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

Find the volume setup: base bounded by $y=2x$ and $y=x^2$ for $0\le x\le 2$ , semicircular cross sections perpendicular to the $x$ -axis.
Answer: $\displaystyle \int_{0}^{2} \frac{\pi}{8}\,(2x-x^2)^2\,dx$ 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.
What integral gives the volume if the base is bounded by $x=0$ and $x=3y$ for $-1\le y\le 2$ , with equilateral triangular cross sections perpendicular to the $y$ -axis?
Answer: $\displaystyle \int_{-1}^{2} \frac{\sqrt{3}}{4}\,(3y)^2\,dy$ 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=0 and x=3y, which is 3y (noting y from -1 to 2, but area squares it). The area of an equilateral triangle is (√3/4) s², where s = 3y. Therefore, the area simplifies to (√3/4) (3y)². The volume is the integral of this area from y=-1 to y=2. A common mistake is to use (√3/4) (3y), 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.
What integral gives the volume if the base is bounded by $y=1$ and $y=\sin x$ on $0\le x\le \pi$ , with equilateral triangular cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{0}^{\pi} \frac{\sqrt{3}}{4}\,(1-\sin x)^2\,dx$ 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.
What integral gives the volume if the base is bounded by $x=1-y$ and $x=1$ for $0\le y\le 1$ , with equilateral triangular cross sections perpendicular to the $y$ -axis?
Answer: $\displaystyle \int_{0}^{1} \frac{\sqrt{3}}{4}\,(1-(1-y))^2\,dy$ 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.
What integral gives the volume when the base is bounded by $y=\cos x$ and $y=0$ on $[0,\tfrac{\pi}{2}]$ , with semicircular cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{0}^{\pi/2} \frac{\pi}{8}\,\cos^2 x\,dx$ This problem requires finding volume when cross-sections are semicircles perpendicular to the x-axis. The base is bounded by y = cos x and y = 0 on [0, π/2], so the diameter at each x equals cos x. For a semicircle with diameter d, the area is A = (π/8)d², which gives A(x) = (π/8)cos²x. The volume integral is ∫_{0}^{π/2} (π/8)cos²x dx. Option E incorrectly uses cos x instead of cos²x, failing to square the diameter in the area formula. Remember that all area formulas for these shapes involve squaring the key dimension: semicircles use (π/8)d², triangles use coefficients times side².
What integral gives the volume if the base is bounded by $y=2x$ and $y=x^2$ for $0\le x\le 2$ , with semicircular cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{0}^{2} \frac{\pi}{8}\,(2x-x^2)^2\,dx$ 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.
What integral gives the volume if the base is bounded by $y=\sqrt{4-x^2}$ and $y=0$ on $-2\le x\le 2$ , with semicircular cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{-2}^{2} \frac{\pi}{8}\,(4-x^2)\,dx$ 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=√(4-x²) and y=0, which is √(4-x²). The area of a semicircle is (π/8) times the square of the diameter, so the cross-sectional area is (π/8)(√(4-x²))² = (π/8)(4 - x²). Thus, the volume is given by integrating this area from x = -2 to x = 2. A tempting distractor is choice E, which uses π(4 - x²), 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.
Find the volume setup: base bounded by $x=0$ , $y=0$ , and $y=6-2x$ ; equilateral triangular cross sections perpendicular to the $x$ -axis.
Answer: $\displaystyle \int_{0}^{3} \frac{\sqrt3}{4}\,(6-2x)^2\,dx$ This problem involves equilateral triangular cross-sections perpendicular to the x-axis. The base is a triangular region bounded by x = 0, y = 0, and y = 6 - 2x, which intersects the x-axis at x = 3. At each x-value between 0 and 3, the triangle's side length equals the height of the region: 6 - 2x. For an equilateral triangle with side s, the area is (√3/4)s², so each cross-section has area (√3/4)(6 - 2x)². The volume integral is ∫_{0}^{3} (√3/4)(6 - 2x)² dx. A common error would be to integrate to x = 6 (choice C) by confusing the y-intercept with the x-intercept, or to use (√3/16) thinking you need to account for the 2 in (6 - 2x) differently. The strategy is to find where the region exists along your integration axis, then apply the correct area formula.
What integral gives the volume if the base is bounded by $x=0$ and $x=4-y^2$ for $-2\le y\le 2$ , with semicircular cross sections perpendicular to the $y$ -axis?
Answer: $\displaystyle \int_{-2}^{2} \frac{\pi}{8}\,(4-y^2)^2\,dy$ This problem involves computing the volume of a solid with semicircular cross-sections perpendicular to the y-axis, a key skill in understanding non-rectangular cross-sections. The diameter of each semicircle is the horizontal distance between the curves x=0 and x=4-y², which is 4-y². The area of a semicircle is (1/2) π r², where r = (4-y²)/2. Therefore, the area simplifies to π (4-y²)² / 8. The volume is the integral of this area from y=-2 to y=2. A common mistake is to use π (4-y²) / 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.
Find the volume setup: base between $y=\sqrt{x}$ and $y=0$ on $0\le x\le 9$ , equilateral triangular cross sections perpendicular to $x$ -axis.
Answer: $\displaystyle \int_{0}^{9} \frac{\sqrt3}{4}\,(\sqrt{x})^2\,dx$ 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.
What integral gives the volume if the base is bounded by $y=2$ and $y=x^2$ on $[-\sqrt{2},\sqrt{2}]$ , with equilateral triangular cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} \frac{\sqrt{3}}{4}\,(2-x^2)^2\,dx$ This problem involves finding volume with equilateral triangular cross-sections perpendicular to the x-axis. The base is bounded by y = 2 and y = x², so at each x-value, the side length of the equilateral triangle is (2 - x²). For an equilateral triangle with side length s, the area is A = (√3/4)s². Therefore, A(x) = (√3/4)(2 - x²)². The volume integral is ∫_{-√2}^{√2} (√3/4)(2 - x²)² dx. Option E incorrectly uses √3/2 instead of √3/4, which would double the area of each cross-section. The key to these problems is knowing the exact area formula for each shape and applying it consistently to the varying dimension from the base region.
What is the correct volume setup if the base is bounded by $y=\sin x$ and $y=0$ on $0\le x\le \pi$ and cross sections perpendicular to the $x$ -axis are semicircles?
Answer: $\displaystyle \int_{0}^{\pi} \frac{\pi}{8}(\sin x)^2\,dx$ This problem requires finding volume with semicircular cross-sections perpendicular to the x-axis. The base region is bounded by y = sin x and y = 0 on [0, π], so the height at each x is sin x. For semicircles, this height is the diameter, making the radius (sin x)/2. The area of each semicircle is (1/2)π[(sin x)/2]² = (π/8)(sin x)². A tempting error is to use (π/4)(sin x)² by forgetting to halve the circle area formula for a semicircle. When working with semicircles, always remember to use half the area of a full circle, and carefully identify whether the base measurement represents the radius or diameter.
What integral gives the volume if the base is bounded by $x=0$ and $x=2-\cos y$ on $0\le y\le \pi$ , with semicircular cross sections perpendicular to the $y$ -axis?
Answer: $\displaystyle \int_{0}^{\pi} \frac{\pi}{8}\,(2-\cos y)^2\,dy$ This problem involves computing the volume of a solid with semicircular cross-sections perpendicular to the y-axis, a key skill in understanding non-rectangular cross-sections. The diameter of each semicircle is the horizontal distance between the curves x=0 and x=2-cos y, which is 2-cos y. The area of a semicircle is (1/2) π r², where r = (2-cos y)/2. Therefore, the area simplifies to π (2-cos y)² / 8. The volume is the integral of this area from y=0 to y=π. A common mistake is to use π (2-cos y) / 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.
What integral gives the volume if the base is bounded by $x=0$ and $x=\sqrt{9-y^2}$ for $-3\le y\le 3$ , with semicircular cross sections perpendicular to the $y$ -axis?
Answer: $\displaystyle \int_{-3}^{3} \frac{\pi}{8}\,(9-y^2)\,dy$ This problem involves computing the volume of a solid with semicircular cross-sections perpendicular to the y-axis, a key skill in understanding non-rectangular cross-sections. The diameter of each semicircle is the horizontal distance between the curves x=0 and x=√(9-y²), which is √(9-y²). The area of a semicircle is (1/2) π r², where r = √(9-y²)/2. Therefore, the area simplifies to π (9-y²) / 8. The volume is the integral of this area from y=-3 to y=3. A common mistake is to use π √(9-y²) / 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.
What integral gives the volume if the base is bounded by $y=x$ and $y=x^3$ for $0\le x\le 1$ , with semicircular cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{0}^{1} \frac{\pi}{8}\,(x-x^3)^2\,dx$ 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=x and y=x³, which is x - x³. The area of a semicircle is (π/8) times the square of the diameter, so the cross-sectional area is (π/8)(x - x³)². Thus, the volume is given by integrating this area from x = 0 to x = 1. A tempting distractor is choice C, which uses π(x - x³)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.
Find the volume setup: base bounded by $x=y^2$ and $x=4$ for $-2\le y\le 2$ , semicircular cross sections perpendicular to the $y$ -axis.
Answer: $\displaystyle \int_{-2}^{2} \frac{\pi}{8}\,(4-y^2)^2\,dy$ 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.
What integral gives the volume if the base is bounded by $y=\sin x$ and $y=0$ on $0\le x\le \pi$ , with semicircular cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{0}^{\pi} \frac{\pi}{8}\,(\sin x)^2\,dx$ 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=sin x and y=0, which is sin x. The area of a semicircle is (π/8) times the square of the diameter, so the cross-sectional area is (π/8)(sin x)². Thus, the volume is given by integrating this area from x = 0 to x = π. A tempting distractor is choice E, which uses π(sin x)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.
What integral gives the volume if the base is bounded by $y=e^x$ and $y=0$ on $0\le x\le 1$ , with semicircular cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{0}^{1} \frac{\pi}{8}\,(e^x)^2\,dx$ 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= $e^x$ and y=0, which is $e^x$ . The area of a semicircle is (π/8) times the square of the diameter, so the cross-sectional area is (π/8)( $e^x$ )². Thus, the volume is given by integrating this area from x = 0 to x = 1. A tempting distractor is choice E, which uses π( $e^x$ )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.
What is the correct volume setup if the base is bounded by $y=1-x$ and $y=0$ on $0\le x\le 1$ and cross sections perpendicular to the $x$ -axis are right isosceles triangles with hypotenuse as the base segment?
Answer: $\displaystyle \int_{0}^{1} \frac{1}{4}(1-x)^2\,dx$ This problem features right isosceles triangles with the hypotenuse as the base segment, perpendicular to the x-axis. The base region has height 1 - x at each x-value, which equals the hypotenuse length. For a right isosceles triangle with hypotenuse h, each leg has length h/√2, and the area is (1/2)(h/√2)² = h²/4. With h = 1 - x, the area is (1 - x)²/4. A common error is using (1/2)(1 - x)² by treating the hypotenuse like a leg in the standard triangle formula. When the hypotenuse is given for a right isosceles triangle, use the relationship leg = hypotenuse/√2 to find the area.
What integral gives the volume if the base is bounded by $y=\frac{x^2}{4}$ and $y=1$ for $-2 \le x \le 2$ , with semicircular cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{-2}^{2} \frac{\pi}{8}\,\left(1-\frac{x^2}{4}\right)^2\,dx$ 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.

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