Volumes with Cross Sections: Squares/Rectangles
Help Questions
AP Calculus BC › Volumes with Cross Sections: Squares/Rectangles
Choose the correct volume setup: base bounded by $x=0$, $x=1$, $y=0$, and $y=3x$, rectangles perpendicular to $x$-axis with height twice the base.
$\displaystyle \int_{0}^{1} (3x)\cdot 2,dx$
$\displaystyle \int_{0}^{3} 2y^2,dy$
$\displaystyle \int_{0}^{1} (3x)^2,dx$
$\displaystyle \int_{0}^{1} 2(3x)^2,dx$
$\displaystyle \int_{0}^{1} \pi,2(3x)^2,dx$
Explanation
This problem requires cross-sectional volume reasoning to compute the volume with rectangular cross sections perpendicular to the x-axis. The base region is bounded by x = 0, x = 1, y = 0, and y = 3x. For each x, the base of the rectangle is the vertical length 3x, and the height is twice that, so 6x. Therefore, the area is 3x * 6x = 18x², or equivalently 2*(3x)², and the integral is from 0 to 1. A tempting distractor is choice E, which includes π, but that's for circular shapes, not rectangles. 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.
Base bounded by $y=2-x$ and $y=0$ on $0,2$; rectangles perpendicular to $x$-axis have height $x$: which setup is correct?
$\displaystyle \int_{0}^{2} (2-x),dx$
$\displaystyle \int_{0}^{2} (x-2)(2-x),dx$
$\displaystyle \int_{0}^{2} x(2-x),dx$
$\displaystyle \int_{0}^{2} x(2-x)^2,dx$
$\displaystyle \int_{0}^{2} x(2-y),dy$
Explanation
This problem requires computing volume with rectangular cross-sections perpendicular to the x-axis, where the base width comes from the region and height varies as x. The base is bounded by y = 2 - x and y = 0 on [0,2]. For any x-value, the width is (2-x) - 0 = 2-x. The cross-sectional area is width × height = (2-x) × x = x(2-x). Choice B incorrectly squares the width, treating cross-sections as squares rather than rectangles with height x. The systematic approach for rectangular cross-sections is to identify the base width from the region, multiply by the specified variable height, then integrate this area function over the given interval.
Choose the correct setup: base bounded by $x=y^2$ and $x=4$, square cross sections perpendicular to the $y$-axis.
$\displaystyle \int_{0}^{4} (4-\sqrt{x})^2,dx$
$\displaystyle \int_{0}^{2} (4-y^2)^2,dy$
$\displaystyle \int_{0}^{2} \pi(4-y^2)^2,dy$
$\displaystyle \int_{0}^{2} (4-y^2),dy$
$\displaystyle \int_{-2}^{2} (4-y^2)^2,dx$
Explanation
This problem involves cross-sectional volume reasoning to determine the volume of a solid with square cross sections perpendicular to the y-axis. The base region is bounded by x = y² and x = 4, considering y from 0 to 2 in the first quadrant. For each y, the side length of the square is the horizontal distance between x = y² and x = 4, which is 4 - y². Therefore, the cross-sectional area is (4 - y²)², and the volume is the integral from 0 to 2. A tempting distractor is choice E, which includes π, but that's incorrect for square cross sections as it suggests a circular shape. 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.
Find the correct volume integral for a solid with base between $y=x$ and $y=x^2$, square cross sections perpendicular to the $x$-axis.
$\displaystyle \int_{0}^{1} (x-x^2),dx$
$\displaystyle \int_{0}^{1} (x-x^2)^2,dx$
$\displaystyle \int_{0}^{1} \pi(x-x^2)^2,dx$
$\displaystyle \int_{0}^{1} (x-x^2)^2,dy$
$\displaystyle \int_{0}^{1} \big(\sqrt{x}-x\big)^2,dx$
Explanation
This problem involves cross-sectional volume reasoning to compute the volume of a solid with square cross sections perpendicular to the x-axis. The base region is bounded by the curves y = x and y = x² from x = 0 to x = 1. At each x in this interval, the side length of the square cross section is the vertical distance between the curves, which is x - x² since y = x is above y = x². Therefore, the area of each cross section is (x - x²)², and the volume is obtained by integrating this area function from 0 to 1. A tempting distractor is choice E, which includes a factor of π, but this is incorrect because π is used for circular cross sections, not squares. In general, when finding volumes with known cross sections, determine the area function based on the shape and integrate it over the base interval.
Find the volume setup: base bounded by $y=2x$ and $y=x^2$ on $0,2$, square cross sections perpendicular to the $x$-axis.
$\displaystyle \int_{0}^{2} \big(\sqrt{y}-\tfrac{y}{2}\big)^2,dy$
$\displaystyle \int_{0}^{2} (2x-x^2),dx$
$\displaystyle \int_{0}^{2} \pi(2x-x^2)^2,dx$
$\displaystyle \int_{0}^{2} (x^2-2x)^2,dx$
$\displaystyle \int_{0}^{2} (2x-x^2)^2,dx$
Explanation
This problem requires cross-sectional volume reasoning to compute the volume with square cross sections perpendicular to the x-axis. The base region is bounded by y = 2x and y = x² from x = 0 to x = 2. At each x, the side length is the vertical distance between y = 2x (upper) and y = x² (lower), which is 2x - x². Thus, the area is (2x - x²)², and the volume integral is from 0 to 2. A tempting distractor is choice E, which adds π, but this is for circular cross sections, not squares. In general, when finding volumes with known cross sections, determine the area function based on the shape and integrate it over the base interval.
Which integral gives the volume when the base is bounded by $y=4-x^2$ and $y=0$, with square cross sections perpendicular to the $x$-axis?
$\displaystyle \int_{-2}^{2} \pi(4-x^2)^2,dx$
$\displaystyle \int_{-2}^{2} (2\sqrt{4-x^2})^2,dx$
$\displaystyle \int_{-2}^{2} (4-x^2),dx$
$\displaystyle \int_{-2}^{2} (4-x^2)^2,dx$
$\displaystyle \int_{0}^{4} \big(\sqrt{4-y}-(-\sqrt{4-y})\big)^2,dy$
Explanation
This problem requires cross-sectional volume reasoning to find the volume of a solid with square cross sections perpendicular to the x-axis. The base region is bounded by y = 4 - x² and y = 0 from x = -2 to x = 2. For each x, the side length of the square is the height of the parabola above the x-axis, which is 4 - x². Thus, the cross-sectional area is (4 - x²)², and the volume is the integral of this from -2 to 2. A tempting distractor is choice D, which adds π, but that's for disks or washers, not square cross sections. 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.
Base is bounded by $x=0$ and $x=1-y$ for $0\le y\le 1$; squares perpendicular to the $y$-axis: which setup?
$\displaystyle \int_{0}^{1} (1+y)^2,dy$
$\displaystyle \int_{0}^{1} (1-y)^2,dy$
$\displaystyle \int_{0}^{1} \left(\frac{1-y}{2}\right)^2,dy$
$\displaystyle \int_{0}^{1} (1-x)^2,dx$
$\displaystyle \int_{0}^{1} (1-y),dy$
Explanation
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.
The base is between $y=\sin x$ and $y=2$ on $0,\pi$; square cross sections perpendicular to the $x$-axis: choose the setup.
$\displaystyle \int_{0}^{\pi} (2-\sin x)^2,dx$
$\displaystyle \int_{0}^{\pi} \left(\frac{2-\sin x}{2}\right)^2,dx$
$\displaystyle \int_{0}^{2} (2-\sin x)^2,dy$
$\displaystyle \int_{0}^{\pi} (2-\sin x),dx$
$\displaystyle \int_{0}^{\pi} (2+\sin x)^2,dx$
Explanation
This problem requires computing volume using square cross-sections perpendicular to the x-axis. The base region is bounded by y = sin x and y = 2 on the interval [0,π]. For any x-value, the vertical distance between these curves is 2 - sin x. Since the cross-sections are squares, this distance serves as the side length, making the area of each square (2 - sin x)². Choice B represents only the side length rather than the area of the square cross-section. The correct method is to find the distance between bounding curves, use it as the square's side length, then integrate the squared expression over the given interval.
Region bounded by $x=1-y$ and $x=0$ for $0\le y\le 1$; rectangular cross sections perpendicular to $y$-axis have height $2$: choose setup.
$\displaystyle \int_{0}^{1} 2(1-x),dx$
$\displaystyle \int_{0}^{1} 2(1-y)^2,dy$
$\displaystyle \int_{0}^{1} \frac{1-y}{2},dy$
$\displaystyle \int_{0}^{1} (1-y),dy$
$\displaystyle \int_{0}^{1} 2(1-y),dy$
Explanation
This problem involves finding volume with rectangular cross-sections perpendicular to the y-axis, where one dimension is the width of the base region and the other is a fixed height of 2. The base is bounded by x = 1 - y and x = 0 for 0 ≤ y ≤ 1. For any y-value, the width is (1-y) - 0 = 1-y. The cross-sectional area is width × height = (1-y) × 2 = 2(1-y). Choice B incorrectly squares the width, treating it as if cross-sections were squares rather than rectangles. The correct approach for rectangular cross-sections is to multiply the base width by the given height, then integrate this area function over the appropriate interval.
Base bounded by $y=1/(x+1)$ and $y=0$ for $0\le x\le 1$; squares perpendicular to $x$-axis: choose the setup.
$\displaystyle \int_{0}^{1} \left(\frac{1}{2(x+1)}\right)^2,dx$
$\displaystyle \int_{0}^{1} \left(\frac{1}{x}\right)^2,dx$
$\displaystyle \int_{0}^{1} \left(\frac{x+1}{1}\right)^2,dx$
$\displaystyle \int_{0}^{1} \frac{1}{x+1},dx$
$\displaystyle \int_{0}^{1} \left(\frac{1}{x+1}\right)^2,dx$
Explanation
This problem requires finding volume with square cross-sections perpendicular to the x-axis. The base is bounded by y = 1/(x+1) and y = 0 for 0 ≤ x ≤ 1. For any x-value, the vertical distance between these curves is 1/(x+1) - 0 = 1/(x+1). Since cross-sections are squares, the side length equals 1/(x+1), so the cross-sectional area is (1/(x+1))². Choice B gives only the side length without squaring, which is insufficient for calculating square cross-sectional areas. The correct strategy is to identify the height of the rational function region, square this distance to get the area of each square cross-section, then integrate over the given interval.