Volumes with Cross Sections: Triangles/Semicircles
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AP Calculus BC › Volumes with Cross Sections: Triangles/Semicircles
Find the correct volume setup: base region between $y=x$ and $y=x^2$ on $0,1$, semicircle cross sections perpendicular to $x$.
$\displaystyle \int_{0}^{1}\frac{\pi}{8}(x^2-x)^2,dx$
$\displaystyle \int_{0}^{1}\frac{1}{2}(x-x^2)^2,dx$
$\displaystyle \int_{0}^{1}\frac{\pi}{2}(x-x^2),dx$
$\displaystyle \int_{0}^{1}\pi(x-x^2)^2,dx$
$\displaystyle \int_{0}^{1}\frac{\pi}{8}(x-x^2)^2,dx$
Explanation
This problem requires finding the volume with semicircular cross sections perpendicular to the x-axis. The base is between y = x and y = x² on [0, 1], where x ≥ x² in this interval, so the base width is (x - x²). For semicircles with diameter (x - x²), the radius is (x - x²)/2, and the area is (1/2)πr² = (1/2)π[(x - x²)/2]² = (π/8)(x - x²)². Choice C incorrectly uses the full circle formula π(x - x²)² instead of the semicircle formula. The key is to use area = (π/8) × (diameter)² for semicircles when diameter equals base width.
Which integral gives the volume: base bounded by $x=y$ and $x=0$ for $0\le y\le 2$; cross sections ⟂ $y$-axis are equilateral triangles?
$V=\displaystyle\int_{0}^{2}\frac{1}{2},y^2,dy$
$V=\displaystyle\int_{0}^{2}\frac{\sqrt{3}}{4},y^2,dy$
$V=\displaystyle\int_{0}^{2}\frac{\sqrt{3}}{4},(2-y)^2,dy$
$V=\displaystyle\int_{0}^{2}\frac{\pi}{8},y^2,dy$
$V=\displaystyle\int_{0}^{2}\frac{\sqrt{3}}{4},y,dy$
Explanation
This problem involves calculating the volume of a solid using cross-sectional areas, specifically equilateral triangles perpendicular to the y-axis. The base region is bounded by x = y and x = 0 from y = 0 to 2, where the width at each y is y, serving as the side length of the equilateral triangle. The area is (√3/4) y². Integrating this from 0 to 2 gives the volume as in choice B. A tempting distractor like choice A uses y instead of y², neglecting the area squaring. For transferable strategy, always express the cross-section's area formula in terms of the integration variable, ensuring quadratic terms for areas like equilateral triangles.
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?
$V=\displaystyle\int_{0}^{3}\frac{\pi}{8},(2x)^2,dx$
$V=\displaystyle\int_{0}^{3}\frac{\sqrt{3}}{4},(2x),dx$
$V=\displaystyle\int_{0}^{6}\frac{\sqrt{3}}{4},y^2,dy$
$V=\displaystyle\int_{0}^{3}\frac{1}{2},(2x)^2,dx$
$V=\displaystyle\int_{0}^{3}\frac{\sqrt{3}}{4},(2x)^2,dx$
Explanation
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.
Base region is between $x=2$ and $x=y^2$ for $-\sqrt2\le y\le\sqrt2$; cross sections perpendicular to the $y$-axis are equilateral triangles. Which integral gives the volume?
$\displaystyle \int_{-\sqrt2}^{\sqrt2}\frac12(2-y^2)^2,dy$
$\displaystyle \int_{0}^{2}\frac{\sqrt3}{4}(2-y^2)^2,dy$
$\displaystyle \int_{-\sqrt2}^{\sqrt2}\frac{\sqrt3}{2}(2-y^2)^2,dy$
$\displaystyle \int_{-\sqrt2}^{\sqrt2}\frac{\sqrt3}{4}(2-y^2)^2,dy$
$\displaystyle \int_{-\sqrt2}^{\sqrt2}\frac{\sqrt3}{4}(2-y^2),dy$
Explanation
This problem requires cross-sectional volume reasoning to find the volume of a solid with equilateral triangular cross-sections perpendicular to the y-axis. The base region is between x=2 and x=y², so the horizontal distance at each y is S = 2 - y², which is the side length of the equilateral triangle. The area of each equilateral triangle is (√3/4) S², resulting in the volume integral ∫ (√3/4) (2 - y²)² dy from -√2 to √2. The symmetric limits ensure the full region is covered without duplication or omission. A tempting distractor like choice C omits the squaring and uses (√3/4)(2 - y²), which mistakes the area for a linear measure rather than quadratic. 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 $y=5$ 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?
$\displaystyle \int_{0}^{\sqrt5}\frac12(5-x^2)^2,dx$
$\displaystyle \int_{-\sqrt5}^{\sqrt5}\frac{\sqrt3}{4}(5-x^2)^2,dx$
$\displaystyle \int_{-\sqrt5}^{\sqrt5}\frac12(5-x^2)^2,dx$
$\displaystyle \int_{-\sqrt5}^{\sqrt5}\frac12(5-x^2),dx$
$\displaystyle \int_{-\sqrt5}^{\sqrt5}\frac14(5-x^2)^2,dx$
Explanation
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.
Which integral represents volume: base between $y=x^3$ and $y=x$ for $0\le x\le1$; ⟂ $x$-axis equilateral triangles.
$\displaystyle \int_{0}^{1} \frac{\sqrt{3}}{4}\big(x-x^3\big),dx$
$\displaystyle \int_{0}^{1} \frac{1}{2}\big(x-x^3\big)^2,dx$
$\displaystyle \int_{0}^{1} \frac{\sqrt{3}}{4}\big(x-x^3\big)^2,dx$
$\displaystyle \int_{0}^{1} \frac{\sqrt{3}}{4}\big(x^3-x\big)^2,dx$
$\displaystyle \int_{0}^{1} \frac{\sqrt{3}}{3}\big(x-x^3\big)^2,dx$
Explanation
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 $x=1$ and $x=1+y$ for $0\le y\le2$; cross sections perpendicular to the $y$-axis are equilateral triangles. Which integral gives the volume?
$\displaystyle \int_{0}^{2}\frac12(y)^2,dy$
$\displaystyle \int_{0}^{2}\frac{\sqrt3}{4}(y)^2,dy$
$\displaystyle \int_{1}^{3}\frac{\sqrt3}{4}(y)^2,dy$
$\displaystyle \int_{0}^{2}\frac{\sqrt3}{4}y,dy$
$\displaystyle \int_{0}^{2}\frac{\sqrt3}{2}(y)^2,dy$
Explanation
This problem requires cross-sectional volume reasoning to find the volume of a solid with equilateral triangular cross-sections perpendicular to the y-axis. The base region is between x=1 and x=1+y, so S = y is the side. The area is (√3/4) S², yielding ∫ (√3/4) y² dy from 0 to 2. This integrates the linear increase. A tempting distractor like choice C uses (√3/4) y without squaring, failing area calculation. 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 $y=2$ and $y=x$ on $0\le x\le2$; cross sections perpendicular to the $x$-axis are equilateral triangles. Which integral gives the volume?
$\displaystyle \int_{0}^{2}\frac{\sqrt3}{4}x^2,dx$
$\displaystyle \int_{0}^{2}\frac12(2-x)^2,dx$
$\displaystyle \int_{0}^{2}\frac{\sqrt3}{4}(2-x),dx$
$\displaystyle \int_{0}^{2}\frac{\sqrt3}{4}(2-x)^2,dx$
$\displaystyle \int_{0}^{2}\frac{\sqrt3}{2}(2-x)^2,dx$
Explanation
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=2 and y=x, so S = 2 - x is the side. The area is (√3/4) S², resulting in ∫ (√3/4) (2 - x)² dx from 0 to 2. This covers the linear decrease. A tempting distractor like choice C integrates (√3/4)(2 - x), 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 enclosed by $x=6$ and $x=2y$ for $0\le y\le3$; cross sections perpendicular to the $y$-axis are semicircles. Which setup gives the volume?
$\displaystyle \int_{0}^{6}\frac{\pi}{8}(6-2y)^2,dy$
$\displaystyle \int_{0}^{3}\pi(6-2y)^2,dy$
$\displaystyle \int_{0}^{3}\frac{\pi}{8}(6-2y)^2,dy$
$\displaystyle \int_{0}^{3}\frac{\pi}{4}(6-2y)^2,dy$
$\displaystyle \int_{0}^{3}\frac{\pi}{8}(6-2y),dy$
Explanation
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=6 and x=2y, so D = 6 - 2y is the diameter. The area is (π/8) D², resulting in ∫ (π/8) (6 - 2y)² dy from 0 to 3. This integrates the linear decrease. A tempting distractor like choice C uses (π/8)(6 - 2y) linearly, mistaking for non-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 between $y=\sqrt{2x}$ and $y=0$ on $0\le x\le2$; cross sections perpendicular to the $x$-axis are equilateral triangles. Which integral gives the volume?
$\displaystyle \int_{0}^{2}\frac12\big(\sqrt{2x}\big)^2,dx$
$\displaystyle \int_{0}^{2}\frac{\sqrt3}{2}\big(\sqrt{2x}\big)^2,dx$
$\displaystyle \int_{0}^{2}\frac{\sqrt3}{4}\sqrt{2x},dx$
$\displaystyle \int_{0}^{2}\frac{\sqrt3}{4}(2-\sqrt{2x})^2,dx$
$\displaystyle \int_{0}^{2}\frac{\sqrt3}{4}\big(\sqrt{2x}\big)^2,dx$
Explanation
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=√(2x) and y=0, so S = √(2x) is the side. The area is (√3/4) S², giving ∫ (√3/4) (√(2x))² dx from 0 to 2. This handles the square root base. A tempting distractor like choice C integrates (√3/4) √(2x), omitting square. Always determine the cross-section's area formula based on the shape and the base dimension, then integrate along the perpendicular axis.