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

What is the correct volume integral for the solid with base between $x=y^2$ and $x=4$ , square cross sections perpendicular to the $y$ -axis?
Answer: $\displaystyle \int_{-2}^{2} \big(4-y^2\big)^2\,dy$ This problem asks for the volume of a solid with square cross-sections perpendicular to the y-axis. The base region lies between x = y² and x = 4, so for each y-value, the width of the base is (4 - y²). Since the cross-sections are squares perpendicular to the y-axis, each square has side length (4 - y²) and area (4 - y²)². The y-values range from -2 to 2 (solving y² = 4 gives y = ±2), so the volume integral is ∫_{-2}^{2} (4 - y²)² dy. Choice C incorrectly uses (4 - y²) without squaring, which would calculate the area of the base region instead of the volume. When cross-sections are perpendicular to the y-axis, integrate with respect to y and ensure the bounds reflect the y-range of the region.
What is the correct volume integral setup if the base is bounded by $y=\sqrt{4-x}$ and $y=0$ on $[0,4]$ with rectangular cross sections perpendicular to the $x$ -axis of height $2$ ?
Answer: $\displaystyle \int_{0}^{4} 2\sqrt{4-x}\,dx$ This problem requires using the method of cross-sectional volumes to find the volume of a solid with rectangular cross sections. The base of the solid is the region bounded by y = √(4 - x) and y = 0 on [0, 4]. For cross sections perpendicular to the x-axis, the width of each rectangle is the vertical distance between the curves, which is √(4 - x). The height of each rectangle is given as 2, so the area is 2 √(4 - x). Integrating this area function from 0 to 4 gives the volume as ∫ from 0 to 4 of 2 √(4 - x) dx. A tempting distractor is choice C, which squares the width as if for squares. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.
What is the correct volume integral setup if the base is bounded by $x=\cos y$ and $x=0$ for $y\in[0,\tfrac{\pi}{2}]$ with square cross sections perpendicular to the $y$ -axis?
Answer: $\displaystyle \int_{0}^{\pi/2} (\cos y)^2\,dy$ This problem requires using the method of cross-sectional volumes to find the volume of a solid with square cross sections. The base of the solid is the region bounded by x = cos y and x = 0 for y in [0, π/2]. For cross sections perpendicular to the y-axis, the side length of each square is the horizontal distance between the curves, which is cos y. Therefore, the area of each cross section is (cos y)². Integrating this area function from 0 to π/2 gives the volume as ∫ from 0 to π/2 of (cos y)² dy. A tempting distractor is choice B, which omits the squaring. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.
What is the correct volume integral setup if the base is bounded by $y=2x$ and $y=x^2$ with square cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{0}^{2} (2x-x^2)^2\,dx$ This problem requires using the method of cross-sectional volumes to find the volume of a solid with square cross sections. The base of the solid is the region bounded by y = 2x and y = x². For cross sections perpendicular to the x-axis from x = 0 to x = 2, the side length of each square is the vertical distance between the curves, which is 2x - x². Therefore, the area of each cross section is (2x - x²)². Integrating this area function from 0 to 2 gives the volume as ∫ from 0 to 2 of (2x - x²)² dx. A common mistake is choice B, which forgets to square the side length and integrates the distance directly. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.
What is the correct volume integral setup if the base is bounded by $y=\sin x$ and $y=0$ on $[0,\pi]$ with square cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{0}^{\pi} (\sin x)^2\,dx$ This problem requires using the method of cross-sectional volumes to find the volume of a solid with square cross sections. The base of the solid is the region bounded by y = sin x and y = 0 on [0, π]. For cross sections perpendicular to the x-axis, the side length of each square is the vertical distance between the curves, which is sin x. Therefore, the area of each cross section is (sin x)². Integrating this area function from 0 to π gives the volume as ∫ from 0 to π of (sin x)² dx. A tempting distractor is choice A, which forgets to square the side length and computes the area under the sine curve instead. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.
What is the correct volume integral setup if the base is bounded by $x=4$ and $x=1+y$ for $y\in[0,3]$ with rectangular cross sections perpendicular to the $y$ -axis of height $y$ ?
Answer: $\displaystyle \int_{0}^{3} y\big(4-(1+y)\big)\,dy$ This problem requires using the method of cross-sectional volumes to find the volume of a solid where the cross-sections are rectangles perpendicular to the y-axis. The base of the solid is the region bounded by x = 4 and x = 1 + y from y = 0 to y = 3, so at each y, the width of the rectangle is 4 - (1 + y). The height of each rectangle is given as y, so the area is y (4 - (1 + y)). The volume is obtained by integrating this area function from y = 0 to y = 3. A tempting distractor is choice D, which integrates (4 - (1 + y)) dy, but that omits the rectangular height and gives the base area. Remember, for volumes with known cross-sections, always determine the area of the cross-section as a function of the variable of integration and integrate that over the interval.
What is the correct volume integral for a solid with base between $y=x^2$ and $y=4$ , square cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{-2}^{2} \big(4-x^2\big)^2\,dx$ This problem requires finding the volume of a solid with cross-sections perpendicular to the x-axis. The base region lies between y = x² (lower curve) and y = 4 (upper curve), which intersect when x² = 4, giving x = ±2. For each x-value, the cross-section is a square whose side length equals the vertical distance between the curves: (4 - x²). Since the area of a square with side length s is s², the cross-sectional area is (4 - x²)². Choice C incorrectly uses (4 - x²) without squaring, which would give the area of a rectangle with height 1, not a square. To solve cross-section problems, always identify the shape's area formula and express its dimensions in terms of the integration variable.
What is the correct volume integral setup if the base is bounded by $x=1$ and $x=1+y^2$ for $y\in[-2,2]$ with rectangular cross sections perpendicular to the $y$ -axis of height $6$ ?
Answer: $\displaystyle \int_{-2}^{2} 6\big((1+y^2)-1\big)\,dy$ This problem requires using the method of cross-sectional volumes to find the volume of a solid with rectangular cross sections. The base of the solid is the region bounded by x = 1 and x = 1 + y² for y in [-2, 2]. For cross sections perpendicular to the y-axis, the width of each rectangle is the horizontal distance between the curves, which is (1 + y²) - 1 = y². The height of each rectangle is given as 6, so the area is 6 y². Integrating this area function from -2 to 2 gives the volume as ∫ from -2 to 2 of 6((1 + y²) - 1) dy. A common mistake is choice C, which squares the width incorrectly for rectangles. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.
Choose the correct volume integral: base region between $y=x^2-1$ and $y=0$ for $-1\le x\le1$ , square cross sections perpendicular to the $x$ -axis.
Answer: $\displaystyle \int_{-1}^{1} \big(1-x^2\big)^2\,dx$ This problem requires finding volume with square cross-sections perpendicular to the x-axis. The base region is between y = x² - 1 and y = 0 on [-1, 1]. Since x² - 1 < 0 on this interval (it equals -1 at x = 0 and 0 at x = ±1), the parabola lies below the x-axis. The vertical distance from the parabola up to y = 0 is 0 - (x² - 1) = 1 - x². For square cross-sections, we square this distance to get area (1 - x²)². Choice A incorrectly uses (x² - 1)², which squares a negative quantity without accounting for the proper orientation. Always identify which function is on top before computing distances for cross-sectional areas.
What is the correct volume integral for the solid with base bounded by $y=x$ and $y=x^2$ , square cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{0}^{1} \big(x-x^2\big)^2\,dx$ This problem requires finding the volume of a solid with square cross-sections perpendicular to the x-axis. The base region is bounded by y = x and y = x², which intersect at x = 0 and x = 1. Between these points, x ≥ x² (since 0 ≤ x ≤ 1), so the height of the base at each x is (x - x²). For square cross-sections, each has side length (x - x²) and area (x - x²)². The volume integral is ∫_{0}^{1} (x - x²)² dx. Choice B incorrectly reverses the order to (x² - x)², which would be negative since x > x² on this interval, making it physically impossible. Always verify which function is on top by testing a point in the interval.
What is the correct volume integral for the solid with base between $y=\sqrt{x}$ and $y=0$ , rectangles perpendicular to the $x$ -axis with height twice the base?
Answer: $\displaystyle \int_{0}^{1} 2(\sqrt{x})^2\,dx$ This problem involves finding the volume of a solid with rectangular cross-sections perpendicular to the x-axis. The base region lies between y = √x and y = 0, so the base dimension of each rectangle is √x. Since the height is twice the base, each rectangle has height 2√x and area (base × height) = √x × 2√x = 2(√x)² = 2x. The volume integral is ∫_{0}^{1} 2x dx, which equals choice D: ∫_{0}^{1} 2(√x)² dx. Choice C incorrectly squares the entire expression (2√x)² = 4x, which would overcount the volume by a factor of 2. For rectangular cross-sections with specified height-to-base ratios, multiply base times height rather than squaring.
Find the correct volume integral: base bounded by $y=1$ and $y=x^3$ on $[0,1]$ , rectangles perpendicular to the $x$ -axis with height $2$ and base in the region.
Answer: $\displaystyle \int_{0}^{1} 2\big(1-x^3\big)\,dx$ This volume problem uses rectangular cross-sections perpendicular to the x-axis. The base region is between y = x³ (lower curve) and y = 1 (upper line) on [0, 1]. Each rectangle has its base in the region with length 1 - x³, and the height is specified as 2. The area of each rectangle is base × height = (1 - x³) × 2 = 2(1 - x³). Choice A incorrectly squares the base length as (1 - x³)², which would apply to square cross-sections, not rectangles with a specified height. When dealing with rectangles, use area = base × height, only squaring when the cross-section is explicitly a square.
What integral gives the volume if cross sections perpendicular to the $x$ -axis are squares over $-1\le x\le1$ between $y=1-x^2$ and $y=0$ ?
Answer: $\displaystyle \int_{-1}^{1} \big(1-x^2\big)^2\,dx$ This problem asks for the volume of a solid with square cross-sections perpendicular to the x-axis. The region lies between y = 1 - x² and y = 0 over [-1, 1], so at each x-value, the side length of the square is (1 - x²) - 0 = 1 - x². Since the cross-sections are squares, the area of each square is (1 - x²)². The volume is obtained by integrating this area from x = -1 to x = 1: ∫₋₁¹ (1 - x²)² dx. Choice A incorrectly uses just (1 - x²), which is the side length rather than the area, while choice E uses √(1 - x²), which would give an even smaller value. Remember that for square cross-sections, you must square the distance between the bounding curves to get the area.
What is the correct volume integral setup if the base is bounded by $y=2$ and $y=x^2$ on $[- \sqrt{2}, \sqrt{2}]$ with rectangular cross sections perpendicular to the $x$ -axis of height $7$ ?
Answer: $\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} 7(2-x^2)\,dx$ This problem requires using the method of cross-sectional volumes to find the volume of a solid where the cross-sections are rectangles perpendicular to the $x$ -axis. The base of the solid is the region bounded by $y=2$ and $y=x^2$ from $x=-\sqrt{2}$ to $x=\sqrt{2}$ , so at each $x$ , the width of the rectangle is $2 - x^2$ . The height of each rectangle is given as 7, so the area is 7 $(2 - x^2)$ . The volume is obtained by integrating this area function from $x=-\sqrt{2}$ to $x=\sqrt{2}$ . A tempting distractor is choice C, which integrates $(2 - x^2)$ dx, but that omits the rectangular height and gives the base area. Remember, for volumes with known cross-sections, always determine the area of the cross-section as a function of the variable of integration and integrate that over the interval.
What is the correct volume integral setup if the base is bounded by $y=2\sin x$ and $y=0$ on $[0,\pi]$ with square cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{0}^{\pi} (2\sin x)^2\,dx$ This problem requires using the method of cross-sectional volumes to find the volume of a solid where the cross-sections are squares perpendicular to the x-axis. The base of the solid is the region bounded by y = 2 sin x and y = 0 from x = 0 to x = π, so at each x, the side length of the square is 2 sin x. Since the cross-sections are squares, the area of each cross-section is (2 sin x)^2. The volume is obtained by integrating this area function from x = 0 to x = π. A tempting distractor is choice B, which integrates 2 sin x dx, but that computes the area of the base rather than the volume with square cross-sections. Remember, for volumes with known cross-sections, always determine the area of the cross-section as a function of the variable of integration and integrate that over the interval.
What integral gives the volume if cross sections perpendicular to the $x$ -axis are squares over $0\le x\le2$ between $y=x^2$ and $y=4$ ?
Answer: $\displaystyle \int_{0}^{2} \big(4-x^2\big)^2\,dx$ This problem requires finding the volume of a solid with square cross-sections perpendicular to the x-axis. The region lies between y = x² and y = 4 over the interval [0, 2], so at each x-value, the side length of the square is the vertical distance between the curves: (4 - x²). Since the cross-sections are squares, the area of each square is the side length squared: (4 - x²)². The volume is found by integrating this area function from x = 0 to x = 2, giving ∫₀² (4 - x²)² dx. Choice A incorrectly uses just (4 - x²), which would be the side length rather than the area. When working with cross-sections, always remember that for squares you must square the linear dimension to get the area.
What is the correct volume integral setup if the base is bounded by $y=\sqrt{1-x^2}$ and $y=0$ on $[-1,1]$ with square cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{-1}^{1} (\sqrt{1-x^2})^2\,dx$ This problem requires using the method of cross-sectional volumes to find the volume of a solid where the cross-sections are squares perpendicular to the x-axis. The base of the solid is the region bounded by y = √(1 - $x^2$ ) and y = 0 from x = -1 to x = 1, so at each x, the side length of the square is √(1 - $x^2$ ). Since the cross-sections are squares, the area of each cross-section is (√(1 - $x^2$ ))^2. The volume is obtained by integrating this area function from x = -1 to x = 1. A tempting distractor is choice B, which integrates √(1 - $x^2$ ) dx without squaring, but that computes the area of the base rather than the volume. Remember, for volumes with known cross-sections, always determine the area of the cross-section as a function of the variable of integration and integrate that over the interval.
What is the correct volume integral setup if the base is bounded by $y=x^2+1$ and $y=5$ with square cross sections perpendicular to the $x$ -axis?
Answer: $\displaystyle \int_{-2}^{2} (5-(x^2+1))^2\,dx$ This problem requires using the method of cross-sectional volumes to find the volume of a solid with square cross sections. The base of the solid is the region bounded by y = x² + 1 and y = 5. For cross sections perpendicular to the x-axis from x = -2 to x = 2, the side length of each square is the vertical distance between the curves, which is 5 - (x² + 1). Therefore, the area of each cross section is (5 - (x² + 1))². Integrating this area function from -2 to 2 gives the volume as ∫ from -2 to 2 of (5 - (x² + 1))² dx. A common mistake is choice B, which forgets to square the side length. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.
What is the correct volume integral for the solid with base between $x=2y$ and $x=y^2$ , rectangles perpendicular to the $y$ -axis with height $3$ ?
Answer: $\displaystyle \int_{0}^{2} 3(2y-y^2)\,dy$ This problem involves finding the volume of a solid with rectangular cross-sections perpendicular to the y-axis. The base region lies between x = 2y and x = y², where these curves intersect at y = 0 and y = 2. For 0 ≤ y ≤ 2, we have 2y ≥ y², so the base of each rectangle is (2y - y²). With constant height 3, each rectangle has area 3(2y - y²). The volume integral is ∫_{0}^{2} 3(2y - y²) dy. Choice C incorrectly uses (2y - y²)² without the height factor, treating the cross-sections as squares instead of rectangles with height 3. For rectangular cross-sections with given height, multiply the base width by the specified height.
What is the correct volume integral setup if the base is bounded by $y=1$ and $y=\sin x$ on $[0,\pi]$ with rectangular cross sections perpendicular to the $x$ -axis of height $4$ ?
Answer: $\displaystyle \int_{0}^{\pi} 4(1-\sin x)\,dx$ This problem requires using the method of cross-sectional volumes to find the volume of a solid with rectangular cross sections. The base of the solid is the region bounded by y = 1 and y = sin x on [0, π]. For cross sections perpendicular to the x-axis, the width of each rectangle is the vertical distance between the curves, which is 1 - sin x. The height of each rectangle is given as 4, so the area is 4(1 - sin x). Integrating this area function from 0 to π gives the volume as ∫ from 0 to π of 4(1 - sin x) dx. A tempting distractor is choice C, which squares the width as if for squares. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.

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