AP Calculus AB ProblemsIn Questions 15–18, the region whose boundaries are given is rotated about the line indicated. Choose the alternative that gives the volume of the solid generated.
y = x2 and y = 9; about the line y = −1.
Rotating about y=−1 (a line below the whole region), each cross-section is a washer. The outer radius (to the line y=9) is 9−(−1)=10. The inner radius (to the curve y=x2) is x2−(−1)=x2+1.
V = π∫−33[102−(x2+1)2]dx.
Expand: (x2+1)2 = x4+2x2+1, so the integrand is 100−x4−2x2−1 = 99−2x2−x4.
The correct setup is π∫−33(99−2x2−x4)dx. The correct answer is A. Choice B forgets the inner radius's cross term entirely, and choice D confuses the shell method's linear radius with the washer method's squared radii.
y = 4x−x2 and y = 0; about the x-axis.
First find the bounds: 4x−x2=0 → x(4−x)=0, so x=0 and x=4.
Since the curve touches the x-axis (the axis of rotation) at both endpoints, this is a simple disk setup with radius equal to the function itself: V = π∫04(4x−x2)2dx.
The correct answer is B. Choice A incorrectly expands the square by dropping the cross term (it should be 16x2−8x3+x4, not just 16x2+x4), and choice C uses the wrong upper bound.
16 down, 4 to go — almost there.
Explore 1,500+ Free AP Calculus AB Problemsy = 4x−x2 and y = 2x; about the x-axis.
Find intersections: 4x−x2=2x → 2x−x2=0 → x(2−x)=0, so x=0 and x=2.
Testing x=1 shows the parabola (4x−x2=3) is above the line (2x=2), so it's the outer radius when rotated about the x-axis, and the line 2x is the inner radius.
Using the washer method (outer2 minus inner2, not the difference squared): V = π∫02[(4x−x2)2−(2x)2]dx = π∫02[(4x−x2)2−4x2]dx.
The correct answer is C. Choice D shows the common error of squaring the difference of the functions instead of taking the difference of their squares.
y = ln x, y = 0, x = e2; about the line x = e2.
Rewrite the curve as x in terms of y: y=ln x → x=ey. The region runs from y=0 (where x=1) to y=2 (where x=e2).
Rotating about the vertical line x=e2, each horizontal slice at height y is a disk with radius equal to the horizontal distance from the curve to that line: radius = e2−ey.
V = π∫02(e2−ey)2dy.
The correct answer is B. This disk-method-in-y approach is often cleaner here than trying to use shells, since the boundary curve is easy to invert (solve for x) but awkward to combine directly with a vertical axis using shells.
Almost done — 2 more.
Explore 1,500+ Free AP Calculus AB ProblemsA sphere of radius R is divided into two parts by a plane at distance d (0 < d < R) from the center. The volume of the larger part equals
The smaller piece (the cap beyond distance d from center) has a known volume formula: Vsmall = (π/3)(2R3+d3−3R2d).
The larger piece is what's left of the whole sphere: Vlarge = (4/3)πR3 − Vsmall.
Substitute and simplify: (4/3)πR3 − (π/3)(2R3+d3−3R2d) = (π/3)[4R3−2R3−d3+3R2d] = (π/3)(2R3+3R2d−d3).
The correct answer is B.
If the curves of f(x) and g(x) intersect for x=a and x=b, and f(x) > g(x) > 0 for all x on (a,b), then the volume obtained when the region bounded by the curves is rotated about the x-axis is equal to
Since f(x) > g(x) > 0 throughout, rotating the region between them about the x-axis produces a washer at each x, with outer radius f(x) and inner radius g(x).
The washer formula is the difference of the squared radii, not the square of the difference: V = π∫ab[f(x)2−g(x)2]dx.
The correct answer is B. Choice A is a very common error — squaring (f−g) instead of computing f2−g2 — and the two expressions are not equal in general. Choice C is the shell-method setup, which would apply to rotation about the y-axis, not the x-axis. Choice D has a sign error (adding instead of subtracting).
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