Volumes with Cross Sections: Triangles/Semicircles - AP Calculus BC

Which integral gives the volume: base bounded by $y=\sqrt{x}$ , $y=0$ , $x=4$ ; cross sections $\perp$ to the $x$ -axis are equilateral triangles?
Answer: $V=\displaystyle\int_{0}^{4}\frac{\sqrt{3}}{4}\,(\sqrt{x})^2\,dx$ This problem involves calculating the volume of a solid using cross-sectional areas, specifically equilateral triangles perpendicular to the x-axis. The base region is bounded by $y = \sqrt{x}$ , $y = 0$ , and $x = 4$ , where the height at each $x$ is $\sqrt{x}$ , serving as the side length of the equilateral triangle. The area of an equilateral triangle with side $s$ is $\frac{\sqrt{3}}{4} s^2$ , so here it is $\frac{\sqrt{3}}{4} (\sqrt{x})^2 = \frac{\sqrt{3}}{4} x$ . Integrating this from 0 to 4 gives the volume as in choice C. A tempting distractor like choice A omits the squaring of the side length, treating it as a linear dimension rather than area. For transferable strategy, always express the cross-section's area formula in terms of the integration variable, using the appropriate geometric formula like $\frac{\sqrt{3}}{4} s^2$ for equilateral triangles.
Base region is between $y=\frac{x}{1+x}$ and $y=0$ on $0\le x\le1$ ; cross sections perpendicular to the $x$ -axis are equilateral triangles. Which integral gives the volume?
Answer: $\displaystyle \int_{0}^{1}\frac{\sqrt3}{4}\left(\frac{x}{1+x}\right)^2\,dx$ This problem requires cross-sectional volume reasoning to find the volume of a solid with equilateral triangular cross-sections perpendicular to the x-axis. The base region is between y=x/(1+x) and y=0, so S = x/(1+x) is the side. The area is (√3/4) S², yielding ∫ (√3/4) (x/(1+x))² dx from 0 to 1. This handles the rational function. A tempting distractor like choice C uses (√3/4)(x/(1+x)) without squaring, incorrect for area. Always determine the cross-section's area formula based on the shape and the base dimension, then integrate along the perpendicular axis.
Base region is bounded by $y=\frac{1}{1+x^2}$ and $y=0$ on $0\le x\le1$ ; cross sections perpendicular to the $x$ -axis are semicircles. Which setup gives the volume?
Answer: $\displaystyle \int_{0}^{1}\frac{\pi}{8}\left(\frac{1}{1+x^2}\right)^2\,dx$ This problem requires cross-sectional volume reasoning to find the volume of a solid with semicircular cross-sections perpendicular to the x-axis. The base region is between y=1/(1+x²) and y=0, so D = 1/(1+x²) is the diameter. The area is (π/8) D², yielding ∫ (π/8) (1/(1+x²))² dx from 0 to 1. This integrates the bell-shaped base. A tempting distractor like choice C uses (π/8)(1/(1+x²)), failing to square. Always determine the cross-section's area formula based on the shape and the base dimension, then integrate along the perpendicular axis.
Select the volume integral: base region between $y=x$ and $y=x^2$ ; cross sections $\perp$ to the $x$ -axis are right isosceles triangles with leg in base.
Answer: $V=\displaystyle\int_{0}^{1}\frac{1}{2}\big(x-x^2\big)^2\,dx$ This problem involves calculating the volume of a solid using cross-sectional areas, specifically right isosceles triangles perpendicular to the $x$ -axis. The base region is between $y = x$ and $y = x^2$ from $x = 0$ to $1$ , where the vertical distance at each $x$ is $x - x^2$ , serving as one leg of the right isosceles triangle. The area is $\frac{1}{2}$ times leg squared, so $\frac{1}{2} (x - x^2)^2$ . Integrating this from $0$ to $1$ gives the volume as in choice A. A tempting distractor like choice B forgets the $1/2$ factor for the triangle area, doubling the result. For transferable strategy, always express the cross-section's area formula in terms of the integration variable, adapting formulas like $\frac{1}{2} l^2$ for right isosceles triangles with equal legs.
Select the correct setup: base between $y=2x$ and $y=x^2$ ; cross sections ⟂ $x$ -axis are right isosceles triangles with leg as vertical distance.
Answer: $\displaystyle \int_{0}^{2} \frac{1}{2}\big(2x-x^2\big)^2\,dx$ This problem requires using the method of cross-sectional volumes to find the volume of a solid where the cross sections are right isosceles triangles perpendicular to the x-axis. The leg length of each right isosceles triangle is the vertical distance between y=2x and y=x², which is 2x - x². The area of a right isosceles triangle with leg length d is (1/2)d², so here the area is (1/2)(2x - x²)². The volume is the integral of this area from x=0 to x=2. A tempting distractor is choice B, which uses (2x - x²)² without the 1/2 factor, overestimating the area by treating it as a square instead of a triangle. Always remember to apply the (1/2) base times height formula for triangular areas and confirm the shape specifications when setting up volume integrals.
Choose the correct integral: base between $y=9-x^2$ and $y=0$ ; cross sections perpendicular to the $x$ -axis are semicircles.
Answer: $V=\displaystyle\int_{-3}^{3}\frac{\pi}{8}\,(9-x^2)^2\,dx$ This problem involves calculating the volume of a solid using cross-sectional areas, specifically semicircles perpendicular to the x-axis. The base region is between y = 9 - x² and y = 0 from x = -3 to 3, where the height at each x is 9 - x², serving as the diameter of the semicircle. The radius is (9 - x²)/2, and the area is (1/2) π [(9 - x²)/2]² = π (9 - x²)² / 8. Integrating this from -3 to 3 gives the volume as in choice C. A tempting distractor like choice E uses the full circle area, forgetting to halve it for semicircles. For transferable strategy, always express the cross-section's area formula in terms of the integration variable, halving for semicircles as needed.
What integral gives the volume when the base is bounded by $y=x^2$ and $y=4$ , with semicircular cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{0}^{2} \frac{\pi}{8}\big(4-x^2\big)^2\,dx$ This problem requires finding the volume of a solid with semicircular cross sections perpendicular to the x-axis. The base region is bounded by y = x² and y = 4, which intersect when x² = 4, giving x = ±2. At each x-value, the diameter of the semicircle equals the vertical distance between the curves: d = 4 - x². For a semicircle with diameter d, the area is A = (π/8)d², so here A = (π/8)(4 - x²)². The common error is using the full circle formula A = πr² = (π/4)d², forgetting to halve it for a semicircle. Remember: semicircle area = (1/2)πr² = (π/8)d² when d is the diameter.
Select the correct volume integral: base bounded by $x=0$ , $y=0$ , and $y=1-x$ ; ⟂ $x$ -axis semicircles.
Answer: $\displaystyle \int_{0}^{1} \frac{\pi}{8}(1-x)^2\,dx$ This problem requires using the method of cross-sectional volumes to find the volume of a solid where the cross sections are semicircles perpendicular to the x-axis. The base of each semicircle is the vertical distance between y=1-x and y=0, which is 1 - x. The radius is (1 - x)/2, so the area is (1/2)π[(1 - x)/2]² = (π/8)(1 - x)². The volume is the integral of this from x=0 to x=1. A tempting distractor is choice D, which uses π(1 - x)² but treats the cross sections as full circles without halving the area. Always remember to adjust for semicircles by using half the circle area and correctly halving the diameter for the radius in volume setups.
Select the correct volume setup: base enclosed by $y=\sin x$ and $y=0$ on $[0,\pi]$ , semicircle cross sections perpendicular to $x$ .
Answer: $\displaystyle \int_{0}^{\pi}\frac{\pi}{8}\sin^2 x\,dx$ This problem requires setting up the volume integral for semicircular cross sections perpendicular to the x-axis. The base is bounded by y = sin x and y = 0 on [0, π], so the base width at each x is sin x. For semicircles with diameter equal to the base width sin x, the radius is (sin x)/2. The area of each semicircle is (1/2)πr² = (1/2)π[(sin x)/2]² = (π/8)sin² x. Choice C incorrectly uses the full circle formula π(sin x)² instead of accounting for the semicircle. Remember that when the diameter equals the base width w, the semicircle area is (π/8)w².
Which integral represents volume: base between $y=x^3$ and $y=x$ for $0\le x\le1$ ; ⟂ $x$ -axis equilateral triangles.
Answer: $\displaystyle \int_{0}^{1} \frac{\sqrt{3}}{4}\big(x-x^3\big)^2\,dx$ This problem requires using the method of cross-sectional volumes to find the volume of a solid where the cross sections are equilateral triangles perpendicular to the x-axis. The base of each equilateral triangle is the vertical distance between y=x and y=x³, which is x - x³ for 0 ≤ x ≤ 1. The area is (√3/4)(x - x³)², capturing the squared side length. The volume is the integral from x=0 to x=1. A tempting distractor is choice B, which uses (√3/4)(x - x³) without squaring, underestimating by treating it as a linear area instead of quadratic. Always remember to square the side length in the (√3/4)s² formula for equilateral triangles when computing cross-sectional volumes.
Base region is between $x=\sin y$ and $x=0$ for $0\le y\le\pi$ ; cross sections perpendicular to the $y$ -axis are semicircles. Which integral gives the volume?
Answer: $\displaystyle \int_{0}^{\pi}\frac{\pi}{8}(\sin y)^2\,dy$ This problem requires cross-sectional volume reasoning to find the volume of a solid with semicircular cross-sections perpendicular to the y-axis. The base region is between x=sin y and x=0, so the horizontal distance at each y is D = sin y, the diameter. The area is (π/8) D², leading to ∫ (π/8) (sin y)² dy from 0 to π. This integrates over the full oscillatory base. A tempting distractor like choice C uses (π/8) sin y linearly, mistaking area for a non-quadratic form. Always determine the cross-section's area formula based on the shape and the base dimension, then integrate along the perpendicular axis.
Base region is enclosed by $y=x^3$ and $y=0$ on $0\le x\le2$ ; cross sections perpendicular to the $x$ -axis are semicircles. Which setup gives the volume?
Answer: $\displaystyle \int_{0}^{2}\frac{\pi}{8}(x^3)^2\,dx$ This problem requires cross-sectional volume reasoning to find the volume of a solid with semicircular cross-sections perpendicular to the x-axis. The base region is between y=x³ and y=0, so D = x³ is the diameter. The area is (π/8) D², leading to ∫ (π/8) (x³)² dx from 0 to 2. This accounts for the cubic growth. A tempting distractor like choice C integrates (π/8) x³, failing to square for area. Always determine the cross-section's area formula based on the shape and the base dimension, then integrate along the perpendicular axis.
Base region is bounded by $y=2x-x^2$ and $y=0$ on $0\le x\le2$ ; cross sections perpendicular to the $x$ -axis are semicircles. Which setup gives the volume?
Answer: $\displaystyle \int_{0}^{2}\frac{\pi}{8}(2x-x^2)^2\,dx$ This problem requires cross-sectional volume reasoning to find the volume of a solid with semicircular cross-sections perpendicular to the x-axis. The base region is between y=2x - x² and y=0, so D = 2x - x² is the diameter. The area is (π/8) D², leading to ∫ (π/8) (2x - x²)² dx from 0 to 2. This captures the parabolic shape. A tempting distractor like choice C integrates (π/8)(2x - x²), incorrect for area. Always determine the cross-section's area formula based on the shape and the base dimension, then integrate along the perpendicular axis.
Which integral gives the volume if the base is $0\le x\le 3$ , $0\le y\le 2x$ , with right-triangle cross sections perpendicular to $x$ ?
Answer: $\displaystyle \int_{0}^{3}\frac{1}{2}(2x)^2\,dx$ This problem involves finding the volume when cross sections perpendicular to the x-axis are right triangles. The base region has 0 ≤ x ≤ 3 and 0 ≤ y ≤ 2x, so at each x-value, the base width is 2x. For right triangle cross sections with legs equal to the base width, both legs have length 2x. The area of a right triangle is (1/2) × base × height = (1/2)(2x)(2x) = (1/2)(2x)². Choice D incorrectly omits the factor of 1/2 from the triangle area formula. The transferable strategy is that for isosceles right triangles with legs equal to the base width w, the area is (1/2)w².
Which integral gives volume: base bounded by $x=y^2$ and $x=4$ ; cross sections ⟂ $y$ -axis are semicircles.
Answer: $\displaystyle \int_{-2}^{2} \frac{\pi}{8}\big(4-y^2\big)^2\,dy$ This problem requires using the method of cross-sectional volumes to find the volume of a solid where the cross sections are semicircles perpendicular to the y-axis. The base of each semicircle is the horizontal distance between x=4 and x=y², which is 4 - y². The radius is half of this distance, or (4 - y²)/2, so the area is (1/2)π[(4 - y²)/2]² = (π/8)(4 - y²)². The volume is the integral of this area from y=-2 to y=2. A tempting distractor is choice D, which uses π(4 - y²)² but incorrectly assumes full circles and mishandles the radius for semicircles. Always remember to halve the circle area and divide the diameter by two for the radius when dealing with semicircular cross sections in volume setups.
Which integral represents the volume: base between $y=3-x$ and $y=0$ on $[0,3]$ , cross sections perpendicular to $x$ are equilateral triangles?
Answer: $\displaystyle \int_{0}^{3}\frac{\sqrt{3}}{4}(3-x)^2\,dx$ This volume problem features equilateral triangle cross sections perpendicular to the x-axis. The base is between y = 3 - x and y = 0 on [0, 3], giving base width (3 - x) at each x. For an equilateral triangle with side length s, the area is (√3/4)s². Since the base width equals the side length, the area is (√3/4)(3 - x)². Choice B incorrectly uses the formula for a right triangle (1/2 × base × height) instead of the equilateral triangle formula. Remember that equilateral triangle area = (√3/4) × (side length)² when the side equals the base width.
Choose the correct integral: base bounded by $y=3$ and $y=x^2$ ; cross sections ⟂ $x$ -axis are equilateral triangles.
Answer: $\displaystyle \int_{-\sqrt{3}}^{\sqrt{3}} \frac{\sqrt{3}}{4}\big(3-x^2\big)^2\,dx$ This problem requires using the method of cross-sectional volumes to find the volume of a solid where the cross sections are equilateral triangles perpendicular to the x-axis. The base of each equilateral triangle is the vertical distance between y=3 and y=x², which is 3 - x². The area is (√3/4)(3 - x²)², integrated from x=-√3 to x=√3. This setup accounts for the full symmetric region bounded by the curves. A tempting distractor is choice D, which uses (√3/3)(3 - x²)² but applies an incorrect coefficient for the equilateral triangle area formula. Always remember to use (√3/4)s² for equilateral triangles and ensure integration limits capture the entire base region for accurate volume calculations.
Base region is bounded by $y=5$ and $y=x^2$ on $-\sqrt5\le x\le\sqrt5$ ; cross sections perpendicular to the $x$ -axis are isosceles right triangles with leg equal to the vertical distance. Which setup gives the volume?
Answer: $\displaystyle \int_{-\sqrt5}^{\sqrt5}\frac12(5-x^2)^2\,dx$ This problem requires cross-sectional volume reasoning to find the volume of a solid with isosceles right triangular cross-sections perpendicular to the x-axis. The base region is bounded by y=5 and y=x², so the vertical distance at each x is L = 5 - x², which serves as the leg length of the triangle. The area of each isosceles right triangle is (1/2) L², leading to the volume integral ∫ (1/2) (5 - x²)² dx from -√5 to √5. Since the region is symmetric about the y-axis, the limits cover the full base appropriately. A tempting distractor like choice D uses (1/2)(5 - x²) without squaring, which incorrectly computes area instead of integrating the triangular cross-sectional areas. Always determine the cross-section's area formula based on the shape and the base dimension, then integrate along the perpendicular axis.
Pick the volume setup: base bounded by $y=\sin x$ and $y=0$ on $0\le x\le \pi$ ; ⟂ $x$ -axis semicircles.
Answer: $\displaystyle \int_{0}^{\pi} \frac{\pi}{8}\big(\sin x\big)^2\,dx$ This problem requires using the method of cross-sectional volumes to find the volume of a solid where the cross sections are semicircles perpendicular to the x-axis. The base of each semicircle is the vertical distance between y=sin x and y=0, which is sin x. The radius is (sin x)/2, so the area is (1/2)π[(sin x)/2]² = (π/8)(sin x)². The volume is the integral of this area from x=0 to x=π. A tempting distractor is choice B, which uses (π/4)(sin x)² but incorrectly doubles the coefficient by mishandling the semicircle halving. Always remember to square the radius and apply the semicircle factor of 1/2 correctly when formulating areas for volume integrals.
Which integral represents the volume: base bounded by $y=2x$ and $y=0$ for $0\le x\le 3$ ; cross sections ⟂ $x$ -axis are equilateral triangles?
Answer: $V=\displaystyle\int_{0}^{3}\frac{\sqrt{3}}{4}\,(2x)^2\,dx$ This problem involves calculating the volume of a solid using cross-sectional areas, specifically equilateral triangles perpendicular to the x-axis. The base region is bounded by y = 2x and y = 0 from x = 0 to 3, where the height at each x is 2x, serving as the side length of the equilateral triangle. The area is (√3/4) (2x)² = √3 x². Integrating this from 0 to 3 gives the volume as in choice B. A tempting distractor like choice A uses the side without squaring, mistaking for a non-area dimension. For transferable strategy, always express the cross-section's area formula in terms of the integration variable, using squared terms for areas like in (√3/4) s².

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