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This deck focuses on Volumes With Cross Sections Triangles Semicircles, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.
Study Volumes With Cross Sections Triangles Semicircles in AP Calculus AB with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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What is the side/leg length s(x) if a cross section uses the base segment between y=f(x) and y=g(x)?
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s(x)=f(x)−g(x). Vertical distance between curves gives side/leg length at each x.
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This deck focuses on Volumes With Cross Sections Triangles Semicircles, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.
Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.
Answer: s(x)=f(x)−g(x). Vertical distance between curves gives side/leg length at each x.
Answer: A(x)=2π(x+1)2. Substitutes r=x+1 into semicircle area formula A=2πr2.
Answer: Volume = 21×b×integral of height dx. Base b is constant, height function is integrated.
Answer: Integrate the area function of the cross sections along the axis. Volume equals integral of cross-sectional areas.
Answer: Integrate the area of cross-sections along the axis. Standard approach for volume by cross-sections.
Answer: Integrate the area function of the cross sections along the axis. Volume equals integral of cross-sectional areas.
Answer: dy. Cross sections perpendicular to y-axis vary with y.
Answer: d(x)=x−x. Top curve minus bottom curve gives vertical distance.
Answer: Volume = 21×integral of Base×Height dx. Triangle cross-sectional area integrated along the axis.
Answer: A=21bh. Standard triangle area formula: half base times height.
Answer: V=∫ab8π(f(x)−g(x))2dx. Combines volume integral with semicircle area using diameter as base.
Answer: d(x)=f(x)−g(x). Vertical distance between curves gives diameter at each x.
Answer: Integrate the area of cross-sections along the axis. Standard approach for volume by cross-sections.
Answer: Volume = 21×b×integral of height dx. Base b is constant, height function is integrated.
Answer: A(x)=21(s(x))2. Substitutes leg function into right isosceles triangle area formula.
Answer: A=2πr2. Half the area of a full circle with radius r.
Answer: A(x)=2πx2. Substitutes d=2x into A=8πd2 to get 8π(2x)2.
Answer: Volume = 21×b×integral of h dx. Triangle area formula converted to integral form.
Answer: Volume = 21×b×integral of h dx. Triangle area formula converted to integral form.
Answer: A(x)=43(s(x))2. Substitutes side function into equilateral triangle area formula.
Answer: A(x)=8π(d(x))2. Substitutes diameter function into semicircle area formula.
Answer: A=43s2. Standard formula for equilateral triangle area using side length.
Answer: A=8πd2. Half of a circle's area πr2 where r=2d.
Answer: x∈[0,1]. Curves intersect where x=x2, so x=0 and x=1.
Answer: A(x)=43x. Substitutes s=x into A=43s2 to get 43x.
Answer: V=∫abA(x)dx. Integrates cross-sectional area along the axis to find total volume.
Answer: A=21s2. For isosceles right triangle, area is half the square of the leg.
Answer: Integrate the area of the cross section along the axis. Integration sums all cross-sectional areas.
Answer: Integrate the area function of the cross sections along the axis. Volume equals integral of cross-sectional areas.
Answer: Volume = 21×∫Base×Height dx. Triangle cross-sectional area integrated along the axis.
Answer: dx. Cross sections perpendicular to x-axis vary with x.
Answer: A(x)=21(3−x)2. Substitutes s=3−x into right isosceles triangle area formula.
Answer: Integrate the area function of the cross sections along the axis. Volume equals integral of cross-sectional areas.
Answer: Integrate the area of the cross section along the axis. Integration sums all cross-sectional areas.