Volumes with Cross Sections: Squares/Rectangles
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AP Calculus BC › Volumes with Cross Sections: Squares/Rectangles
Base bounded by $x=2$ and $x=y^2$ for $-\sqrt{2}\le y\le \sqrt{2}$; squares perpendicular to $y$-axis: which setup?
$\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} (2-y^2),dy$
$\displaystyle \int_{-2}^{2} (2-y^2)^2,dy$
$\displaystyle \int_{0}^{2} (2-y^2)^2,dy$
$\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} (2+y^2)^2,dy$
$\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} (2-y^2)^2,dy$
Explanation
This problem requires finding volume with square cross-sections perpendicular to the y-axis. The region is bounded by x = 2 and x = y² for -√2 ≤ y ≤ √2. These boundaries intersect where y² = 2, giving y = ±√2. For any y-value in this interval, the horizontal distance is 2 - y². Since cross-sections are squares, the side length equals 2 - y², making the cross-sectional area (2-y²)². Choice C gives only the side length without squaring, which would not provide the correct area for square cross-sections. The systematic 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.
Base bounded by $y=\ln x$ and $y=1$ for $1\le x\le e$; rectangles perpendicular to $x$-axis have height $2$: which setup?
$\displaystyle \int_{0}^{1} 2(1-\ln x),dy$
$\displaystyle \int_{1}^{e} (1-\ln x)^2,dx$
$\displaystyle \int_{1}^{e} 2(1-\ln x),dx$
$\displaystyle \int_{1}^{e} 2(\ln x)^2,dx$
$\displaystyle \int_{1}^{e} 2(\ln x-1),dx$
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 is fixed at 2. The base is bounded by y = ln x and y = 1 for 1 ≤ x ≤ e. For any x-value, we need to determine which function is on top: at x = e, ln e = 1, so they meet at the right endpoint. For 1 ≤ x ≤ e, we have ln x ≤ 1, so the width is 1 - ln x. The cross-sectional area is width × height = (1 - ln x) × 2 = 2(1 - ln x). Choice C incorrectly squares the width, treating cross-sections as squares rather than rectangles with height 2. The key strategy for rectangular cross-sections is to identify which boundary is upper, find the width, multiply by the specified height, then integrate.
The base is bounded by $y=\sin x$ and $y=0$ on $0,\pi$; cross sections perpendicular to the $x$-axis are squares. Which integral gives the volume?
$\displaystyle \int_{0}^{\pi} (\sin x)^2,dx$
$\displaystyle \int_{0}^{\pi} (\pi-\sin x)^2,dx$
$\displaystyle \int_{0}^{1} (\arcsin y)^2,dy$
$\displaystyle \int_{0}^{\pi} \sin x,dx$
$\displaystyle \int_{0}^{\pi} (\sin x)^3,dx$
Explanation
This problem tests cross-sectional volume reasoning for solids with square cross-sections. The base region is bounded by y = sin x and y = 0 from x = 0 to π. The cross-sections are perpendicular to the x-axis, so the side length of each square is sin x. The area of each cross-section is (sin x)². A tempting distractor is choice B, which is ∫ sin x dx, but that computes the area of the base rather than the volume. A transferable strategy for cross-section volumes is to identify the integration variable, express the side length(s) in terms of that variable based on the base boundaries, compute the cross-sectional area, and integrate over the interval.
The base is bounded by $x=1$, $x=9$, $y=0$, and $y=\sqrt{x}$; cross sections perpendicular to the $x$-axis are squares. Which setup gives the volume?
$\displaystyle \int_{1}^{9} (x)^2,dx$
$\displaystyle \int_{1}^{9} \sqrt{x},dx$
$\displaystyle \int_{1}^{9} (9-x)^2,dx$
$\displaystyle \int_{0}^{3} (y^2)^2,dy$
$\displaystyle \int_{1}^{9} (\sqrt{x})^2,dx$
Explanation
This problem tests cross-sectional volume reasoning for solids with square cross-sections. The base region is bounded by x = 1, x = 9, y = 0, and y = √x from x = 1 to x = 9. The cross-sections are perpendicular to the x-axis, so the side length of each square is √x. The area of each cross-section is (√x)² = x. A tempting distractor is choice C, which is ∫ √x dx, but that computes the area of the base rather than the volume. A transferable strategy for cross-section volumes is to identify the integration variable, express the side length(s) in terms of that variable based on the base boundaries, compute the cross-sectional area, and integrate over the interval.
The base is the region between $y=x^2$ and $y=2$; cross sections perpendicular to the $x$-axis are rectangles with height $5$. Which integral gives the volume?
$\displaystyle \int_{0}^{2} 5(2-y^2),dy$
$\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} (2-x^2),dx$
$\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} 5(2-x^2),dx$
$\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} (5-x^2)(2-x^2),dx$
$\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} 5(2-x^2)^2,dx$
Explanation
This problem tests cross-sectional volume reasoning for solids with rectangular cross-sections. The base region is between y = x² and y = 2 from x = -√2 to √2. The cross-sections are perpendicular to the x-axis, so the base length of each rectangle is 2 - x². The height is a constant 5, so the area is 5(2 - x²). A tempting distractor is choice D, which omits the height factor of 5 and computes something closer to the base area. A transferable strategy for cross-section volumes is to identify the integration variable, express the side length(s) in terms of that variable based on the base boundaries, compute the cross-sectional area, and integrate over the interval.
What integral gives the volume if the base is bounded by $y=\ln(x)$ and $y=0$ for $1\le x\le e$, squares perpendicular to the $x$-axis?
$\displaystyle \int_{1}^{e} \big(\ln(x)\big)^2,dx$
$\displaystyle \int_{1}^{e} \big(\ln(x)\big)^3,dx$
$\displaystyle \int_{1}^{e} (e-x)^2,dx$
$\displaystyle \int_{0}^{1} \ln(x)^2,dx$
$\displaystyle \int_{1}^{e} \ln(x),dx$
Explanation
This problem asks for the volume when square cross-sections are perpendicular to the x-axis. The base is bounded by y = ln(x) and y = 0 for 1 ≤ x ≤ e, so the side length of each square equals ln(x) - 0 = ln(x). Since we have square cross-sections, the area is (side length)² = (ln(x))². The integral is ∫_${1}^{e}$ (ln(x))² dx. Choice A incorrectly uses just ln(x) without squaring, which would give the area under the curve rather than the volume. For square cross-sections, always square the distance between boundary curves to obtain the cross-sectional area.
What is the correct volume setup if the base is bounded by $y=6-x$ and $y=0$ on $0,6$, squares perpendicular to the $x$-axis?
$\displaystyle \int_{0}^{6} (6-x),dx$
$\displaystyle \int_{0}^{6} (6-x)^3,dx$
$\displaystyle \int_{0}^{6} (6-x)^2,dx$
$\displaystyle \int_{0}^{6} (6)^2,dx$
$\displaystyle \int_{0}^{6} x^2,dx$
Explanation
This problem asks for the volume when the base is bounded by y = 6 - x and y = 0, with square cross-sections perpendicular to the x-axis. At each x-value between 0 and 6, the side length of the square equals the vertical distance (6 - x) - 0 = 6 - x. Since we have square cross-sections, the area of each square is (side length)² = (6 - x)². The integral becomes ∫_${0}^{6}$ (6 - x)² dx. Choice B incorrectly uses just the linear expression without squaring it, which would give the area under the curve rather than the volume. Remember that for square cross-sections, you must square the distance between boundaries to find the cross-sectional area.
Select the correct volume setup: base bounded by $y=2x$ and $y=x^2$; squares perpendicular to the $x$-axis.
$\displaystyle \int_{0}^{2}\big(2x-x^2\big)^2,dx$
$\displaystyle \int_{0}^{2}\big(2x-x^2\big),dx$
$\displaystyle \int_{0}^{2}\big(2x-x^2\big)^2,dy$
$\displaystyle \int_{0}^{2}\big(2- x\big)^2,dx$
$\displaystyle \int_{0}^{2}\big(x^2-2x\big)^2,dx$
Explanation
This problem involves finding the volume of a solid with square cross-sections perpendicular to the x-axis. The base region is bounded by y = 2x and y = x², which intersect when 2x = x², giving x = 0 and x = 2. For each x-value between 0 and 2, since 2x ≥ x² in this interval, the side length of the square equals 2x - x². The area of each square cross-section is (side length)² = (2x - x²)². Choice C incorrectly uses just the side length without squaring it, which would give the area of the base region rather than the volume. Remember that for square cross-sections, you must square the expression for the side length before integrating.
The base is bounded by $y=2x$ and $y=x^2$ for $0\le x\le 2$; cross sections perpendicular to the $x$-axis are squares. Which integral gives the volume?
$\displaystyle \int_{0}^{2} \big(2x^2-x\big)^2,dx$
$\displaystyle \int_{0}^{2} (x^2-2x)^2,dx$
$\displaystyle \int_{0}^{2} (2x-x^2),dx$
$\displaystyle \int_{0}^{2} (2x-x^2)^2,dx$
$\displaystyle \int_{0}^{2} (2x-x^2)^2,dy$
Explanation
This problem involves finding the volume when square cross sections are perpendicular to the x-axis. The base region is bounded by y = 2x (upper curve) and y = x² (lower curve) for 0 ≤ x ≤ 2. At each x-value, the vertical distance between curves is (2x - x²), which becomes the side length of each square cross section. The area of each square is therefore (2x - x²)², and integrating this from 0 to 2 gives the volume. Choice B shows just (2x - x²) without squaring, which would give the area of the base region, not the volume. Remember that for square cross sections, you must square the distance between boundary curves to get the cross-sectional area.
Find the correct volume setup: base between $x=y^2$ and $x=4$ for $0\le y\le2$, squares perpendicular to the $y$-axis.
$\displaystyle \int_{0}^{2}\big(4-y^2\big)^2,dx$
$\displaystyle \int_{0}^{2}\big(4-y^2\big)^2,dy$
$\displaystyle \int_{0}^{2}\big(2-\sqrt{y}\big)^2,dy$
$\displaystyle \int_{0}^{2}\big(4-y^2\big),dy$
$\displaystyle \int_{0}^{4}\big(4-y^2\big)^2,dy$
Explanation
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.