Volumes with Cross Sections: Squares/Rectangles - AP Calculus BC

Region bounded by $x=1$ and $x=y^2+1$ for $0\le y\le 2$ ; rectangles perpendicular to $y$ -axis have height $3y$ : choose setup.
Answer: $\displaystyle \int_{0}^{2} 3y\big((y^2+1)-1\big)\,dy$ This problem involves finding volume with rectangular cross-sections perpendicular to the y-axis, where the width comes from the base region and height varies as 3y. The base is bounded by x = 1 and x = y² + 1 for 0 ≤ y ≤ 2. For any y-value, the width is (y²+1) - 1 = y². The cross-sectional area is width × height = y² × 3y = 3y³. However, examining choice A: 3y((y²+1)-1) = 3y(y²) = 3y³, which matches our calculation. Choice D incorrectly squares the width, treating cross-sections as squares rather than rectangles. The correct method for rectangular cross-sections is to find the base width, multiply by the variable height function, then integrate along the appropriate axis.
Base is bounded by $x=0$ and $x=1-y$ for $0\le y\le 1$ ; squares perpendicular to the $y$ -axis: which setup?
Answer: $\displaystyle \int_{0}^{1} (1-y)^2\,dy$ This problem involves computing volume with square cross-sections perpendicular to the y-axis. The base region is bounded by x = 0 and x = 1 - y for 0 ≤ y ≤ 1. For any y-value, the horizontal distance between these boundaries is (1 - y) - 0 = 1 - y. Since cross-sections are squares, this distance serves as the side length, so the area is (1 - y)². Choice B represents only the side length rather than the area of the square. The key method is to find the width of the base region at each y-value, use this as the square's side length, then integrate the squared expression along the y-axis.
Region bounded by $x=y^2$ and $x=4$ ; rectangular cross sections perpendicular to $y$ -axis have height $y$ : select correct volume setup.
Answer: $\displaystyle \int_{-2}^{2} y(4-y^2)\,dy$ This problem involves finding volume with rectangular cross-sections perpendicular to the y-axis, where the width comes from the base region and height varies as y. The base is bounded by x = y² and x = 4 for -2 ≤ y ≤ 2. For any y-value, the width is 4 - y². The cross-sectional area is width × height = (4-y²) × y = y(4-y²). Choice B incorrectly squares the width, treating it as square cross-sections rather than rectangles with height y. The correct approach for rectangular cross-sections is to multiply the base width by the specified variable height, then integrate this area function along the appropriate axis.
For the region bounded by $y=x^2$ and $y=4$ , cross sections perpendicular to the $x$ -axis are squares; which integral gives the volume?
Answer: $\displaystyle \int_{-2}^{2} \big(4-x^2\big)^2\,dx$ This problem requires finding the volume of a solid with square cross-sections perpendicular to the x-axis. The region is bounded by y = x² and y = 4, so the curves intersect at x = ±2. For any x-value in [-2,2], the vertical distance between the curves is 4 - x². Since cross-sections are squares, the side length equals this vertical distance, so the area of each square is (4 - x²)². Choice D represents the side length rather than the area, making it a common distractor. The correct strategy is: identify the side length of the square cross-section, square it to get area, then integrate over the appropriate interval.
Which integral represents the volume when the base is bounded by $x=1$ and $x=y+3$ for $0\le y\le2$ , square cross sections perpendicular to the $y$ -axis?
Answer: $\displaystyle \int_{0}^{2} (y+3-1)^2\,dy$ This problem involves cross-sectional volume reasoning to determine the volume with square cross sections perpendicular to the y-axis. The base region is bounded by x = 1 and x = y + 3 from y = 0 to y = 2. For each y, the side length of the square is the horizontal distance (y + 3) - 1 = y + 2. Therefore, the area is (y + 2)², and the volume is the integral from 0 to 2. A tempting distractor is choice E, which includes π, but that's for circular cross sections, not squares. In general, for volumes with cross sections perpendicular to an axis, identify the varying dimension as the side length or radius depending on the shape, square it for area, and integrate along the axis.
Region between $x=0$ and $x=\ln(y+1)$ for $0\le y\le e-1$ ; squares perpendicular to $y$ -axis: choose the integral.
Answer: $\displaystyle \int_{0}^{e-1} \big(\ln(y+1)\big)^2\,dy$ This problem involves finding volume with square cross-sections perpendicular to the y-axis. The region is between x = 0 and x = ln(y+1) for 0 ≤ y ≤ e-1. For any y-value in this interval, the horizontal distance between these boundaries is ln(y+1) - 0 = ln(y+1). Since cross-sections are squares, this distance serves as the side length, so the cross-sectional area is (ln(y+1))². Choice B gives only the side length without squaring, which would not provide the correct area for square cross-sections. The fundamental method is to determine the width of the base region at each y-value, square this distance to obtain the area of each square cross-section, then integrate along the y-axis.
Find the correct volume setup: base between $x=y^2$ and $x=4$ for $0\le y\le2$ , squares perpendicular to the $y$ -axis.
Answer: $\displaystyle \int_{0}^{2}\big(4-y^2\big)^2\,dy$ This problem requires finding the volume of a solid with square cross-sections perpendicular to the y-axis. The base region lies between x = y² and x = 4 for 0 ≤ y ≤ 2, so for each y-value, the side length of the square equals the horizontal distance: 4 - y². Since we need square cross-sections, the area of each square is (side length)² = (4 - y²)². The integration variable must be y since we're slicing perpendicular to the y-axis, and the limits are from y = 0 to y = 2. Choice B incorrectly uses just the side length without squaring it, while choice D incorrectly uses dx instead of dy. When slicing perpendicular to the y-axis, always integrate with respect to y and square the side length for square cross-sections.
Base bounded by $x=4-y^2$ and $x=0$ for $-2\le y\le 2$ ; squares perpendicular to $y$ -axis: choose the volume integral.
Answer: $\displaystyle \int_{-2}^{2} (4-y^2)^2\,dy$ This problem requires finding volume with square cross-sections perpendicular to the y-axis. The base is bounded by x = 4 - y² and x = 0 for -2 ≤ y ≤ 2. For any y-value in this interval, the horizontal distance between these boundaries is (4-y²) - 0 = 4 - y². Since cross-sections are squares, the side length equals 4 - y², making the cross-sectional area (4-y²)². Choice B represents only the side length rather than the area of the square cross-section. The systematic method is to determine the width of the parabolic region at each y-value, square this distance to obtain the area of each square cross-section, then integrate along the y-axis.
Set up the volume integral: base bounded by $y=\cos x$ and $y=\sin x$ for $0\le x\le \tfrac{\pi}{2}$ , squares perpendicular to $x$ -axis.
Answer: $\displaystyle \int_{0}^{\pi/2} (\cos x-\sin x)^2\,dx$ This problem involves cross-sectional volume reasoning to set up the volume integral with square cross sections perpendicular to the x-axis. The base region is bounded by y = cos x and y = sin x from x = 0 to x = π/2. At each x, the side length is the absolute difference |cos x - sin x|, but since it's squared, (cos x - sin x)² works. Thus, the volume is the integral of (cos x - sin x)² from 0 to π/2. A tempting distractor is choice E, which adds π, but this is incorrect for squares as it implies circular cross sections. In general, when finding volumes with known cross sections, determine the area function based on the shape and integrate it over the base interval.
Region bounded by $x=0$ and $x=2y$ for $0\le y\le 3$ ; rectangles perpendicular to $y$ -axis have height $5$ : choose setup.
Answer: $\displaystyle \int_{0}^{3} 5(2y)\,dy$ This problem involves finding volume with rectangular cross-sections perpendicular to the y-axis, where the width comes from the base region and height is fixed at 5. The base is bounded by x = 0 and x = 2y for 0 ≤ y ≤ 3. For any y-value, the width is 2y - 0 = 2y. The cross-sectional area is width × height = 2y × 5 = 5(2y) = 10y. However, looking at the choices, choice A matches the pattern 5(2y), which represents the structure but could be written more simply. Choice B incorrectly squares the width, treating cross-sections as squares rather than rectangles. The correct strategy for rectangular cross-sections is to multiply the base width by the specified height, then integrate over the appropriate interval.
Which integral represents the volume if the base is bounded by $y=\cos x$ and $y=\sin x$ on $[0,\pi/4]$ , squares perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{0}^{\pi/4} (\cos x-\sin x)^2\,dx$ This problem involves finding the volume with square cross-sections perpendicular to the x-axis. The base is bounded by y = cos x and y = sin x on [0, π/4], and since cos x > sin x on this interval, the side length is cos x - sin x. For square cross-sections, the area is (side length)² = (cos x - sin x)². The integral becomes ∫_{0}^{π/4} (cos x - sin x)² dx. Choice B incorrectly reverses the order to sin x - cos x, which would give negative values since sin x < cos x on this interval. When setting up cross-sectional volumes, always verify which function is larger to ensure positive side lengths.
Region bounded by $x=3$ and $x=3y$ for $0\le y\le 1$ ; square cross sections perpendicular to $y$ -axis: which setup is correct?
Answer: $\displaystyle \int_{0}^{1} (3-3y)^2\,dy$ This problem involves computing volume with square cross-sections perpendicular to the y-axis. The region is bounded by x = 3 and x = 3y for 0 ≤ y ≤ 1. For any y-value, the horizontal distance between these boundaries is 3 - 3y. Since cross-sections are squares, this distance serves as the side length, so the cross-sectional area is (3 - 3y)². Choice C gives only the side length without squaring, which is insufficient for calculating the area of square cross-sections. The fundamental approach is to determine the width of the base region at each y-value, square this distance to obtain the area of each square cross-section, then integrate along the y-axis.
Select the correct volume setup: base between $y=\ln x$ and $y=0$ for $1\le x\le e$ ; rectangles perpendicular to $x$ -axis with height $x$ .
Answer: $\displaystyle \int_{1}^{e}x\ln x\,dx$ This problem requires finding the volume of a solid with rectangular cross-sections perpendicular to the x-axis. The base region lies between y = ln x and y = 0 for 1 ≤ x ≤ e, so for each x-value, the base of the rectangle has length ln x (since ln x ≥ 0 on this interval). Each rectangle has height x (varying with position), making the area of each rectangle: base × height = ln x × x = x ln x. We integrate from x = 1 to x = e as specified. Choice B incorrectly uses (ln x)/x, which doesn't represent the correct area calculation for these rectangles. For rectangular cross-sections where both dimensions vary, multiply the base length by the height at each x-value.
Find the correct volume setup: base bounded by $x=0$ , $x=2$ , $y=0$ , $y=1+x$ ; squares perpendicular to the $x$ -axis.
Answer: $\displaystyle \int_{0}^{2}(1+x)^2\,dx$ This problem involves finding the volume of a solid with square cross-sections perpendicular to the x-axis. The base region is a rectangle bounded by x = 0, x = 2, y = 0, and y = 1 + x, so for each x-value, the vertical distance (side length of the square) is (1 + x) - 0 = 1 + x. Since we need square cross-sections, the area of each square is (side length)² = (1 + x)². We integrate from x = 0 to x = 2 based on the given x-bounds. Choice A incorrectly uses just the side length without squaring it, which would give a different solid's volume. The key insight for square cross-sections is that the area equals the square of the side length, not just the side length itself.
Base bounded by $x=\cos y$ and $x=0$ for $0\le y\le \pi/2$ ; squares perpendicular to $y$ -axis: which integral is correct?
Answer: $\displaystyle \int_{0}^{\pi/2} (\cos y)^2\,dy$ This problem requires finding volume with square cross-sections perpendicular to the y-axis. The base is bounded by x = cos y and x = 0 for 0 ≤ y ≤ π/2. For any y-value in this interval, the horizontal distance between these boundaries is cos y - 0 = cos y. Since cross-sections are squares, the side length equals cos y, making the cross-sectional area (cos y)². Choice B gives only the side length without squaring, which would not provide the correct area for square cross-sections. The systematic approach is to determine the width of the cosine region at each y-value, square this distance to obtain the area of each square cross-section, then integrate along the y-axis.
Base region is bounded by $y=\cos x$ and $y=-1$ on $[0,\pi]$ ; squares perpendicular to the $x$ -axis: which integral?
Answer: $\displaystyle \int_{0}^{\pi} (\cos x+1)^2\,dx$ This problem involves finding volume with square cross-sections perpendicular to the x-axis. The base is bounded by y = cos x and y = -1 on [0,π]. For any x-value in this interval, the vertical distance between the curves is cos x - (-1) = cos x + 1. Since cross-sections are squares, this distance becomes the side length, so the area is (cos x + 1)². Choice D gives only the side length without squaring, which is incorrect for computing the area of square cross-sections. The systematic approach is to determine the vertical separation between boundary curves, square this distance to get cross-sectional area, then integrate over the specified interval.
Base bounded by $x=\sin y$ and $x=1$ for $0\le y\le \pi/2$ ; squares perpendicular to $y$ -axis: select the integral.
Answer: $\displaystyle \int_{0}^{\pi/2} (1-\sin y)^2\,dy$ This problem requires computing volume with square cross-sections perpendicular to the y-axis. The base is bounded by x = sin y and x = 1 for 0 ≤ y ≤ π/2. For any y-value in this interval, the horizontal distance between these curves is 1 - sin y. Since cross-sections are squares, this distance becomes the side length, so the area is (1 - sin y)². Choice B gives only the side length without squaring, which is insufficient for calculating the area of square cross-sections. The systematic approach is to find the width of the base region at each y-value, use this as the square's side length, then integrate the squared expression along the y-axis.
For $0\le x\le 3$ , the base is bounded by $y=3$ and $y=x$ ; cross sections perpendicular to the $x$ -axis are rectangles with height twice the base in the region. Which integral gives the volume?
Answer: $\displaystyle \int_{0}^{3} 2(3-x)^2\,dx$ This problem involves rectangular cross sections perpendicular to the x-axis, where each rectangle has height twice its base. The base region is bounded by y = 3 (upper) and y = x (lower) for 0 ≤ x ≤ 3. At each x-value, the base of the rectangle equals the vertical distance (3 - x), and the height is 2(3 - x). The area of each rectangular cross section is base × height = (3 - x) × 2(3 - x) = 2(3 - x)². Choice B shows only (3 - x)², which would be correct for square cross sections but misses the factor of 2 for rectangular height. For rectangular cross sections with specified height-to-base ratios, multiply the squared base distance by the ratio coefficient.
Base bounded by $x=y^3$ and $x=8$ for $0\le y\le 2$ ; square cross sections perpendicular to $y$ -axis: select the setup.
Answer: $\displaystyle \int_{0}^{2} (8-y^3)^2\,dy$ This problem requires finding volume with square cross-sections perpendicular to the y-axis. The base is bounded by x = y³ and x = 8 for 0 ≤ y ≤ 2. For any y-value in this interval, the horizontal distance between these boundaries is 8 - y³. Since cross-sections are squares, the side length equals 8 - y³, making the cross-sectional area (8-y³)². Choice E gives only the side length without squaring, which would not provide the correct area for square cross-sections. The fundamental approach is to determine the width of the base region at each y-value, square this distance to obtain the area of each square cross-section, then integrate along the y-axis.
Region bounded by $y=2-x^2$ and $y=0$ where they intersect; square cross sections perpendicular to $x$ -axis: which integral gives volume?
Answer: $\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} (2-x^2)^2\,dx$ This problem involves computing volume with square cross-sections perpendicular to the x-axis. The region is bounded by $y = 2 - x^2$ and $y = 0$ where they intersect. Setting $2 - x^2 = 0$ gives $x^2 = 2$ , so $x = \pm \sqrt{2}$ . For any x-value in $[-\sqrt{2}, \sqrt{2}]$ , the vertical distance is $(2 - x^2) - 0 = 2 - x^2$ . Since cross-sections are squares, this distance serves as the side length, so the cross-sectional area is $(2 - x^2)^2$ . Choice C gives only the side length without squaring, which is insufficient for calculating square cross-sectional areas. The key strategy is to find the intersection points to determine integration limits, then use the height of the parabolic region as the square's side length and integrate the squared expression.

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