Cross Sections: Triangles & Semicircles - Calculus 2

A solid has a base in the first quadrant bounded by $y = x^3$ and $y = x$ . If cross-sections perpendicular to the y-axis are equilateral triangles, which integral represents the volume?
Answer: $\frac{\sqrt{3}}{4} \int_{0}^{1} (\sqrt[3]{y} - y)^2 dy$ Since cross-sections are perpendicular to the y-axis, we must express the boundaries as functions of $y$ . The curves are $x = y$ and $x = \sqrt[3]{y}$ . The intersection points are (0,0) and (1,1). For $y \in [0, 1]$ , $\sqrt[3]{y} \ge y$ , so the right boundary is $x = \sqrt[3]{y}$ and the left is $x = y$ . The side length of the equilateral triangle is $s(y) = \sqrt[3]{y} - y$ . The area is $A(y) = \frac{\sqrt{3}}{4}s(y)^2 = \frac{\sqrt{3}}{4}(\sqrt[3]{y} - y)^2$ . The volume is the integral of this area from $y=0$ to $y=1$ .
The base of a solid is the region enclosed by $x^2 + 4y^2 = 4$ . Cross sections of the solid cut by planes perpendicular to the x-axis are isosceles right triangles. The legs of each triangle are equal in length and lie in the plane of the cross section, with one leg parallel to the y-axis. If the vertex of the right angle lies on the xy-plane, what is the volume of the solid?
Answer: $\frac{16}{3}$ The base region is an ellipse: x²/4 + y² = 1, so x ranges from -2 to 2. At each x, y ranges from -√(1 - x²/4) to √(1 - x²/4), giving a total width of 2√(1 - x²/4) = √(4 - x²). For the isosceles right triangle at position x, one leg is parallel to the y-axis with length √(4 - x²), and the other leg is parallel to the z-axis with the same length √(4 - x²). The area of this triangle is (1/2) × √(4 - x²) × √(4 - x²) = (4 - x²)/2. The volume is ∫[-2 to 2] (4 - x²)/2 dx = (1/2)∫[-2 to 2] (4 - x²) dx = (1/2)[4x - x³/3][-2 to 2] = (1/2)[8 - 8/3 - (-8 + 8/3)] = (1/2)[16 - 16/3] = (1/2)(32/3) = 16/3.
The integral $V = \int_{0}^{4} \frac{\sqrt{3}}{4}(4-y)^2 dy$ represents the volume of a solid. Which of the following could be the solid?
Answer: A solid with a base bounded by $x = 4-y$ , $x = 0$ , and $y = 0$ , with equilateral triangle cross-sections perpendicular to the y-axis. The integral is with respect to $y$ , so the cross-sections are perpendicular to the y-axis. The area of a cross-section is $A(y) = \frac{\sqrt{3}}{4}(4-y)^2$ . The term $\frac{\sqrt{3}}{4}s^2$ is the area of an equilateral triangle with side $s$ . Thus, the cross-sections are equilateral triangles and the side length is $s(y) = 4-y$ . This side length represents the horizontal width of the base at height $y$ . The expression $s(y) = 4-y$ can be interpreted as the distance from a right boundary $x = 4-y$ to a left boundary $x=0$ . The limits of integration $y=0$ to $y=4$ define the vertical extent of the base. The base is therefore the triangular region bounded by $x=4-y$ , $x=0$ , and $y=0$ .
Let $S_1$ be a solid with a base region R and cross-sections perpendicular to the x-axis that are semicircles. Let $S_2$ be a solid with the same base region R and cross-sections perpendicular to the x-axis that are equilateral triangles. What is the ratio of the volume of $S_1$ to the volume of $S_2$ ?
Answer: $\frac{\pi}{2\sqrt{3}}$ Let the side length of a cross-section at $x$ be $s(x)$ . For the semicircular cross-sections of $S_1$ , the area is $A_1(x) = \frac{\pi}{8}(s(x))^2$ . For the equilateral triangle cross-sections of $S_2$ , the area is $A_2(x) = \frac{\sqrt{3}}{4}(s(x))^2$ . The volume of each solid is the integral of its cross-sectional area. The ratio of the volumes is the ratio of the area formulas: $\frac{V(S_1)}{V(S_2)} = \frac{\int A_1(x) dx}{\int A_2(x) dx} = \frac{\int \frac{\pi}{8}(s(x))^2 dx}{\int \frac{\sqrt{3}}{4}(s(x))^2 dx} = \frac{\frac{\pi}{8}}{\frac{\sqrt{3}}{4}} = \frac{\pi}{8} \cdot \frac{4}{\sqrt{3}} = \frac{\pi}{2\sqrt{3}}$ .
The base of a solid is the region under the curve $y = 1/x$ from $x=1$ to $x=3$ . Cross-sections perpendicular to the x-axis are semicircles. Calculate the volume of the solid.
Answer: $\frac{\pi}{12}$ The diameter of the semicircle at a given $x$ is $s(x) = 1/x$ . The area of a semicircle with diameter $s$ is $A = \frac{\pi}{8}s^2$ . So, $A(x) = \frac{\pi}{8}(\frac{1}{x})^2 = \frac{\pi}{8x^2}$ . The volume is the integral of this area from $x=1$ to $x=3$ : $V = \int_{1}^{3} \frac{\pi}{8x^2} dx = \frac{\pi}{8} \int_{1}^{3} x^{-2} dx = \frac{\pi}{8} \left[ -x^{-1} \right]_{1}^{3} = \frac{\pi}{8} \left[ -\frac{1}{x} \right]_{1}^{3} = \frac{\pi}{8} \left( -\frac{1}{3} - (-1) \right) = \frac{\pi}{8} \left( \frac{2}{3} \right) = \frac{2\pi}{24} = \frac{\pi}{12}$ .
The base of a solid is the region enclosed by $y=2$ and $y=\sec(x)$ for $x$ in the interval $[-\pi/3, \pi/3]$ . Cross-sections perpendicular to the x-axis are equilateral triangles. Which integral represents the volume of the solid?
Answer: $\frac{\sqrt{3}}{4} \int_{-\pi/3}^{\pi/3} (2 - \sec(x))^2 dx$ In the interval $[-\pi/3, \pi/3]$ , $\sec(x)$ ranges from $\sec(0)=1$ to $\sec(\pi/3)=2$ . Thus, $2 \ge \sec(x)$ in this region. The side length of the equilateral triangle cross-section is $s(x) = \text{top} - \text{bottom} = 2 - \sec(x)$ . The area of an equilateral triangle is $A(s) = \frac{\sqrt{3}}{4}s^2$ . Therefore, the cross-sectional area is $A(x) = \frac{\sqrt{3}}{4}(2 - \sec(x))^2$ . The volume is the integral of this area over the given interval: $V = \int_{-\pi/3}^{\pi/3} \frac{\sqrt{3}}{4}(2 - \sec(x))^2 dx$ .
The base of a solid is the region bounded by the graph of $y = 4 - |x|$ and the x-axis. Cross-sections perpendicular to the x-axis are semicircles. What is the volume of the solid?
Answer: $\frac{16\pi}{3}$ The base is an isosceles triangle with vertices at (-4,0), (4,0), and (0,4). The diameter of a semicircular cross-section at $x$ is $s(x) = 4 - |x|$ . The area is $A(x) = \frac{\pi}{8}s(x)^2 = \frac{\pi}{8}(4 - |x|)^2$ . Due to symmetry, we can integrate from 0 to 4 and double the result. For $x \ge 0$ , $|x| = x$ . So, $V = 2 \int_{0}^{4} \frac{\pi}{8}(4-x)^2 dx = \frac{\pi}{4} \int_{0}^{4} (4-x)^2 dx$ . Let $u = 4-x$ , $du = -dx$ . The integral becomes $\frac{\pi}{4} \int_{4}^{0} u^2 (-du) = \frac{\pi}{4} \int_{0}^{4} u^2 du = \frac{\pi}{4} \left[ \frac{u^3}{3} \right]_{0}^{4} = \frac{\pi}{4} \left( \frac{64}{3} \right) = \frac{16\pi}{3}$ .
The base of a solid is the triangular region bounded by the line $x+2y=2$ and the coordinate axes. The cross-sections perpendicular to the x-axis are isosceles right triangles with their hypotenuse on the base. What is the volume of the solid?
Answer: $\frac{1}{6}$ This problem tests your ability to find volumes of solids with known cross-sections, a key application of integration in Calculus 2. When you see "cross-sections perpendicular to an axis," think about setting up an integral where each slice contributes an area that depends on position. First, you need to understand the base region. The triangle is bounded by $x + 2y = 2$ , the x-axis ( $y = 0$ ), and the y-axis ( $x = 0$ ). This gives vertices at $(0,0)$ , $(2,0)$ , and $(0,1)$ . For any x-value from 0 to 2, the height of the triangular base at that position is $y = \frac{2-x}{2}$ . Each cross-section is an isosceles right triangle with hypotenuse lying on the base. Since the hypotenuse has length $\frac{2-x}{2}$ , and in an isosceles right triangle the legs have length $\frac{\text{hypotenuse}}{\sqrt{2}}$ , each leg has length $\frac{2-x}{2\sqrt{2}}$ . The area of each triangular cross-section is $\frac{1}{2} \times \text{leg}^2 = \frac{1}{2} \times \left(\frac{2-x}{2\sqrt{2}}\right)^2 = \frac{(2-x)^2}{16}$ . The volume is $\int_0^2 \frac{(2-x)^2}{16} dx = \frac{1}{16} \int_0^2 (2-x)^2 dx = \frac{1}{16} \left[-\frac{(2-x)^3}{3}\right]_0^2 = \frac{1}{16} \times \frac{8}{3} = \frac{1}{6}$ . Choice A ( $\frac{1}{12}$ ) likely comes from incorrectly using the full triangle area instead of the isosceles right triangle. Choice B ( $\frac{2}{3}$ ) and C ( $\frac{1}{3}$ ) represent computational errors in the integration. Remember: always visualize the cross-section geometry carefully and double-check your area formula before integrating.
The base of a solid is the region enclosed by the ellipse $\frac{x^2}{9} + \frac{y^2}{16} = 1$ . Cross-sections perpendicular to the y-axis are equilateral triangles. Which integral represents the volume of the solid?
Answer: $\frac{\sqrt{3}}{4} \int_{-4}^{4} 36\left(1 - \frac{y^2}{16}\right) dy$ Cross-sections are perpendicular to the y-axis, so we solve the ellipse equation for $x$ . $\frac{x^2}{9} = 1 - \frac{y^2}{16} \implies x^2 = 9(1 - \frac{y^2}{16}) \implies x = \pm 3\sqrt{1 - \frac{y^2}{16}}$ . The horizontal side length of the triangle is $s(y) = 3\sqrt{1 - \frac{y^2}{16}} - (-3\sqrt{1 - \frac{y^2}{16}}) = 6\sqrt{1 - \frac{y^2}{16}}$ . The area of an equilateral triangle is $A(y) = \frac{\sqrt{3}}{4}s(y)^2 = \frac{\sqrt{3}}{4} \left( 6\sqrt{1 - \frac{y^2}{16}} \right)^2 = \frac{\sqrt{3}}{4} (36)(1 - \frac{y^2}{16})$ . The y-values for the ellipse range from -4 to 4. So the volume integral is $V = \int_{-4}^{4} \frac{\sqrt{3}}{4} (36)(1 - \frac{y^2}{16}) dy$ .
The base of a solid is the region bounded by the parabola $x = y^2$ and the line $x = 4$ . The cross-sections perpendicular to the y-axis are isosceles right triangles with their hypotenuse on the base. Which integral gives the volume of this solid?
Answer: $\frac{1}{4} \int_{-2}^{2} (4 - y^2)^2 dy$ The cross-sections are perpendicular to the y-axis, so we integrate with respect to $y$ . The region is bounded by $x = y^2$ (right boundary for $y<0$ , left for $y>0$ ) and $x=4$ (right boundary). The intersections are at $y^2=4$ , so $y = \pm 2$ . The length of a horizontal cross-section of the base is $s(y) = \text{right} - \text{left} = 4 - y^2$ . For an isosceles right triangle with hypotenuse $s$ , the legs are $s/\sqrt{2}$ , and the area is $A = \frac{1}{2}(\frac{s}{\sqrt{2}})^2 = \frac{1}{4}s^2$ . So, $A(y) = \frac{1}{4}(4 - y^2)^2$ . The volume is $V = \int_{-2}^{2} \frac{1}{4}(4 - y^2)^2 dy$ .
The base of a solid is the region bounded by $y=k$ and $y=x^2$ for some constant $k>0$ . Cross-sections perpendicular to the y-axis are equilateral triangles. If the volume of the solid is $8\sqrt{3}$ , what is the value of $k$ ?
Answer: 4 We integrate with respect to $y$ from $y=0$ to $y=k$ . At a given height $y$ , $x = \pm\sqrt{y}$ . The horizontal side length of the triangular cross-section is $s(y) = \sqrt{y} - (-\sqrt{y}) = 2\sqrt{y}$ . The area of an equilateral triangle is $A = \frac{\sqrt{3}}{4}s^2$ . So, $A(y) = \frac{\sqrt{3}}{4}(2\sqrt{y})^2 = \frac{\sqrt{3}}{4}(4y) = \sqrt{3}y$ . The volume is $V = \int_{0}^{k} \sqrt{3}y \,dy = \sqrt{3} \left[ \frac{y^2}{2} \right]_{0}^{k} = \frac{\sqrt{3}}{2}k^2$ . We are given $V = 8\sqrt{3}$ . So, $\frac{\sqrt{3}}{2}k^2 = 8\sqrt{3} \implies \frac{k^2}{2} = 8 \implies k^2 = 16 \implies k=4$ .
The base of a solid is the region in the first quadrant bounded by $y = e^x$ , $y = 1$ , and $x = 2$ . Cross-sections perpendicular to the x-axis are semicircles. Which integral represents the volume of the solid?
Answer: $\frac{\pi}{8} \int_{0}^{2} (e^x - 1)^2 dx$ The region is bounded above by $y=e^x$ and below by $y=1$ . The limits of integration are from $x=0$ (where $e^x=1$ ) to $x=2$ . The diameter of a semicircular cross-section is $s(x) = e^x - 1$ . The area of a semicircle with diameter $s$ is $A = \frac{\pi}{8}s^2$ . Thus, the area of a cross-section is $A(x) = \frac{\pi}{8}(e^x - 1)^2$ . The volume is given by the integral $V = \int_{0}^{2} \frac{\pi}{8}(e^x - 1)^2 dx$ .
The base of a solid is the circular region in the xy-plane defined by $x^2 + y^2 \le 9$ . Cross-sections perpendicular to the x-axis are semicircles. What is the volume of the solid?
Answer: $18\pi$ The base is a circle of radius 3. For a given $x$ , the base extends from $y = -\sqrt{9-x^2}$ to $y = \sqrt{9-x^2}$ . The diameter of the semicircular cross-section is $s(x) = \sqrt{9-x^2} - (-\sqrt{9-x^2}) = 2\sqrt{9-x^2}$ . The area of a semicircle with diameter $s$ is $A = \frac{\pi}{8}s^2$ . So, $A(x) = \frac{\pi}{8}(2\sqrt{9-x^2})^2 = \frac{\pi}{8}(4(9-x^2)) = \frac{\pi}{2}(9-x^2)$ . The volume is $V = \int_{-3}^{3} \frac{\pi}{2}(9-x^2) dx = \frac{\pi}{2} \left[ 9x - \frac{x^3}{3} \right]_{-3}^{3} = \frac{\pi}{2} \left( (27-9) - (-27+9) \right) = \frac{\pi}{2}(18 - (-18)) = \frac{\pi}{2}(36) = 18\pi$ . (This solid is half of a sphere of radius 3).
The base of a solid is the region enclosed by $y = \sin(x)$ and the x-axis for $0 \le x \le \pi$ . The cross-sections perpendicular to the x-axis are isosceles right triangles with one leg lying on the base. Find the volume of the solid.
Answer: $\frac{\pi}{4}$ The length of the leg on the base at a given $x$ is $s(x) = \sin(x) - 0 = \sin(x)$ . For an isosceles right triangle with leg $s$ , the area is $A = \frac{1}{2}s^2$ . So, the area of a cross-section is $A(x) = \frac{1}{2}(\sin(x))^2 = \frac{1}{2}\sin^2(x)$ . The volume is $V = \int_{0}^{\pi} \frac{1}{2}\sin^2(x) dx$ . Using the identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ , we get $V = \frac{1}{2} \int_{0}^{\pi} \frac{1 - \cos(2x)}{2} dx = \frac{1}{4} \left[ x - \frac{1}{2}\sin(2x) \right]_{0}^{\pi} = \frac{1}{4} \left( (\pi - 0) - (0 - 0) \right) = \frac{\pi}{4}$ .
A solid with volume $V$ is generated from a base region R with semicircular cross-sections perpendicular to the x-axis. A new solid is created using the same base region R, but with cross-sections that are equilateral triangles. What is the volume of the new solid in terms of $V$ ?
Answer: $\frac{2\sqrt{3}}{\pi}V$ Let $s(x)$ be the width of the base R at $x$ . The original volume is $V = V_{semi} = \int \frac{\pi}{8}(s(x))^2 dx$ . The new volume is $V_{new} = V_{tri} = \int \frac{\sqrt{3}}{4}(s(x))^2 dx$ . We can write $V_{new} = \frac{\sqrt{3}}{4} \int (s(x))^2 dx$ . From the first equation, $\int (s(x))^2 dx = \frac{8}{\pi}V$ . Substituting this into the second equation gives $V_{new} = \frac{\sqrt{3}}{4} \left( \frac{8}{\pi}V \right) = \frac{2\sqrt{3}}{\pi}V$ .
The base of a solid is the region bounded by $y = \ln(x)$ , the x-axis, and the line $x = e$ . Cross-sections perpendicular to the x-axis are isosceles right triangles with one leg on the base. Which of the following integrals represents the volume of the solid?
Answer: $\frac{1}{2} \int_{1}^{e} (\ln x)^2 dx$ The region is bounded by $y=\ln(x)$ and $y=0$ , from $x=1$ (where $\ln(x)=0$ ) to $x=e$ . The length of the leg on the base is $s(x) = \ln(x) - 0 = \ln(x)$ . The area of an isosceles right triangle with leg $s$ is $A = \frac{1}{2}s^2$ . Thus, the cross-sectional area is $A(x) = \frac{1}{2}(\ln x)^2$ . The volume is the integral of this area over the interval $[1, e]$ , which is $V = \int_{1}^{e} \frac{1}{2}(\ln x)^2 dx$ .
The volume of a solid is described by the integral $V = \frac{1}{4} \int_{1}^{5} (g(y))^2 dy$ . If the cross-sections are perpendicular to the y-axis, which of the following accurately describes the solid's cross-sections?
Answer: Isosceles right triangles with hypotenuse $g(y)$ on the base. The volume of a solid with cross-sections perpendicular to the y-axis is given by $V = \int_{c}^{d} A(y) dy$ , where $A(y)$ is the area of the cross-section at $y$ . From the given integral, we can identify the area function as $A(y) = \frac{1}{4}(g(y))^2$ . We must match this to the area formula for the given shapes, assuming $s = g(y)$ is the side length on the base. Semicircle: $A = \frac{\pi}{8}s^2$ . Equilateral triangle: $A = \frac{\sqrt{3}}{4}s^2$ . Isosceles right triangle with leg on base: $A = \frac{1}{2}s^2$ . Isosceles right triangle with hypotenuse on base: $A = \frac{1}{4}s^2$ . The formula $A(y) = \frac{1}{4}(g(y))^2$ matches that of an isosceles right triangle with its hypotenuse on the base.
A solid's base is enclosed by $x = y^{1/2}$ , the y-axis, and the line $y = 4$ . If cross-sections taken perpendicular to the y-axis are isosceles right triangles with their hypotenuse on the base, what is the volume?
Answer: 2 Cross-sections are perpendicular to the y-axis, so we integrate with respect to $y$ . The horizontal length of the base at height $y$ is $s(y) = y^{1/2} - 0 = \sqrt{y}$ . The cross-sections are isosceles right triangles with the hypotenuse $s(y)$ on the base. The area of such a triangle is $A(s) = \frac{1}{4}s^2$ . So, $A(y) = \frac{1}{4}(\sqrt{y})^2 = \frac{1}{4}y$ . The volume is found by integrating from $y=0$ to $y=4$ : $V = \int_{0}^{4} \frac{1}{4}y \,dy = \frac{1}{4} \left[ \frac{y^2}{2} \right]_{0}^{4} = \frac{1}{4} \left( \frac{16}{2} - 0 \right) = \frac{1}{4}(8) = 2$ .
A solid has its base in the xy-plane, bounded by the curve $y = \sqrt{x}$ , the x-axis, and the line $x = 4$ . If the cross-sections of the solid perpendicular to the x-axis are equilateral triangles, what is the volume of the solid?
Answer: $2\sqrt{3}$ The side length $s$ of an equilateral triangle at a given $x$ is the height of the region, so $s(x) = \sqrt{x} - 0 = \sqrt{x}$ . The area of an equilateral triangle with side $s$ is $A = \frac{\sqrt{3}}{4}s^2$ . Thus, the area of a cross-section is $A(x) = \frac{\sqrt{3}}{4}(\sqrt{x})^2 = \frac{\sqrt{3}}{4}x$ . The volume is found by integrating this area from $x=0$ to $x=4$ : $V = \int_{0}^{4} \frac{\sqrt{3}}{4}x \,dx = \frac{\sqrt{3}}{4} \left[ \frac{x^2}{2} \right]_{0}^{4} = \frac{\sqrt{3}}{4} \left( \frac{16}{2} - 0 \right) = \frac{\sqrt{3}}{4}(8) = 2\sqrt{3}$ .
Let R be a planar region. Let $V_S$ be the volume of a solid with base R and semicircular cross-sections. Let $V_T$ be the volume of a solid with the same base R and isosceles right triangle cross-sections with the hypotenuse on the base. What is the value of the ratio $\frac{V_S}{V_T}$ ?
Answer: $\frac{\pi}{2}$ This problem tests your understanding of volumes with cross-sections, where you integrate the area of each cross-section along the base region. The key insight is comparing how the areas of semicircular and triangular cross-sections relate when they share the same base width. Let's say the width of the base at any point is $w$ . For the semicircular cross-sections, the diameter equals $w$ , so the radius is $w/2$ . The area of a semicircle is $\frac{1}{2}\pi r^2 = \frac{1}{2}\pi \left(\frac{w}{2}\right)^2 = \frac{\pi w^2}{8}$ . For the isosceles right triangle with hypotenuse $w$ on the base, you need to find the legs. In an isosceles right triangle, if the hypotenuse is $w$ , each leg has length $\frac{w}{\sqrt{2}}$ . The area is $\frac{1}{2} \times \frac{w}{\sqrt{2}} \times \frac{w}{\sqrt{2}} = \frac{w^2}{4}$ . The ratio of areas at each cross-section is: $\frac{\frac{\pi w^2}{8}}{\frac{w^2}{4}} = \frac{\pi w^2}{8} \times \frac{4}{w^2} = \frac{\pi}{2}$ Since this ratio is constant for every cross-section, the volume ratio $\frac{V_S}{V_T} = \frac{\pi}{2}$ , which is choice D. Choice A ( $\frac{\pi}{4}$ ) likely comes from incorrectly using the full circle area instead of semicircle. Choice B ( $\frac{4}{\pi}$ ) is the reciprocal of A. Choice C ( $\frac{2}{\pi}$ ) is the reciprocal of the correct answer. Remember: when comparing cross-sectional volumes, focus on the area ratio at each slice—the volume ratio will be identical.

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