Cross Sections: Triangles & Semicircles - Calculus 1

1. A solid has as its base the region bounded by the curve $y=\sin(x)$ and the x-axis from $x=0$ to $x=\pi$. The cross-sections of the solid perpendicular to the x-axis are isosceles right triangles with their hypotenuse lying on the base. What is the volume of the solid?
   Answer: $\frac{\pi}{8}$ The base of each triangular cross-section has length $s(x) = \sin(x)$. If this length is the hypotenuse of an isosceles right triangle, its legs have length $s/\sqrt{2}$. The area of the triangle is $A(x) = \frac{1}{2} \cdot (\frac{s}{\sqrt{2}}) \cdot (\frac{s}{\sqrt{2}}) = \frac{1}{4}s^2$. Thus, $A(x) = \frac{1}{4}\sin^2(x)$. The volume is $V = \int_{0}^{\pi} \frac{1}{4}\sin^2(x) dx$. Using the power-reducing identity $\sin^2(x) = \frac{1-\cos(2x)}{2}$, we get $V = \frac{1}{4} \int_{0}^{\pi} \frac{1-\cos(2x)}{2} dx = \frac{1}{8} [x - \frac{1}{2}\sin(2x)]_{0}^{\pi} = \frac{1}{8}((\pi - 0) - (0 - 0)) = \frac{\pi}{8}$.
2. Let $R$ be the region enclosed by the graphs of $y=4-x^2$ and $y=x+2$. The region $R$ is the base of a solid. For the solid, each cross-section perpendicular to the x-axis is a semicircle. Which of the following integrals gives the volume of the solid?
   Answer: $\int_{-2}^{1} \frac{\pi}{8}(2-x-x^2)^2 dx$ The points of intersection are found by setting $4-x^2 = x+2$, which gives $x^2+x-2=0$, or $(x+2)(x-1)=0$. The bounds of integration are $x=-2$ and $x=1$. For a cross-section at a given $x$, the diameter of the semicircle is the vertical distance between the curves, $s(x) = (4-x^2) - (x+2) = 2-x-x^2$. The radius is $r(x) = s(x)/2$. The area of a semicircular cross-section is $A(x) = \frac{1}{2}\pi r(x)^2 = \frac{1}{2}\pi (\frac{s(x)}{2})^2 = \frac{\pi}{8}s(x)^2 = \frac{\pi}{8}(2-x-x^2)^2$. The volume is the integral of this area from $x=-2$ to $x=1$, which is $\int_{-2}^{1} \frac{\pi}{8}(2-x-x^2)^2 dx$.
3. Let $R$ be a region in the xy-plane that forms the base of two different solids. For the first solid, the cross-sections perpendicular to the x-axis are isosceles triangles with a height equal to their base, and its volume is $V_T$. For the second solid, the cross-sections are semicircles, and its volume is $V_S$. Which of the following statements must be true for any such region $R$?
   Answer: $V_T = \frac{4}{\pi}V_S$ Let the length of the cross-section base at a given $x$ be $s(x)$. For the first solid, the cross-section is an isosceles triangle with base $s(x)$ and height $s(x)$. Its area is $A_T(x) = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2}s(x)^2$. For the second solid, the cross-section is a semicircle with diameter $s(x)$. Its area is $A_S(x) = \frac{1}{2}\pi (\frac{s(x)}{2})^2 = \frac{\pi}{8}s(x)^2$. The ratio of the areas at any $x$ is constant: $\frac{A_T(x)}{A_S(x)} = \frac{\frac{1}{2}s(x)^2}{\frac{\pi}{8}s(x)^2} = \frac{1/2}{\pi/8} = \frac{4}{\pi}$. Since the volumes are the integrals of these area functions over the same interval, the ratio of the volumes is also constant: $\frac{V_T}{V_S} = \frac{4}{\pi}$. Therefore, $V_T = \frac{4}{\pi}V_S$.
4. The base of a solid is the region enclosed by the upper half of the circle $x^2+y^2=9$ and the x-axis. Cross-sections perpendicular to the x-axis are equilateral triangles. What is the volume of the solid?
   Answer: $9\sqrt{3}$ The base is the region defined by $y = \sqrt{9-x^2}$ for $x$ from $-3$ to $3$. At a given $x$, the side length of the equilateral triangle is $s(x) = y_{top} - y_{bottom} = \sqrt{9-x^2} - 0 = \sqrt{9-x^2}$. The area of an equilateral triangle with side $s$ is $A(x) = \frac{\sqrt{3}}{4}s(x)^2 = \frac{\sqrt{3}}{4}(\sqrt{9-x^2})^2 = \frac{\sqrt{3}}{4}(9-x^2)$. The volume is $V = \int_{-3}^{3} \frac{\sqrt{3}}{4}(9-x^2) dx$. The integrand is an even function, so we can write $V = 2 \cdot \frac{\sqrt{3}}{4} \int_{0}^{3} (9-x^2) dx = \frac{\sqrt{3}}{2} [9x - \frac{x^3}{3}]_{0}^{3} = \frac{\sqrt{3}}{2}((27-9)-0) = \frac{\sqrt{3}}{2}(18) = 9\sqrt{3}$.
5. The base of a solid, Solid 1, is the region enclosed by the circle $x^2+y^2=4$. Its cross-sections perpendicular to the x-axis are equilateral triangles. A second solid, Solid 2, has the same base, but its cross-sections perpendicular to the x-axis are semicircles. What is the ratio of the volume of Solid 1 to the volume of Solid 2?
   Answer: $\frac{2\sqrt{3}}{\pi}$ For any $x$ in $[-2, 2]$, the length of the base of a cross-section is $s(x) = \sqrt{4-x^2} - (-\sqrt{4-x^2}) = 2\sqrt{4-x^2}$. For Solid 1, the cross-sectional area is an equilateral triangle, $A_1(x) = \frac{\sqrt{3}}{4}s(x)^2 = \frac{\sqrt{3}}{4}(4(4-x^2)) = \sqrt{3}(4-x^2)$. For Solid 2, the cross-sectional area is a semicircle, $A_2(x) = \frac{\pi}{8}s(x)^2 = \frac{\pi}{8}(4(4-x^2)) = \frac{\pi}{2}(4-x^2)$. The volume is the integral of the area function. The ratio of the volumes is the ratio of the integrals: $\frac{V_1}{V_2} = \frac{\int_{-2}^{2} A_1(x) dx}{\int_{-2}^{2} A_2(x) dx} = \frac{\int_{-2}^{2} \sqrt{3}(4-x^2) dx}{\int_{-2}^{2} \frac{\pi}{2}(4-x^2) dx}$. Since the integrands differ only by a constant factor, the ratio is $\frac{\sqrt{3}}{\pi/2} = \frac{2\sqrt{3}}{\pi}$.
6. The base of a solid is the region in the first quadrant bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$. Cross-sections perpendicular to the y-axis are isosceles triangles with a height that is twice the length of their base. Which of the following integrals represents the volume of the solid?
   Answer: $\int_{0}^{3} \frac{16}{9}(9-y^2) dy$ Cross-sections are perpendicular to the y-axis, so we integrate with respect to $y$. The bounds are from $y=0$ to $y=3$. We need to solve the ellipse equation for $x$: $\frac{x^2}{16} = 1 - \frac{y^2}{9} = \frac{9-y^2}{9} \implies x^2 = \frac{16}{9}(9-y^2)$. Since we are in the first quadrant, $x = \frac{4}{3}\sqrt{9-y^2}$. This is the length of the base of the triangle, so $s(y) = \frac{4}{3}\sqrt{9-y^2}$. The height is given as $h(y) = 2s(y)$. The area of the triangular cross-section is $A(y) = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} s(y) (2s(y)) = s(y)^2$. So, $A(y) = (\frac{4}{3}\sqrt{9-y^2})^2 = \frac{16}{9}(9-y^2)$. The volume is the integral of the area: $V = \int_{0}^{3} \frac{16}{9}(9-y^2) dy$.
7. The base of a solid is the region enclosed by the graphs of $y=x$ and $y=x^3$. Cross-sections perpendicular to the x-axis are semicircles. What is the volume of the solid?
   Answer: $\frac{2\pi}{105}$ This is a solid of revolution problem where you need to set up an integral to find volume using cross-sectional areas. When cross-sections perpendicular to an axis have a known shape, the volume equals the integral of the cross-sectional area over the interval. First, find where the curves intersect by solving $x = x^3$, which gives $x(1-x^2) = 0$, so $x = 0, 1, -1$. Since we want the enclosed region, we integrate from $x = 0$ to $x = 1$, where $y = x$ is above $y = x^3$. The diameter of each semicircular cross-section is the distance between the curves: $x - x^3$. So the radius is $\frac{x - x^3}{2}$, and the area of each semicircle is $\frac{1}{2}\pi r^2 = \frac{1}{2}\pi\left(\frac{x - x^3}{2}\right)^2 = \frac{\pi(x - x^3)^2}{8}$. The volume is: $V = \int_0^1 \frac{\pi(x - x^3)^2}{8} dx = \frac{\pi}{8}\int_0^1 (x^2 - 2x^4 + x^6) dx = \frac{\pi}{8}\left[\frac{x^3}{3} - \frac{2x^5}{5} + \frac{x^7}{7}\right]_0^1 = \frac{\pi}{8}\left(\frac{1}{3} - \frac{2}{5} + \frac{1}{7}\right) = \frac{\pi}{8} \cdot \frac{4}{105} = \frac{2\pi}{105}$ Choice A) $\frac{\pi}{105}$ likely forgot the factor of 2 in the final calculation. Choice B) $\frac{\pi}{21}$ probably used the full circle area instead of semicircle. Choice C) $\frac{4\pi}{105}$ missed the $\frac{1}{8}$ factor when squaring the radius expression. Always carefully track your factors when setting up cross-sectional area formulas—semicircles, quarters of the radius expression, and integration constants can easily be missed.
8. The base of a solid is the region in the first quadrant bounded by the graph of $y=e^x$, the x-axis, the y-axis, and the line $x=k$ for some $k>0$. The cross-sections perpendicular to the x-axis are semicircles. If the volume of the solid is $\frac{\pi}{16}(e^{10}-1)$, what is the value of $k$?
   Answer: $5$ The diameter of a semicircular cross-section at a given $x$ is $s(x) = e^x$. The area of the semicircle is $A(x) = \frac{\pi}{8}s(x)^2 = \frac{\pi}{8}(e^x)^2 = \frac{\pi}{8}e^{2x}$. The volume is given by the integral $V = \int_{0}^{k} \frac{\pi}{8}e^{2x} dx$. Evaluating the integral gives $V = \frac{\pi}{8} [\frac{1}{2}e^{2x}]_{0}^{k} = \frac{\pi}{16}[e^{2k} - e^0] = \frac{\pi}{16}(e^{2k}-1)$. We are given that the volume is $\frac{\pi}{16}(e^{10}-1)$. Setting the two expressions for the volume equal, we have $e^{2k}-1 = e^{10}-1$, which simplifies to $e^{2k} = e^{10}$. Therefore, $2k=10$, and $k=5$.
9. The base of a solid is the region enclosed by the parabola $x=y^2$ and the line $x=4$. Each cross-section perpendicular to the y-axis is an equilateral triangle. What is the volume of the solid?
   Answer: $\frac{128\sqrt{3}}{15}$ Since cross-sections are perpendicular to the y-axis, we integrate with respect to $y$. The region extends from $y=-2$ to $y=2$ (where $y^2=4$). At a given $y$, the side length of the equilateral triangle is the horizontal distance $s(y) = x_{right} - x_{left} = 4 - y^2$. The area of an equilateral triangle with side $s$ is $A = \frac{\sqrt{3}}{4}s^2$. So, $A(y) = \frac{\sqrt{3}}{4}(4-y^2)^2 = \frac{\sqrt{3}}{4}(16 - 8y^2 + y^4)$. The volume is $V = \int_{-2}^{2} \frac{\sqrt{3}}{4}(16 - 8y^2 + y^4) dy$. By symmetry, $V = 2 \cdot \frac{\sqrt{3}}{4} \int_{0}^{2} (16 - 8y^2 + y^4) dy = \frac{\sqrt{3}}{2} [16y - \frac{8}{3}y^3 + \frac{1}{5}y^5]_{0}^{2} = \frac{\sqrt{3}}{2} (32 - \frac{64}{3} + \frac{32}{5}) = \frac{\sqrt{3}}{2} (\frac{480-320+96}{15}) = \frac{\sqrt{3}}{2}(\frac{256}{15}) = \frac{128\sqrt{3}}{15}$.
10. The base of a solid is the region bounded by the graph of $y = \ln(x)$, the line $x=e$, and the x-axis. The cross-sections of the solid perpendicular to the y-axis are isosceles right triangles with one leg in the xy-plane. Which integral represents the volume of the solid?
    Answer: $\int_{0}^{1} \frac{1}{2} (e - e^y)^2 dy$ Since the cross-sections are perpendicular to the y-axis, we must integrate with respect to $y$. We need to express the boundaries in terms of $y$. The curve $y=\ln(x)$ becomes $x=e^y$. The line $x=e$ is a right boundary. The x-axis ($y=0$) is the lower bound for $y$. The upper bound for $y$ is where $x=e^y$ meets $x=e$, so $e=e^y \implies y=1$. For a given $y$ between 0 and 1, the length of the base of the triangle is the horizontal distance $s(y) = x_{right} - x_{left} = e - e^y$. Since the cross-sections are isosceles right triangles with a leg in the base, the area is $A(y) = \frac{1}{2}s(y)^2 = \frac{1}{2}(e-e^y)^2$. The volume is the integral of this area from $y=0$ to $y=1$, which is $\int_{0}^{1} \frac{1}{2} (e - e^y)^2 dy$.

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