AP Calculus AB Practice Quiz: Volumes with Cross Sections: Triangles and Semicircles
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 9 questions to check your progress.
Question 1 of 9
All Questions (9)
A) ∫[a, b] A'(x) dx
B) π ∫[a, b] A(x) dx
C) ∫[a, b] A(x) dx
D) ∫[a, b] (A(x))^2 dx
Correct Answer: C
The volume of a solid with a known cross-sectional area A(x) from x=a to x=b is found by integrating the area function over that interval. This is the fundamental application of definite integrals for calculating volumes of solids with known cross-sections. [cite: 2913, 2919]
A) π(4 - x)^2
B) (π/2)(4 - x)^2
C) (π/8)(4 - x)^2
D) (π/4)(4 - x)^2
Correct Answer: C
The diameter of the semicircle at a given x is the length of the vertical segment from the x-axis to the line y = 4 - x. So, the diameter d = 4 - x. The radius is r = d/2 = (4 - x)/2. The area of a semicircle is (1/2)πr^2. Substituting the radius, we get A(x) = (1/2)π * ((4 - x)/2)^2 = (π/8)(4 - x)^2. [cite: 2921]
A) ∫[0, 4] (√3/4)x dx
B) ∫[0, 4] (√3/2)x dx
C) ∫[0, 4] (1/2)x dx
D) ∫[0, 4] (√3/4)√x dx
Correct Answer: A
The side length 's' of the equilateral triangle at a given x is the value of the function, s = √x. The area of an equilateral triangle with side s is A = (s^2 * √3) / 4. Substituting s = √x, we get A(x) = ((√x)^2 * √3) / 4 = (x * √3) / 4. The volume is the integral of this area function from x=0 to x=4, which is ∫[0, 4] (√3/4)x dx. [cite: 2920]
A) ∫[0, 2] (π/8)(2x - x^2)^2 dx
B) ∫[0, 2] (π/2)(2x - x^2)^2 dx
C) ∫[0, 2] (π/8)(x^2 - 2x)^2 dx
D) ∫[0, 2] π(2x - x^2)^2 dx
Correct Answer: A
First, find the intersection points: 2x = x^2 gives x=0 and x=2. The diameter 'd' of a semicircle at a given x is the vertical distance between the curves, d = (top curve) - (bottom curve) = 2x - x^2. The radius is r = d/2 = (2x - x^2)/2. The area of a semicircle is A(x) = (1/2)πr^2 = (1/2)π * ((2x - x^2)/2)^2 = (π/8)(2x - x^2)^2. The volume is the integral of A(x) from 0 to 2. [cite: 2919, 2921]
A) ∫[0, 3] (1/2)y^2 dy
B) ∫[0, 3] y^4 dy
C) ∫[0, 3] (1/2)y^4 dy
D) ∫[0, 9] (1/2)x dx
Correct Answer: C
Since the cross-sections are perpendicular to the y-axis, we integrate with respect to y. The length of the base of the triangle at a given y is the horizontal distance from the y-axis to the curve x = y^2. This length is s = y^2. Since the cross-sections are isosceles right triangles with a leg on the base, both legs have length s. The area of the triangle is A(y) = (1/2) * base * height = (1/2) * s * s = (1/2)s^2. Substituting s = y^2, we get A(y) = (1/2)(y^2)^2 = (1/2)y^4. The volume is the integral of A(y) from y=0 to y=3. [cite: 2920]
A) ∫[-1, 1] (1/2)(1 - x^2)^2 dx
B) ∫[-1, 1] (1/4)(1 - x^2)^2 dx
C) ∫[-1, 1] (1 - x^2)^2 dx
D) ∫[-1, 1] (1/4)(1 - x^2) dx
Correct Answer: B
The base of the solid is on the x-axis from x=-1 to x=1. The length of the hypotenuse 'h' of the triangle at a given x is h = y = 1 - x^2. For an isosceles right triangle with hypotenuse h, the two equal legs 's' satisfy s^2 + s^2 = h^2, so 2s^2 = h^2. The area of the triangle is A = (1/2) * s * s = (1/2)s^2 = (1/2)(h^2/2) = h^2/4. Substituting h = 1 - x^2, the area is A(x) = (1/4)(1 - x^2)^2. The volume is the integral of this area function from x=-1 to x=1. [cite: 2920]
A) The height of a single triangular cross-section at position x.
B) The length of the base of a single triangular cross-section at position x.
C) The volume of a single, infinitesimally thin triangular slice at position x.
D) The area of a single triangular cross-section at position x.
Correct Answer: D
In the formula for the volume of a solid with known cross-sections, V = ∫[a, b] A(x) dx, the function A(x) represents the area of the cross-section at a specific value of x. The integral sums the volumes of infinitesimally thin slices, where each slice has a volume of A(x)dx. [cite: 2913, 2920]
A) V1 = ∫ (√3/4)s(x)^2 dx and V2 = ∫ (π/8)s(x)^2 dx. Since √3/4 > π/8, V1 > V2.
B) V1 = ∫ (√3/4)s(x)^2 dx and V2 = ∫ (π/4)s(x)^2 dx. Since π/4 > √3/4, V2 > V1.
C) V1 = ∫ (1/2)s(x)^2 dx and V2 = ∫ (π/2)s(x)^2 dx. Since π/2 > 1/2, V2 > V1.
D) V1 = ∫ (√3/4)s(x)^2 dx and V2 = ∫ (π/8)s(x)^2 dx. Since π/8 > √3/4, V2 > V1.
Correct Answer: A
For S1 (equilateral triangle), the side is s(x), so the area is A1(x) = (s(x)^2 * √3)/4. For S2 (semicircle), the diameter is s(x), so the radius is s(x)/2, and the area is A2(x) = (1/2)π(s(x)/2)^2 = (π/8)s(x)^2. The volumes are the integrals of these area functions. To compare them, we compare the constants: √3/4 ≈ 1.732/4 = 0.433 and π/8 ≈ 3.141/8 = 0.393. Since √3/4 > π/8, the volume V1 will be greater than V2. [cite: 2920, 2921]
A) The radius of the semicircular cross-section.
B) The area of the semicircular cross-section.
C) The circumference of the semicircular cross-section.
D) The diameter of the semicircular cross-section.
Correct Answer: D
The term f(x) - g(x) represents the vertical distance between the two bounding curves, which is the length of the segment on the base of the solid. For a semicircular cross-section with its flat side on the base, this length is defined as the diameter. The radius would be (f(x) - g(x))/2, and the area would be (π/8)(f(x) - g(x))^2. [cite: 2919, 2921]