AP Calculus AB Quiz: Volumes With Cross Sections Triangles Semicircles

What this quiz covers

This quiz focuses on Volumes With Cross Sections Triangles Semicircles, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\displaystyle \int_{-3}^{3} \frac{\sqrt3}{4}\,(9-x^2)^2\,dx$
2. $\displaystyle \int_{0}^{9} \frac{\sqrt3}{4}\,(2\sqrt{9-y})^2\,dy$ (correct answer)
3. $\displaystyle \int_{0}^{9} \frac12\,(2\sqrt{9-y})^2\,dy$
4. $\displaystyle \int_{0}^{9} \frac{\sqrt3}{16}\,(2\sqrt{9-y})^2\,dy$
5. $\displaystyle \int_{0}^{9} \frac{\pi}{8}\,(2\sqrt{9-y})^2\,dy$

Explanation: This problem involves equilateral triangular cross-sections perpendicular to the y-axis. The base is bounded by y = 9 - x² and y = 0, which gives x² = 9 - y, so x = ±√(9 - y) for 0 ≤ y ≤ 9. At each y-value, the triangle's side length equals the horizontal distance between the curves: 2√(9 - y). For an equilateral triangle with side s, the area is (√3/4)s², so each cross-section has area (√3/4)(2√(9 - y))² = (√3/4) · 4(9 - y) = √3(9 - y). The volume integral is ∫{0}^{9} √3(9 - y) dy, which equals ∫{0}^{9} (√3/4)(2√(9 - y))² dy. A common error is using the wrong triangle formula, such as (1/2) for right triangles (choice C). When working with cross-sections perpendicular to a different axis, express all dimensions in terms of that integration variable.

Question 2

1. $\displaystyle \int_{0}^{2} \frac{\pi}{8}\,(2x-x^2)\,dx$
2. $\displaystyle \int_{0}^{2} \frac{\pi}{8}\,(2x-x^2)^2\,dx$ (correct answer)
3. $\displaystyle \int_{0}^{2} \pi\,(2x-x^2)^2\,dx$
4. $\displaystyle \int_{0}^{2} \frac12\,(2x-x^2)^2\,dx$
5. $\displaystyle \int_{0}^{2} \frac{\sqrt3}{4}\,(2x-x^2)^2\,dx$

Explanation: This problem requires finding volume when semicircular cross-sections are perpendicular to the x-axis. The base is bounded by y = 2x and y = x² for 0 ≤ x ≤ 2, where these curves intersect at x = 0 and x = 2. At each x-value, the semicircle has diameter equal to the vertical distance between curves: 2x - x². For a semicircle with diameter d, the area is (π/8)d², so each cross-section has area (π/8)(2x - x²)². The volume integral is ∫_{0}^{2} (π/8)(2x - x²)² dx. Common errors include using the full circle formula π(r²) = (π/4)d² (choice C) or using formulas for other shapes like triangles. Remember that semicircle area = (π/8) × (diameter)², not (π/2) × radius² or other variations.

Question 3

1. $\displaystyle \int_{0}^{2} \frac{\pi}{8}\,(4-2x)^2\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{2} \frac{\pi}{8}\,(4-2x)\,dx$
3. $\displaystyle \int_{0}^{2} \frac{\pi}{4}\,(4-2x)^2\,dx$
4. $\displaystyle \int_{0}^{2} \frac{1}{2}\,(4-2x)^2\,dx$
5. $\displaystyle \int_{0}^{2} \pi\,(4-2x)^2\,dx$

Explanation: This problem involves computing the volume of a solid with semicircular cross-sections perpendicular to the x-axis, a key skill in understanding non-rectangular cross-sections. The diameter of each semicircle is the vertical distance between the curves y=4-x and y=x, which is 4-2x. The area of a semicircle is (1/2) π r², where r = (4-2x)/2. Therefore, the area simplifies to π (4-2x)² / 8. The volume is the integral of this area from x=0 to x=2. A common mistake is to use π (4-2x) / 8, as in choice B, which forgets to square the diameter in the formula. Remember, for any cross-section shape, determine the base length, apply the area formula correctly, and integrate along the perpendicular axis.

Question 4

1. $\displaystyle \int_{0}^{1} \frac{\sqrt{3}}{4}\,\ln(1+x)\,dx$
2. $\displaystyle \int_{0}^{1} \frac{\pi}{8}\,(\ln(1+x))^2\,dx$
3. $\displaystyle \int_{0}^{1} \frac{\sqrt{3}}{4}\,(\ln(1+x))^2\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{1} \frac{1}{2}\,(\ln(1+x))^2\,dx$
5. $\displaystyle \int_{0}^{1} \frac{\sqrt{3}}{4}\,(1+x)^2\,dx$

Explanation: This problem involves computing the volume of a solid with equilateral triangular cross-sections perpendicular to the x-axis, a key skill in understanding non-rectangular cross-sections. The side length of each triangle is the vertical distance between the curves y=ln(1+x) and y=0, which is ln(1+x). The area of an equilateral triangle is (√3/4) s², where s = ln(1+x). Therefore, the area simplifies to (√3/4) (ln(1+x))². The volume is the integral of this area from x=0 to x=1. A common mistake is to use (√3/4) ln(1+x), as in choice A, which forgets to square the side length. Remember, for any cross-section shape, determine the base length, apply the area formula correctly, and integrate along the perpendicular axis.

Question 5

1. $\displaystyle \int_{0}^{\pi} \frac{\pi}{4}\,\sin^2 x\,dx$
2. $\displaystyle \int_{0}^{\pi} \frac{\pi}{8}\,\sin^2 x\,dx$ (correct answer)
3. $\displaystyle \int_{0}^{\pi} \frac{1}{2}\,\sin^2 x\,dx$
4. $\displaystyle \int_{0}^{\pi} \pi\,\sin^2 x\,dx$
5. $\displaystyle \int_{0}^{\pi} \frac{\pi}{8}\,\sin x\,dx$

Explanation: This problem requires finding volume when cross-sections are semicircles perpendicular to the x-axis. The base is between y = sin x and y = 0 on [0,π], so the diameter at each x equals sin x. For a semicircle with diameter d, the area is A = (π/8)d², which gives A(x) = (π/8)sin²x. The volume integral is ∫_{0}^{π} (π/8)sin²x dx. Option A incorrectly uses π/4, which would result from confusing the semicircle formula with A = (π/2)r² where r = d/2. The consistent approach is to identify the cross-section shape, determine its area formula based on the appropriate dimension from the base region, then integrate.

Question 6

1. $\displaystyle \int_{0}^{2} \frac{\pi}{4}\,(2x-x^2)^2\,dx$
2. $\displaystyle \int_{0}^{2} \pi\,(2x-x^2)^2\,dx$
3. $\displaystyle \int_{0}^{2} \frac{\pi}{8}\,(2x-x^2)^2\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{2} \frac{1}{2}\,(2x-x^2)^2\,dx$
5. $\displaystyle \int_{0}^{2} \frac{\pi}{4}\,(2x-x^2)\,dx$

Explanation: This problem requires finding volume with semicircular cross-sections perpendicular to the x-axis. The base lies between y = 2x and y = x², so the diameter at each x is (2x - x²). For a semicircle with diameter d, the area is A = (π/8)d², giving us A(x) = (π/8)(2x - x²)². The volume integral is ∫_{0}^{2} (π/8)(2x - x²)² dx. Option A incorrectly uses π/4 instead of π/8, which would be the coefficient for a semicircle if we mistakenly used A = (π/2)r² with r = d/2. The strategy remains consistent: identify the cross-section shape, apply its area formula using the base dimension as the key measurement, then integrate.

Question 7

1. $\displaystyle \int_{-2}^{2} \frac{\pi}{8}\,(4-y^2)^2\,dy$ (correct answer)
2. $\displaystyle \int_{-2}^{2} \frac{\pi}{8}\,(4-y)^2\,dy$
3. $\displaystyle \int_{-2}^{2} \pi\,(4-y^2)^2\,dy$
4. $\displaystyle \int_{-2}^{2} \frac12\,(4-y^2)^2\,dy$
5. $\displaystyle \int_{0}^{4} \frac{\pi}{8}\,(4-x)^2\,dx$

Explanation: This problem requires finding volume with semicircular cross-sections perpendicular to the y-axis. The base is bounded by x = y² and x = 4 for -2 ≤ y ≤ 2. At each y-value, the semicircle has diameter equal to the horizontal distance from the parabola to the vertical line: 4 - y². For a semicircle with diameter d, the area is (π/8)d², so each cross-section has area (π/8)(4 - y²)². The volume integral is ∫_{-2}^{2} (π/8)(4 - y²)² dy. A common mistake would be to use 4 - y instead of 4 - y² (choice B), forgetting that x = y² not x = y, or using the full circle formula (choice C). Always carefully identify which curve gives which boundary and express the diameter in terms of the integration variable.

Question 8

1. $\displaystyle \int_{0}^{9} \frac{\sqrt3}{4}\,x\,dx$
2. $\displaystyle \int_{0}^{9} \frac12\,(\sqrt{x})^2\,dx$
3. $\displaystyle \int_{0}^{9} \frac{\sqrt3}{4}\,(\sqrt{x})^2\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{9} \pi(\sqrt{x})^2\,dx$
5. $\displaystyle \int_{0}^{9} \frac{\sqrt3}{16}\,(\sqrt{x})^2\,dx$

Explanation: This problem involves finding the volume of a solid with equilateral triangular cross-sections perpendicular to the x-axis. The base region lies between y = √x and y = 0 on the interval [0, 9]. At each x-value, the cross-section is an equilateral triangle with side length equal to the height of the region, which is √x. For an equilateral triangle with side length s, the area is (√3/4)s², so each cross-section has area (√3/4)(√x)². The volume integral is ∫{0}^{9} (√3/4)(√x)² dx = ∫{0}^{9} (√3/4)x dx. A tempting error would be to use the formula for a right triangle (½ base × height), giving choice B. Remember that for non-rectangular cross-sections, you must use the specific area formula for that shape, then express it in terms of the varying dimension from the base region.

Question 9

1. $\displaystyle \int_{0}^{\pi} \frac{\pi}{8}\,(\sin y)^2\,dy$ (correct answer)
2. $\displaystyle \int_{0}^{\pi} \frac{\pi}{8}\,\sin y\,dy$
3. $\displaystyle \int_{0}^{\pi} \frac{\pi}{4}\,(\sin y)^2\,dy$
4. $\displaystyle \int_{0}^{\pi} \frac{1}{2}\,(\sin y)^2\,dy$
5. $\displaystyle \int_{0}^{\pi} \pi\,(\sin y)^2\,dy$

Explanation: The skill here is to compute volumes of solids where cross-sections perpendicular to an axis are non-rectangular shapes like semicircles or equilateral triangles. The diameter of each semicircular cross-section is the distance between x=0 and x=sin y, which is sin y. The area of a semicircle is (π/8) times the square of the diameter, so the cross-sectional area is (π/8)(sin y)². Thus, the volume is given by integrating this area from y = 0 to y = π. A tempting distractor is choice E, which uses π(sin y)², treating the cross-sections as full circles instead of semicircles. In general, to find volumes with non-rectangular cross-sections, express the area of the shape in terms of the varying dimension and integrate along the axis of orientation.

Question 10

1. $\displaystyle \int_{-2}^{2} \frac{\sqrt{3}}{4}\, (4-x^2)\, dx$
2. $\displaystyle \int_{-2}^{2} \frac{\sqrt{3}}{4}\, (4-x^2)^2\, dx$ (correct answer)
3. $\displaystyle \int_{-2}^{2} \frac{1}{2}\, (4-x^2)^2\, dx$
4. $\displaystyle \int_{-2}^{2} \frac{\pi}{8}\, (4-x^2)^2\, dx$
5. $\displaystyle \int_{-2}^{2} \frac{\sqrt{3}}{4}\, (5-x^2)^2\, dx$

Explanation: The skill here is to compute volumes of solids where cross-sections perpendicular to an axis are non-rectangular shapes like semicircles or equilateral triangles. The side length of each equilateral triangular cross-section is the distance between $y=5-x^2$ and $y=1$, which is $5 - x^2 - 1 = 4 - x^2$. The area of an equilateral triangle is $\frac{\sqrt{3}}{4}$ times the square of the side length, so the cross-sectional area is $\frac{\sqrt{3}}{4} (4 - x^2)^2$. Thus, the volume is given by integrating this area from $x = -2$ to $x = 2$. A tempting distractor is choice C, which uses $\frac{1}{2} (4 - x^2)^2$, confusing the area formula with that of a right triangle. In general, to find volumes with non-rectangular cross-sections, express the area of the shape in terms of the varying dimension and integrate along the axis of orientation.

Question 11

1. $\displaystyle \int_{0}^{1} \frac{\pi}{8}\,\frac{1}{1+x^2}\,dx$
2. $\displaystyle \int_{0}^{1} \frac{\pi}{8}\,\left(\frac{1}{1+x^2}\right)^2\,dx$ (correct answer)
3. $\displaystyle \int_{0}^{1} \frac{\pi}{4}\,\left(\frac{1}{1+x^2}\right)^2\,dx$
4. $\displaystyle \int_{0}^{1} \frac{1}{2}\,\left(\frac{1}{1+x^2}\right)^2\,dx$
5. $\displaystyle \int_{0}^{1} \pi\,\left(\frac{1}{1+x^2}\right)^2\,dx$

Explanation: This problem involves computing the volume of a solid with semicircular cross-sections perpendicular to the x-axis, a key skill in understanding non-rectangular cross-sections. The diameter of each semicircle is the vertical distance between the curves y=1/(1+x²) and y=0, which is 1/(1+x²). The area of a semicircle is (1/2) π r², where r = (1/(1+x²))/2. Therefore, the area simplifies to π (1/(1+x²))² / 8. The volume is the integral of this area from x=0 to x=1. A common mistake is to use π (1/(1+x²)) / 8, as in choice A, which forgets to square the diameter in the formula. Remember, for any cross-section shape, determine the base length, apply the area formula correctly, and integrate along the perpendicular axis.

Question 12

1. $\displaystyle \int_{0}^{1} \frac{\sqrt{3}}{4}\,(1-y)^2\,dy$
2. $\displaystyle \int_{0}^{1} \frac{\sqrt{3}}{4}\,(1-(1-y))^2\,dy$ (correct answer)
3. $\displaystyle \int_{0}^{1} \frac{\sqrt{3}}{4}\,(1-(1-y))\,dy$
4. $\displaystyle \int_{0}^{1} \frac{\pi}{8}\,(y)^2\,dy$
5. $\displaystyle \int_{0}^{1} \frac{1}{2}\,(y)^2\,dy$

Explanation: This problem involves computing the volume of a solid with equilateral triangular cross-sections perpendicular to the y-axis, a key skill in understanding non-rectangular cross-sections. The side length of each triangle is the horizontal distance between the curves x=1 and x=1-y, which is y. The area of an equilateral triangle is (√3/4) s², where s = y. Therefore, the area simplifies to (√3/4) y². The volume is the integral of this area from y=0 to y=1. A common mistake is to use (√3/4) (1-y)², as in choice A, which reverses the distance calculation. Remember, for any cross-section shape, determine the base length, apply the area formula correctly, and integrate along the perpendicular axis.

Question 13

1. $\displaystyle \int_{0}^{1} \frac{\pi}{8}\,(1-x)\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{1} \frac{\pi}{8}\,\sqrt{1-x}\,dx$
3. $\displaystyle \int_{0}^{1} \frac{\pi}{4}\,(1-x)\,dx$
4. $\displaystyle \int_{0}^{1} \frac{1}{2}\,(1-x)\,dx$
5. $\displaystyle \int_{0}^{1} \pi\,(1-x)\,dx$

Explanation: This problem involves computing the volume of a solid with semicircular cross-sections perpendicular to the x-axis, a key skill in understanding non-rectangular cross-sections. The diameter of each semicircle is the vertical distance between the curves y=√(1-x) and y=0, which is √(1-x). The area of a semicircle is (1/2) π r², where r = √(1-x)/2. Therefore, the area simplifies to π (1-x) / 8. The volume is the integral of this area from x=0 to x=1. A common mistake is to use π √(1-x) / 8, as in choice B, which forgets to square the radius properly. Remember, for any cross-section shape, determine the base length, apply the area formula correctly, and integrate along the perpendicular axis.

Question 14

1. $\displaystyle \int_{0}^{\pi} (\sin x)^2\,dx$
2. $\displaystyle \int_{0}^{\pi} \frac12\,(\sin x)^2\,dx$ (correct answer)
3. $\displaystyle \int_{0}^{\pi} \frac{\sqrt3}{4}\,(\sin x)^2\,dx$
4. $\displaystyle \int_{0}^{\pi} \frac{\pi}{8}\,(\sin x)^2\,dx$
5. $\displaystyle \int_{0}^{\pi} \frac12\,\sin x\,dx$

Explanation: This problem involves finding volume with right isosceles triangular cross-sections perpendicular to the x-axis, where the legs are in the base. The base region is between y = sin x and y = 0 on [0, π]. At each x-value, the triangle has legs of length sin x (since it's isosceles with legs in the base). For a right triangle with legs of length ℓ, the area is (1/2)ℓ², so each cross-section has area (1/2)(sin x)². The volume integral is ∫_{0}^{π} (1/2)(sin x)² dx. A common error would be to use the equilateral triangle formula (√3/4)s², giving choice C, or to forget to square the leg length, giving choice E. The key insight is that "right isosceles triangle with legs in the base" means both legs equal the base dimension, and the area formula is half the square of the leg length.

Question 15

1. $\displaystyle \int_{0}^{\pi/4} \frac{\sqrt3}{4}\,(\cos x-\sin x)^2\,dx$
2. $\displaystyle \int_{0}^{\pi/4} \frac{\pi}{8}\,(\cos x-\sin x)^2\,dx$
3. $\displaystyle \int_{0}^{\pi/4} \frac12\,(\cos x-\sin x)^2\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{\pi/4} \frac12\,(\cos x+\sin x)^2\,dx$
5. $\displaystyle \int_{0}^{\pi/4} (\cos x-\sin x)^2\,dx$

Explanation: This problem involves right isosceles triangular cross-sections perpendicular to the x-axis. The base is between y = cos x and y = sin x on [0, π/4], where cos x ≥ sin x throughout this interval. At each x-value, the triangle has legs equal to the vertical distance: cos x - sin x. For a right isosceles triangle with leg length ℓ, the area is (1/2)ℓ², so each cross-section has area (1/2)(cos x - sin x)². The volume integral is ∫_{0}^{π/4} (1/2)(cos x - sin x)² dx. A common error would be to add instead of subtract (choice D), or to use a different triangle formula like (√3/4) for equilateral triangles. The key is recognizing that right isosceles triangles have area = (1/2) × (leg length)², where the leg length is the base dimension.

Question 16

1. $\displaystyle \int_{0}^{2} \frac{\pi}{8}\,(3x)^2\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{2} \frac{\pi}{8}\,(3x)\,dx$
3. $\displaystyle \int_{0}^{2} \frac{\pi}{4}\,(3x)^2\,dx$
4. $\displaystyle \int_{0}^{2} \frac{1}{2}\,(3x)^2\,dx$
5. $\displaystyle \int_{0}^{2} \pi\,(3x)^2\,dx$

Explanation: This problem involves computing the volume of a solid with semicircular cross-sections perpendicular to the x-axis, a key skill in understanding non-rectangular cross-sections. The diameter of each semicircle is the vertical distance between the curves y=3x and y=0, which is 3x. The area of a semicircle is (1/2) π r², where r = (3x)/2. Therefore, the area simplifies to π (3x)² / 8. The volume is the integral of this area from x=0 to x=2. A common mistake is to use π (3x) / 8, as in choice B, which forgets to square the diameter in the formula. Remember, for any cross-section shape, determine the base length, apply the area formula correctly, and integrate along the perpendicular axis.

Question 17

1. $\displaystyle \int_{0}^{2} \frac{\pi}{8}\,(2x-x^2)\,dx$
2. $\displaystyle \int_{0}^{2} \frac{\pi}{8}\,(2x-x^2)^2\,dx$ (correct answer)
3. $\displaystyle \int_{0}^{2} \pi\,(2x-x^2)^2\,dx$
4. $\displaystyle \int_{0}^{2} \frac{1}{2}\,(2x-x^2)^2\,dx$
5. $\displaystyle \int_{0}^{2} \frac{\pi}{4}\,(2x-x^2)^2\,dx$

Explanation: The skill here is to compute volumes of solids where cross-sections perpendicular to an axis are non-rectangular shapes like semicircles or equilateral triangles. The diameter of each semicircular cross-section is the distance between y=2x and y=x², which is 2x - x². The area of a semicircle is (π/8) times the square of the diameter, so the cross-sectional area is (π/8)(2x - x²)². Thus, the volume is given by integrating this area from x = 0 to x = 2. A tempting distractor is choice C, which uses π(2x - x²)², treating the cross-sections as full circles instead of semicircles. In general, to find volumes with non-rectangular cross-sections, express the area of the shape in terms of the varying dimension and integrate along the axis of orientation.

Question 18

1. $\displaystyle \int_{-2}^{2} \frac{\pi}{8}\,\left(1-\frac{x^2}{4}\right)^2\,dx$ (correct answer)
2. $\displaystyle \int_{-2}^{2} \frac{\pi}{8}\,\left(1-\frac{x^2}{4}\right)\,dx$
3. $\displaystyle \int_{-2}^{2} \frac{\pi}{4}\,\left(1-\frac{x^2}{4}\right)^2\,dx$
4. $\displaystyle \int_{-2}^{2} \frac{1}{2}\,\left(1-\frac{x^2}{4}\right)^2\,dx$
5. $\displaystyle \int_{-2}^{2} \pi\,\left(1-\frac{x^2}{4}\right)^2\,dx$

Explanation: This problem involves computing the volume of a solid with semicircular cross-sections perpendicular to the x-axis, a key skill in understanding non-rectangular cross-sections. The diameter of each semicircle is the vertical distance between the curves $y=1$ and $y=\frac{x^2}{4}$, which is $1 - \frac{x^2}{4}$. The area of a semicircle is $\frac{1}{2} \pi r^2$, where $r = \frac{1 - \frac{x^2}{4}}{2}$. Therefore, the area simplifies to $\pi \left(1 - \frac{x^2}{4}\right)^2 / 8$. The volume is the integral of this area from x=-2 to x=2. A common mistake is to use $\pi \left(1 - \frac{x^2}{4}\right) / 8$, as in choice B, which forgets to square the diameter in the formula. Remember, for any cross-section shape, determine the base length, apply the area formula correctly, and integrate along the perpendicular axis.

Question 19

1. $\displaystyle \int_{1}^{e} \frac{\sqrt{3}}{4}\,\ln(x)\,dx$
2. $\displaystyle \int_{1}^{e} \frac{\pi}{8}\,(\ln x)^2\,dx$
3. $\displaystyle \int_{1}^{e} \frac{\sqrt{3}}{4}\,(\ln x)^2\,dx$ (correct answer)
4. $\displaystyle \int_{0}^{1} \frac{\sqrt{3}}{4}\,(e^y)^2\,dy$
5. $\displaystyle \int_{1}^{e} \frac{1}{2}\,(\ln x)^2\,dx$

Explanation: The skill here is to compute volumes of solids where cross-sections perpendicular to an axis are non-rectangular shapes like semicircles or equilateral triangles. The side length of each equilateral triangular cross-section is the distance between y=ln x and y=0, which is ln x. The area of an equilateral triangle is (√3/4) times the square of the side length, so the cross-sectional area is (√3/4)(ln x)². Thus, the volume is given by integrating this area from x = 1 to x = e. A tempting distractor is choice E, which uses (1/2)(ln x)², confusing the area formula with that of a right triangle. In general, to find volumes with non-rectangular cross-sections, express the area of the shape in terms of the varying dimension and integrate along the axis of orientation.

Question 20

1. $\displaystyle \int_{0}^{\pi} \frac{\sqrt{3}}{4}\,(1-\sin x)^2\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{\pi} \frac{\sqrt{3}}{4}\,(1-\sin x)\,dx$
3. $\displaystyle \int_{0}^{\pi} \frac{1}{2}\,(1-\sin x)^2\,dx$
4. $\displaystyle \int_{0}^{\pi} \frac{\pi}{8}\,(1-\sin x)^2\,dx$
5. $\displaystyle \int_{0}^{\pi} \frac{\sqrt{3}}{4}\,(\sin x)^2\,dx$

Explanation: This problem involves computing the volume of a solid with equilateral triangular cross-sections perpendicular to the x-axis, a key skill in understanding non-rectangular cross-sections. The side length of each triangle is the vertical distance between the curves y=1 and y=sin x, which is 1 - sin x. The area of an equilateral triangle is (√3/4) s², where s = 1 - sin x. Therefore, the area simplifies to (√3/4) (1 - sin x)². The volume is the integral of this area from x=0 to x=π. A common mistake is to use (√3/4) (1 - sin x), as in choice B, which forgets to square the side length. Remember, for any cross-section shape, determine the base length, apply the area formula correctly, and integrate along the perpendicular axis.