Cross Sections: Squares & Rectangles - Calculus 2

1. Let R be the region bounded by $y = 1/x^2$, $y=0$, $x=1$, and $x=2$. A solid has base R and cross-sections perpendicular to the x-axis that are squares. What is the volume of the solid?
   Answer: $\frac{7}{24}$ The limits of integration are from $x=1$ to $x=2$. The side of the square cross-section is $s(x) = 1/x^2$. The area is $A(x) = (1/x^2)^2 = 1/x^4 = x^{-4}$. The volume is $V = \int_1^2 x^{-4} dx = [\frac{x^{-3}}{-3}]_1^2 = [-\frac{1}{3x^3}]_1^2 = (-\frac{1}{3(2^3)}) - (-\frac{1}{3(1^3)}) = -\frac{1}{24} + \frac{1}{3} = \frac{-1+8}{24} = \frac{7}{24}$.
2. The base of a solid is the region enclosed by $x = y^2$ and $x = y+2$. If the cross-sections perpendicular to the y-axis are squares, which of the following integrals gives the volume of the solid?
   Answer: $\int_{-1}^{2} (y+2-y^2)^2 dy$ To find the limits, set $y^2 = y+2$, which yields $y^2 - y - 2 = 0$, or $(y-2)(y+1)=0$. The limits are $y=-1$ to $y=2$. For cross-sections perpendicular to the y-axis, we integrate with respect to $y$. The rightmost curve is $x = y+2$ and the leftmost is $x = y^2$. The side length of the square is $s(y) = (y+2) - y^2$. The area is $A(y) = s(y)^2 = (y+2-y^2)^2$. The volume is the integral of the area, $V = \int_{-1}^{2} (y+2-y^2)^2 dy$.
3. A solid has a base in the xy-plane bounded by $x=y^2$ and $x=1$. Cross-sections perpendicular to the x-axis are squares. Which integral represents the volume of this solid?
   Answer: $\int_{0}^{1} 4x dx$ For cross-sections perpendicular to the x-axis, we integrate with respect to $x$ from 0 to 1. The curves bounding the base are $y=\sqrt{x}$ (top) and $y=-\sqrt{x}$ (bottom). The side length of the square at $x$ is $s(x) = \sqrt{x} - (-\sqrt{x}) = 2\sqrt{x}$. The area of the square is $A(x) = s(x)^2 = (2\sqrt{x})^2 = 4x$. The volume is given by the integral $V = \int_{0}^{1} 4x dx$.
4. The base of a solid is the region enclosed by $y = \sqrt{x}$, $y = x - 2$, and the x-axis. Cross-sections perpendicular to the x-axis are squares. What is the volume of the solid?
   Answer: $\int_{0}^{2} x dx + \int_{2}^{4} (\sqrt{x} - x + 2)^2 dx$ The region has two parts. The intersection of $y=\sqrt{x}$ and $y=x-2$ is at $x=4, y=2$. The line $y=x-2$ intersects the x-axis at $x=2$. From $x=0$ to $x=2$, the base of the square is $s(x) = \sqrt{x} - 0 = \sqrt{x}$, so the area is $A(x) = (\sqrt{x})^2 = x$. From $x=2$ to $x=4$, the base is $s(x) = \sqrt{x} - (x-2)$, so the area is $A(x) = (\sqrt{x} - x + 2)^2$. The total volume is the sum of the two integrals: $V = \int_{0}^{2} x dx + \int_{2}^{4} (\sqrt{x} - x + 2)^2 dx$.
5. Let the base of a solid be the region bounded by the parabola $y = 4 - x^2$ and the x-axis. The cross-sections perpendicular to the x-axis are rectangles whose height is twice the length of their base. What is the volume of the solid?
   Answer: $\frac{1024}{15}$ The base extends from $x=-2$ to $x=2$. The base of a rectangular cross-section at $x$ is $s(x) = (4-x^2) - 0 = 4-x^2$. The height is given as $h(x) = 2s(x) = 2(4-x^2)$. The area of the cross-section is $A(x) = s(x)h(x) = (4-x^2) \cdot 2(4-x^2) = 2(4-x^2)^2$. The volume is $V = \int_{-2}^{2} 2(4-x^2)^2 dx = 2\int_{-2}^{2} (16-8x^2+x^4) dx = 2[16x - \frac{8x^3}{3} + \frac{x^5}{5}]_{-2}^{2} = \frac{1024}{15}$.
6. The base of a solid is the region in the first quadrant bounded by $y=\cos(x)$, $y=\sin(x)$, and the y-axis. Cross-sections perpendicular to the x-axis are squares. What is the volume of the solid?
   Answer: $\frac{\pi}{4} - \frac{1}{2}$ The curves intersect when $\cos(x) = \sin(x)$, which is at $x = \pi/4$ in the first quadrant. From $x=0$ to $x=\pi/4$, $\cos(x) \ge \sin(x)$. The side of the square is $s(x) = \cos(x) - \sin(x)$. The area is $A(x) = (\cos(x) - \sin(x))^2 = \cos^2(x) - 2\sin(x)\cos(x) + \sin^2(x) = 1 - \sin(2x)$. The volume is $V = \int_{0}^{\pi/4} (1 - \sin(2x)) dx = [x + \frac{1}{2}\cos(2x)]_{0}^{\pi/4} = (\frac{\pi}{4} + \frac{1}{2}\cos(\frac{\pi}{2})) - (0 + \frac{1}{2}\cos(0)) = (\frac{\pi}{4} + 0) - (\frac{1}{2}) = \frac{\pi}{4} - \frac{1}{2}$.
7. A solid has its base as the region between $y = \sqrt{x}$ and $y = x^2$ for $0 \leq x \leq 1$. Each cross section perpendicular to the x-axis is a rectangle whose height is 3 times its base. Which expression gives the volume?
   Answer: $3\int_{0}^{1} (\sqrt{x} - x^2)^2 dx$ For $0 \leq x \leq 1$, we have $\sqrt{x} \geq x^2$. The base width of each rectangle is $\sqrt{x} - x^2$, and the height is $3(\sqrt{x} - x^2)$. The area of each rectangle is base × height = $(\sqrt{x} - x^2) \cdot 3(\sqrt{x} - x^2) = 3(\sqrt{x} - x^2)^2$. The volume is $\int_{0}^{1} 3(\sqrt{x} - x^2)^2 dx = 3\int_{0}^{1} (\sqrt{x} - x^2)^2 dx$. Choice A gives the area between the curves, not the cross-sectional area. Choice C gives 3 times the area between curves. Choice D omits the factor of 3 from the height condition.
8. Consider a solid whose base is bounded by $x = y^2$ and $x = 4$. Cross sections perpendicular to the x-axis are squares. If we set up the volume integral in the form $\int_{c}^{d} g(x) dx$, what is the correct expression for $g(x)$?
   Answer: $g(x) = 4x$ The parabola $x = y^2$ intersects the line $x = 4$ when $y^2 = 4$, so $y = \pm 2$. For a given $x$-value between 0 and 4, the $y$-values range from $-\sqrt{x}$ to $\sqrt{x}$, giving a total distance of $2\sqrt{x}$. This is the side length of the square cross section. The area of the square is $(2\sqrt{x})^2 = 4x$, so $g(x) = 4x$. Choice A gives twice the distance from the x-axis to one branch. Choice B gives the side length, not the area. Choice D gives half the side length.
9. A solid is constructed with base region defined by $|x| + |y| \leq 2$. Cross sections perpendicular to the x-axis are rectangles with height equal to half the base width. What is the volume of this solid?
   Answer: $\int_{-2}^{2} (2-|x|)^2 dx$ The region $|x| + |y| \leq 2$ is a diamond with vertices at $(\pm 2, 0)$ and $(0, \pm 2)$. For a given $x$, the $y$-values satisfy $|y| \leq 2 - |x|$, so $y$ ranges from $-(2-|x|)$ to $(2-|x|)$. The base width is $2(2-|x|)$ and the height is half this: $(2-|x|)$. The cross-sectional area is base × height = $2(2-|x|) \cdot (2-|x|) = (2-|x|)^2$. Choice B doubles the correct area. Choice C gives base width times height but missing the factor of 2 in base width. Choice D gives half the correct area.
10. The base of a solid is the region in the xy-plane bounded by the x-axis, the y-axis, and the line $x+2y=6$. Cross-sections perpendicular to the y-axis are squares. What is the volume of the solid?
    Answer: $36$ For cross-sections perpendicular to the y-axis, we need to express $x$ in terms of $y$: $x = 6-2y$. The line intersects the y-axis at $y=3$ and the x-axis at $x=6$. We integrate with respect to $y$ from $y=0$ to $y=3$. The side length of the square is the horizontal distance from the y-axis to the line: $s(y) = 6-2y$. The area is $A(y) = (6-2y)^2 = 36 - 24y + 4y^2$. The volume is $V = \int_{0}^{3} (36 - 24y + 4y^2) dy = [36y - 12y^2 + \frac{4}{3}y^3]_{0}^{3} = 36(3) - 12(3^2) + \frac{4}{3}(3^3) = 108 - 108 + 36 = 36$.
11. The base of a solid is the region in the xy-plane enclosed by the graphs of $y = x^2$ and $y = x+2$. The cross-sections of the solid perpendicular to the x-axis are squares. What is the volume of the solid?
    Answer: $\frac{81}{10}$ The points of intersection are found by setting $x^2 = x+2$, which gives $x^2 - x - 2 = 0$, or $(x-2)(x+1) = 0$. So, the limits of integration are from $x=-1$ to $x=2$. The upper curve is $y = x+2$ and the lower curve is $y = x^2$. The side length of a square cross-section at $x$ is $s(x) = (x+2) - x^2$. The area of the cross-section is $A(x) = s(x)^2 = (x+2-x^2)^2$. The volume is the integral of the area: $V = \int_{-1}^{2} (x+2-x^2)^2 dx = \int_{-1}^{2} (x^4 - 2x^3 - 3x^2 + 4x + 4) dx = [\frac{x^5}{5} - \frac{x^4}{2} - x^3 + 2x^2 + 4x]_{-1}^{2} = \frac{81}{10}$.
12. A solid is formed with base in the first quadrant bounded by $y = \sin x$, $y = \cos x$, and the y-axis. Cross sections perpendicular to the x-axis are rectangles with height twice the base width. What integral represents the volume?
    Answer: $\int_{0}^{\pi/4} 2(\cos x - \sin x)^2 dx$ The curves intersect at $x = \pi/4$ where $\sin x = \cos x$. For $0 \leq x \leq \pi/4$, we have $\cos x \geq \sin x$. The base width of each rectangle is $\cos x - \sin x$, and the height is twice this: $2(\cos x - \sin x)$. The area of each rectangle is base × height = $(\cos x - \sin x) \cdot 2(\cos x - \sin x) = 2(\cos x - \sin x)^2$. Choice A gives the area under one curve. Choice B gives twice the base width, not the rectangle area. Choice D omits the factor of 2 from the height condition.
13. The base of a solid is the region enclosed by the parabola $x = 4 - y^2$ and the y-axis. Cross-sections perpendicular to the y-axis are rectangles with a constant height of 5. What is the volume of the solid?
    Answer: $\frac{160}{3}$ The region is bounded by $x=4-y^2$ and $x=0$. The limits of integration are found by setting $4-y^2=0$, which gives $y=\pm 2$. The base of a rectangular cross-section at $y$ is $s(y) = (4-y^2)-0 = 4-y^2$. The height is a constant $h=5$. The area is $A(y) = s(y)h = 5(4-y^2)$. The volume is $V = \int_{-2}^{2} 5(4-y^2) dy = 5[4y - \frac{y^3}{3}]_{-2}^{2} = 5[(8-\frac{8}{3}) - (-8+\frac{8}{3})] = 5[\frac{32}{3}] = \frac{160}{3}$.
14. The base of a solid is the region bounded by $y = \sqrt{x}$, $x=9$, and the x-axis. The cross-sections perpendicular to the x-axis are squares whose diagonals lie in the xy-plane. What is the volume of the solid?
    Answer: $\frac{81}{4}$ The length of the diagonal of a square cross-section at $x$ is $d(x) = \sqrt{x} - 0 = \sqrt{x}$. The area of a square with diagonal $d$ is $A = \frac{d^2}{2}$. Therefore, the area of a cross-section is $A(x) = \frac{(\sqrt{x})^2}{2} = \frac{x}{2}$. The volume is $V = \int_{0}^{9} \frac{x}{2} dx = [\frac{x^2}{4}]_{0}^{9} = \frac{81}{4}$.
15. A solid is formed with base region bounded by $y = \sec x$ and $y = 2$ for $-\frac{\pi}{3} \leq x \leq \frac{\pi}{3}$. Cross sections perpendicular to the x-axis are rectangles with height equal to $\frac{3}{4}$ times the base width. What integral represents the volume?
    Answer: $\int_{-\pi/3}^{\pi/3} \frac{3}{4}(2 - \sec x)^2 dx$ When you encounter a volume problem with cross-sections, you need to visualize the solid's construction. Here, you're building a solid where each vertical slice at position x is a rectangle, and you integrate these slices to find total volume. First, identify the base width of each rectangular cross-section. The region is bounded above by $y = 2$ and below by $y = \sec x$. Since $\sec x \leq 2$ in the given interval, the base width at any x-value is $(2 - \sec x)$. The problem states that the height equals $\frac{3}{4}$ times this base width, so the height is $\frac{3}{4}(2 - \sec x)$. The area of each rectangular cross-section is base × height = $(2 - \sec x) \times \frac{3}{4}(2 - \sec x) = \frac{3}{4}(2 - \sec x)^2$. Integrating from $x = -\frac{\pi}{3}$ to $x = \frac{\pi}{3}$ gives the volume. Option A gives $\frac{3}{4}(2 - \sec x)$ without squaring—this would be the height alone, not the area of the cross-section. Option B incorrectly uses $(\sec x - 2)^2$, which assumes $\sec x > 2$, but in our interval $\sec x < 2$, making this expression negative. Option C uses $(\sec x - 2)$ without squaring, again missing the area calculation and using the wrong sign. Remember: for volume by cross-sections, always multiply the dimensions to get the cross-sectional area, then integrate. Pay careful attention to which function is on top to avoid sign errors.
16. The base of a solid is the region enclosed by $y=x^2$ and $y=k$, where $k>0$. Cross-sections perpendicular to the y-axis are squares. If the volume of the solid is 50, what is the value of $k$?
    Answer: $5$ For cross-sections perpendicular to the y-axis, we integrate with respect to $y$ from $y=0$ to $y=k$. The curves are $x = \sqrt{y}$ and $x = -\sqrt{y}$. The side length of the square is $s(y) = \sqrt{y} - (-\sqrt{y}) = 2\sqrt{y}$. The area is $A(y) = (2\sqrt{y})^2 = 4y$. The volume is $V = \int_{0}^{k} 4y dy = [2y^2]_{0}^{k} = 2k^2$. Given that $V=50$, we have $2k^2 = 50 \Rightarrow k^2 = 25 \Rightarrow k=5$.
17. Let $R$ be the region in the first quadrant bounded by $y=x$ and $y=x^2$. Let $V_S$ be the volume of a solid with base $R$ and square cross-sections perpendicular to the x-axis. Let $V_R$ be the volume of a solid with the same base $R$ but with rectangular cross-sections perpendicular to the x-axis whose height is half their base. What is the ratio $\frac{V_R}{V_S}$?
    Answer: $\frac{1}{2}$ For any cross-section at a given $x$, the base is $s(x) = x - x^2$. For the solid with squares, the area is $A_S(x) = s(x)^2 = (x-x^2)^2$. The volume is $V_S = \int_0^1 (x-x^2)^2 dx$. For the solid with rectangles, the area is $A_R(x) = s(x) \cdot h(x)$, where $h(x) = \frac{1}{2}s(x)$. So, $A_R(x) = s(x) \cdot \frac{1}{2}s(x) = \frac{1}{2}s(x)^2$. The volume is $V_R = \int_0^1 \frac{1}{2}(x-x^2)^2 dx = \frac{1}{2} \int_0^1 (x-x^2)^2 dx = \frac{1}{2}V_S$. Therefore, the ratio $\frac{V_R}{V_S} = \frac{1}{2}$.
18. The base of a solid is the region bounded by $y = \frac{1}{x}$, $x=1$, $x=4$, and $y=0$. Cross-sections perpendicular to the x-axis are rectangles whose height is always 3. What is the volume of the solid?
    Answer: $3 \ln 4$ The base of a rectangular cross-section at $x$ is $s(x) = \frac{1}{x} - 0 = \frac{1}{x}$. The height is given as a constant, $h=3$. The area of the cross-section is $A(x) = s(x)h = \frac{1}{x} \cdot 3 = \frac{3}{x}$. The volume is $V = \int_{1}^{4} \frac{3}{x} dx = 3[\ln|x|]_{1}^{4} = 3(\ln 4 - \ln 1) = 3 \ln 4$.
19. The base of a solid is the region enclosed by $y = 1 - x^2/4$ and the x-axis. Cross-sections perpendicular to the x-axis are rectangles whose perimeter is 12. What is the volume of the solid?
    Answer: $\frac{208}{15}$ The base extends from $x=-2$ to $x=2$. The base of a rectangular cross-section is $s(x) = 1 - x^2/4$. The perimeter is $P = 2s + 2h = 12$, so $s+h=6$. The height is $h(x) = 6 - s(x) = 6 - (1 - x^2/4) = 5 + x^2/4$. The area is $A(x) = s(x)h(x) = (1 - x^2/4)(5 + x^2/4) = 5 - x^2 - x^4/16$. The volume is $V = \int_{-2}^{2} (5 - x^2 - x^4/16) dx = 2 \int_{0}^{2} (5 - x^2 - x^4/16) dx = 2[5x - \frac{x^3}{3} - \frac{x^5}{80}]_{0}^{2} = 2(10 - \frac{8}{3} - \frac{32}{80}) = 2(10 - \frac{8}{3} - \frac{2}{5}) = 2(\frac{150-40-6}{15}) = 2(\frac{104}{15}) = \frac{208}{15}$.
20. The base of a solid is the region in the first quadrant bounded by $y = x^3$ and $y = x$. The cross-sections perpendicular to the x-axis are rectangles whose height is equal to the y-coordinate of the upper bounding curve. What is the volume of the solid?
    Answer: $\frac{2}{15}$ This is a volumes by cross-sections problem, where you need to set up an integral based on the geometry of the solid. When cross-sections are perpendicular to the x-axis, you integrate with respect to x, and each cross-section's area becomes your integrand. First, find where the curves intersect by solving $x^3 = x$, which gives $x(x^2 - 1) = 0$, so $x = 0, 1, -1$. Since we're in the first quadrant, the region spans from $x = 0$ to $x = 1$. Between these points, $y = x$ is above $y = x^3$ (since $x > x^3$ when $0 < x < 1$). Each rectangular cross-section has width equal to the distance between the curves: $x - x^3$. The height equals the y-coordinate of the upper curve, which is $y = x$. So each cross-section has area $(x - x^3) \cdot x = x^2 - x^4$. The volume is: $V = \int_0^1 (x^2 - x^4) dx = \left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \frac{1}{3} - \frac{1}{5} = \frac{5-3}{15} = \frac{2}{15}$ Choice A ($\frac{4}{21}$) likely comes from incorrectly using $x^3$ as the height instead of $x$. Choice B ($\frac{8}{105}$) might result from errors in setting up the cross-sectional area. Choice C ($\frac{2}{35}$) could come from computational mistakes in the integration. Remember: for cross-sections perpendicular to the x-axis, carefully identify the width and height of each cross-section, then integrate the area function over the appropriate interval.

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