Volumes with Cross Sections: Triangles/Semicircles
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AP Calculus BC › Volumes with Cross Sections: Triangles/Semicircles
Choose the correct integral: base between $y=9-x^2$ and $y=0$; cross sections perpendicular to the $x$-axis are semicircles.
$V=\displaystyle\int_{-3}^{3}\frac{\pi}{4},(9-x^2)^2,dx$
$V=\displaystyle\int_{-3}^{3}\pi,(9-x^2)^2,dx$
$V=\displaystyle\int_{-3}^{3}\frac{\pi}{8},(9-x^2)^2,dx$
$V=\displaystyle\int_{0}^{9}\frac{\pi}{8},(9-y^2)^2,dy$
$V=\displaystyle\int_{-3}^{3}\frac{\pi}{8},(9-x^2),dx$
Explanation
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.
Find the correct volume setup: base bounded by $y=1-x^2$ and $y=0$ on $-1,1$, semicircle cross sections perpendicular to $x$.
$\displaystyle \int_{-1}^{1}\pi(1-x^2)^2,dx$
$\displaystyle \int_{-1}^{1}\frac{1}{2}(1-x^2)^2,dx$
$\displaystyle \int_{-1}^{1}\frac{\pi}{2}(1-x^2),dx$
$\displaystyle \int_{0}^{1}\frac{\pi}{8}(1-y)^2,dy$
$\displaystyle \int_{-1}^{1}\frac{\pi}{8}(1-x^2)^2,dx$
Explanation
This volume problem features semicircular cross sections perpendicular to the x-axis. The base is bounded by y = 1 - x² and y = 0 on [-1, 1], giving base width (1 - x²) at each x. For semicircles with diameter (1 - x²), the radius is (1 - x²)/2, and the area is (1/2)πr² = (1/2)π[(1 - x²)/2]² = (π/8)(1 - x²)². Choice B incorrectly uses a linear expression (π/2)(1 - x²) instead of the squared term needed for area. The consistent strategy is that semicircle area = (π/8) × (diameter)² when diameter equals base width.
Find the correct volume setup: base region between $y=x$ and $y=x^2$ on $0,1$, semicircle cross sections perpendicular to $x$.
$\displaystyle \int_{0}^{1}\frac{1}{2}(x-x^2)^2,dx$
$\displaystyle \int_{0}^{1}\pi(x-x^2)^2,dx$
$\displaystyle \int_{0}^{1}\frac{\pi}{8}(x-x^2)^2,dx$
$\displaystyle \int_{0}^{1}\frac{\pi}{8}(x^2-x)^2,dx$
$\displaystyle \int_{0}^{1}\frac{\pi}{2}(x-x^2),dx$
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.
Find the correct volume setup: base bounded by $y=x^2$ and $y=4$; cross sections perpendicular to the $x$-axis are semicircles.
$V=\displaystyle\int_{0}^{4}\frac{\pi}{8}\big(\sqrt{y}-y\big)^2,dy$
$V=\displaystyle\int_{-2}^{2}\pi\big(4-x^2\big)^2,dx$
$V=\displaystyle\int_{-2}^{2}\frac{\pi}{2}\big(4-x^2\big),dx$
$V=\displaystyle\int_{-2}^{2}\frac{\pi}{4}\big(4-x^2\big)^2,dx$
$V=\displaystyle\int_{-2}^{2}\frac{\pi}{8}\big(4-x^2\big)^2,dx$
Explanation
This problem involves calculating the volume of a solid using cross-sectional areas, specifically semicircles perpendicular to the x-axis. The base region is bounded by y = x² and y = 4, extending from x = -2 to x = 2, where the vertical distance at each x is 4 - x², serving as the diameter of the semicircle. The radius is then (4 - x²)/2, and the area of each semicircular cross-section is (1/2) π [(4 - x²)/2]² = π (4 - x²)² / 8. Integrating this area from -2 to 2 gives the volume setup in choice B. A tempting distractor like choice A uses the full circle area by omitting the 1/8 factor, overestimating the volume. For transferable strategy, always express the cross-section's area formula in terms of the integration variable, ensuring to account for shapes like semicircles by halving the full circle area.
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?
$V=\displaystyle\int_{0}^{2}\frac{\pi}{8},y^2,dy$
$V=\displaystyle\int_{0}^{2}\frac{\sqrt{3}}{4},y,dy$
$V=\displaystyle\int_{0}^{2}\frac{\sqrt{3}}{4},(2-y)^2,dy$
$V=\displaystyle\int_{0}^{2}\frac{1}{2},y^2,dy$
$V=\displaystyle\int_{0}^{2}\frac{\sqrt{3}}{4},y^2,dy$
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?
$V=\displaystyle\int_{0}^{3}\frac{1}{2},(2x)^2,dx$
$V=\displaystyle\int_{0}^{3}\frac{\pi}{8},(2x)^2,dx$
$V=\displaystyle\int_{0}^{6}\frac{\sqrt{3}}{4},y^2,dy$
$V=\displaystyle\int_{0}^{3}\frac{\sqrt{3}}{4},(2x)^2,dx$
$V=\displaystyle\int_{0}^{3}\frac{\sqrt{3}}{4},(2x),dx$
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?
$\displaystyle \int_{-\sqrt2}^{\sqrt2}\frac12(2-y^2)^2,dy$
$\displaystyle \int_{-\sqrt2}^{\sqrt2}\frac{\sqrt3}{2}(2-y^2)^2,dy$
$\displaystyle \int_{-\sqrt2}^{\sqrt2}\frac{\sqrt3}{4}(2-y^2),dy$
$\displaystyle \int_{0}^{2}\frac{\sqrt3}{4}(2-y^2)^2,dy$
$\displaystyle \int_{-\sqrt2}^{\sqrt2}\frac{\sqrt3}{4}(2-y^2)^2,dy$
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?
$\displaystyle \int_{-\sqrt5}^{\sqrt5}\frac12(5-x^2)^2,dx$
$\displaystyle \int_{-\sqrt5}^{\sqrt5}\frac{\sqrt3}{4}(5-x^2)^2,dx$
$\displaystyle \int_{-\sqrt5}^{\sqrt5}\frac14(5-x^2)^2,dx$
$\displaystyle \int_{0}^{\sqrt5}\frac12(5-x^2)^2,dx$
$\displaystyle \int_{-\sqrt5}^{\sqrt5}\frac12(5-x^2),dx$
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.
Choose the correct setup: base bounded by $x=1$ and $x=1+y^2$ for $-1\le y\le1$; ⟂ $y$-axis right isosceles triangles.
$\displaystyle \int_{-1}^{1} \big((1+y^2)-1\big)^2,dy$
$\displaystyle \int_{-1}^{1} \frac{1}{2}\big((1+y^2)-1\big),dy$
$\displaystyle \int_{-1}^{1} \frac{1}{2}\big((1+y^2)-1\big)^2,dy$
$\displaystyle \int_{0}^{1} \frac{1}{2}\big((1+y^2)-1\big)^2,dy$
$\displaystyle \int_{-1}^{1} \frac{\sqrt{3}}{4}\big((1+y^2)-1\big)^2,dy$
Explanation
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 y-axis. The leg length of each right isosceles triangle is the horizontal distance between x=1+y² and x=1, which is y². The area is (1/2)(y²)² = (1/2)(y⁴), simplified from (1/2)d² where d = y². The volume is the integral from y=-1 to y=1. A tempting distractor is choice A, which uses (y²)² without the 1/2, mistaking the shape for a square rather than a triangle. Always remember to apply the (1/2) factor for the area of right isosceles triangles and verify the distance orientation matches the axis of integration.
Find the correct volume setup: base bounded by $y=x^2$ and $y=4$; cross sections ⟂ $x$-axis are semicircles.
$\displaystyle \int_{-2}^{2} \frac{1}{2}\big(4-x^2\big)^2,dx$
$\displaystyle \int_{-2}^{2} \frac{\pi}{8}\big(4-x^2\big)^2,dx$
$\displaystyle \int_{-2}^{2} \frac{\pi}{2}\big(4-x^2\big),dx$
$\displaystyle \int_{-2}^{2} \pi\big(4-x^2\big)^2,dx$
$\displaystyle \int_{0}^{4} \frac{\pi}{8}\big(4-y\big)^2,dy$
Explanation
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 the curves y=4 and y=x², which is 4 - x². Since the cross sections are semicircles, the radius is half of this distance, or (4 - x²)/2. Therefore, the area of each cross section is (1/2)π[(4 - x²)/2]², simplifying to (π/8)(4 - x²)², and the volume is the integral of this area from x=-2 to x=2. A tempting distractor is choice A, which uses π(4 - x²)² but forgets to account for the semicircle by halving the full circle area and properly adjusting for the radius. Always remember to use the correct formula for the cross-sectional shape and ensure the dimensions match the base length when setting up volume integrals.