BC-level geometry applications go beyond the area-and-volume toolkit from AB: this unit adds arc length for curves defined parametrically, improper integrals (where a bound is infinite or the integrand blows up somewhere in the interval), and polar area. These topics connect in a very BC-specific way — computing the volume of a solid that extends infinitely far, for instance, is really an improper integral wearing a volume-of-revolution costume, and whether that volume is finite at all depends entirely on whether the improper integral converges.
Arc length for a parametric curve comes from the formula L = ∫√[(dx/dt)2+(dy/dt)2] dt — the algebra almost always simplifies dramatically once you expand and combine the two squared terms, often collapsing several trig terms down to a single clean expression via a Pythagorean identity. For improper integrals, the two situations to watch for are an infinite bound (like ∫1∞) and a vertical asymptote hiding inside or at the edge of the interval (like a denominator that hits zero somewhere between the limits) — both require rewriting the integral as a limit and checking whether that limit actually exists. A classic surprise in this unit is that a region with infinite length or infinite area can still produce a finite volume when rotated, depending on how quickly the cross-sections shrink.
Below are 18 AP Calculus BC-style practice problems covering arc length of parametric curves, identifying and evaluating improper integrals, volumes of revolution over unbounded regions, the area enclosed by polar curves (including a rose and a limaçon's inner loop), and the volume of a sphere cut by a plane. 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 BC QBank.
The length of one arch of the cycloid x = 2t−2sin(t), y = 2−2cos(t) equals
One full arch of a cycloid corresponds to t running from 0 to 2π. Find dx/dt = 2−2cos t and dy/dt = 2sin t.
Combine: (dx/dt)2+(dy/dt)2 = (2−2cos t)2+4sin2t = 4−8cos t+4cos2t+4sin2t = 4−8cos t+4 = 8−8cos t.
Arc length = ∫02π√(8−8cos t) dt. The correct answer is D.
The length of the curve of the parabola 4x = y2 cut off by the line x=3 is given by the integral
Since the parabola is naturally written as x in terms of y (x=y2/4), it's easiest to integrate with respect to y. Find dx/dy = y/2.
At x=3, y2=12 → y=±2√3. Arc length = ∫−2√32√3√(1+y2/4) dy = ∫−2√32√3√(4+y2)/2 dy.
By symmetry, this equals twice the integral from 0 to 2√3, and the resulting factor of 2 exactly cancels the /2: ∫02√3√(4+y2) dy.
The correct answer is D.
The length of x = e2tcos t, y = e2tsin t from t=1 to t=2 is equal to
Find the derivatives using the product rule: dx/dt = e2t(2cos t−sin t) and dy/dt = e2t(2sin t+cos t).
Combine and simplify: (dx/dt)2+(dy/dt)2 = e4t[(2cos t−sin t)2+(2sin t+cos t)2]. Expanding, the cross terms cancel and the identity sin2t+cos2t=1 simplifies the bracket to exactly 5.
So the speed is √(5e4t) = √5 e2t. Arc length = ∫12√5 e2tdt = √5[e2t/2]12 = (√5/2)(e4−e2).
Evaluating numerically: ≈ 52.781. The correct answer is B.
Keep the momentum going.
Explore 1,500+ Free AP Calculus BC ProblemsWhich one of the following is an improper integral?
An integral is improper when the interval is infinite, or when the integrand is undefined (blows up) somewhere within or at the edge of the interval of integration.
In choice A, x+2 ≥ 2 > 0 throughout, so no issue. In choice B, the denominator 4+x2 is never zero. In choice D, sin(x)≠0 for all x in [π/6,π/3], so no issue there either.
In choice C, the denominator 4−x2 equals zero at x=2, which lies inside the interval [0,3] — the integrand is undefined there.
The correct answer is C.
Which one of the following improper integrals diverges?
Choice A converges: for ∫1∞x−pdx, convergence at infinity requires p>1, and here p=3.
Choice B converges (exponential decay always produces a finite area over an infinite interval). Choice C converges: for ∫01x−pdx, convergence near 0 requires p<1, and here p=2/3<1.
Choice D is the problem: the interval [−2,2] contains x=0, where 1/x4 is undefined and blows up faster than any convergent power allows near an interior singularity.
The correct answer is D.
Which one of the following improper integrals diverges?
Choice A converges (the denominator grows fast enough at infinity, behaving like x3, well above the p=1 threshold). Choice B converges: for a singularity at 0, p=2/5<1 is convergent. Choice C converges (exponential growth in the denominator).
Choice D is ∫1∞x−1/2dx. For convergence at an infinite bound, the exponent p must satisfy p>1, but here p=1/2<1.
The correct answer is D.
6 down, 12 to go — keep practicing.
Explore 1,500+ Free AP Calculus BC ProblemsA sphere of radius r is divided into two parts by a plane at distance h (0<h<r) from the center. Using the disk method with a coordinate axis through the center, the volume of the larger part is given by the integral
Set up coordinates with the center of the sphere at the origin, so the sphere is bounded by x2+y2 = r2, and the sphere spans x from −r to r. Each disk (slicing perpendicular to the x-axis) has radius √(r2−x2), so its area is π(r2−x2).
The cutting plane at distance h from the center corresponds to x=h. The larger piece (which contains the center) is the part from x=−r up to x=h.
The volume is ∫−rhπ(r2−x2)dx.
The correct answer is A. Choice D integrates over the entire sphere (giving the total volume, not just the larger piece), and choice B describes the smaller cap instead.
If the curves of f(x) and g(x) intersect for x=a and x=b (0≤a<b), and if 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 y-axis is equal to
Since the axis of rotation (the y-axis) is parallel to the vertical strips of the region (rather than being a boundary of the region itself), the shell method is the natural choice here, not the washer method.
Each cylindrical shell at position x has radius x, height f(x)−g(x), and thickness dx, giving volume 2πx[f(x)−g(x)]dx per shell.
Summing all the shells: V = 2π∫abx[f(x)−g(x)]dx.
The correct answer is B. Choice A is the washer-method formula, which would be correct for rotation about the x-axis instead.
Find the volume of the solid generated when the region bounded above by y = 2/x, at the left by x=1, and below by y=0, is rotated about the x-axis.
This region extends infinitely (from x=1 to x=∞), so the volume is an improper integral: V = π∫1∞(2/x)2dx = 4π∫1∞x−2dx.
This is a p-integral with p=2>1, which converges: 4π[−1/x]1∞ = 4π[0−(−1)] = 4π.
Even though the region has infinite length, the cross-sections shrink fast enough (as 1/x2) that the total volume is finite. The correct answer is B.
9 down, 9 to go — halfway there.
Explore 1,500+ Free AP Calculus BC ProblemsFind the volume of the solid generated when the first-quadrant region under y = e−2x is rotated about the x-axis.
The first-quadrant region under this curve runs from x=0 to x=∞. The volume is V = π∫0∞(e−2x)2dx = π∫0∞e−4xdx.
This converges since the exponent is negative: π[−(1/4)e−4x]0∞ = π[0−(−1/4)].
= π/4. The correct answer is B.
The area A enclosed by the three-leaved rose r = cos(3θ) equals, to three decimal places,
One petal is traced as θ ranges over an interval where r goes from 0, up to its max, and back to 0. For r=cos(3θ), this happens as θ runs from −π/6 to π/6.
Area of one petal: ∫−π/6π/6(1/2)cos2(3θ)dθ. Using the identity cos2u=(1+cos2u)/2: (1/4)[θ+sin(6θ)/6]−π/6π/6 = (1/4)(π/3) = π/12.
Since r=cos(nθ) with n=3 (an odd number) produces exactly n petals (not 2n), the total area is 3 × π/12 = π/4.
π/4 ≈ 0.785. The correct answer is B.
The area bounded by the small loop of the limaçon r = 2−4sinθ is given by the definite integral
The small inner loop occurs where r≤0. Find where r=0: 2−4sinθ=0 → sinθ=1/2 → θ=π/6 or 5π/6.
Between these two angles (from π/6 to 5π/6), r is negative throughout — this middle range traces out the small inner loop.
Using the polar area formula A = ∫(1/2)r2dθ over this range: ∫π/65π/6(1/2)(2−4sinθ)2dθ.
The correct answer is B. Choice A is missing the required 1/2 factor from the polar area formula, choice C describes the outer loop's extra region instead, and choice D swaps the coefficients inside the formula for r.
12 down, 6 to go — over halfway there.
Explore 1,500+ Free AP Calculus BC Problems∫0e2 du/u2 =
The integrand 1/u2 is undefined at u=0, so rewrite as a limit: limε→0+∫εe2u−2du = limε→0+[−1/u]εe2.
= limε→0+[−1/e2 + 1/ε].
As ε→0+, 1/ε → ∞, so the limit does not exist as a finite number.
The correct answer is D.
∫19 dt/∛(t−1) =
The integrand is undefined at t=1 (the lower bound). Let u = t−1, so the integral becomes ∫08u−1/3du (a p-integral with p=1/3<1, which converges near 0).
Find the antiderivative: ∫u−1/3du = (3/2)u2/3.
Evaluate: (3/2)[u2/3]08 = (3/2)(82/3−0) = (3/2)(4) = 6 (since 81/3=2, so 82/3=4).
The correct answer is B.
∫110 dx/(x−2)2/3 =
The integrand is undefined at x=2, which lies inside [1,10] — split the integral there. Let u=x−2 throughout.
Left piece: ∫12dx/(x−2)2/3 = ∫−10u−2/3du = [3u1/3]−10 = 0−3(−1) = 3 (since exponent 2/3<1, this converges).
Right piece: ∫210dx/(x−2)2/3 = ∫08u−2/3du = [3u1/3]08 = 3(2)−0 = 6.
Total: 3+6 = 9. The correct answer is B.
15 down, 3 to go — almost there.
Explore 1,500+ Free AP Calculus BC Problems∫15 dx/(x−2)2 =
The integrand is undefined at x=2, which lies inside [1,5]. Check the piece from 1 to 2: with u=x−2, this becomes ∫−10u−2du, a p-integral with p=2.
Since p=2 ≥ 1, this piece diverges near the singularity: limε→0−[−1/u]−1ε blows up to infinity as ε→0.
Since one piece of the split integral already diverges, the entire integral diverges — there's no need to check the other piece.
The correct answer is D.
∫0π/2 sin(x) dx/√(2−2cos x) =
The integrand is undefined at x=0 (since 2−2cos(0)=0), making this improper at the lower bound.
Let u = 2−2cos x, so du = 2sin x dx → sin x dx = du/2. Convert the bounds: at x=0, u=0; at x=π/2, u=2.
The integral becomes ∫02(du/2)/√u = (1/2)[2√u]02 = √2−0.
Since this limit exists as a finite number, the integral converges to √2. The correct answer is C.
Find the area in the first quadrant under the curve y = e−3x.
The first-quadrant region under this curve runs from x=0 to x=∞: Area = ∫0∞e−3xdx.
Find the antiderivative: ∫e−3xdx = −(1/3)e−3x.
Evaluate as a limit: [−(1/3)e−3x]0∞ = 0−(−1/3).
= 1/3. The correct answer is A.
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