Disc Method: Other Axes - Calculus 2

1. A region is bounded by $y = \ln x$, $x = 1$, $x = e^2$, and $y = 0$. When this region is revolved around the horizontal line $y = -3$, the volume using the disc method is:
   Answer: $\pi \int_1^{e^2} (\ln x + 3)^2 \, dx$ When revolving around $y = -3$, each vertical slice at $x$ creates a disc with radius equal to the distance from $y = -3$ to $y = \ln x$, which is $\ln x - (-3) = \ln x + 3$. We integrate with respect to $x$ from $x = 1$ to $x = e^2$. The volume is $\pi \int_1^{e^2} (\ln x + 3)^2 \, dx$. Choice A uses the wrong variable (should integrate with respect to $x$, not $y$). Choices C and D incorrectly subtract $9$, applying washer method when there's no inner hole.
2. The region bounded by $y = \sin x$ and $y = 0$ for $0 \leq x \leq \pi$ is revolved around the line $y = -2$. Which integral represents the volume?
   Answer: $\pi \int_0^\pi (\sin x + 2)^2 \, dx$ When you're finding the volume of a solid of revolution using the disk/washer method, you need to carefully identify the outer and inner radii from the axis of rotation to the boundaries of your region. Here, you're rotating the region between $y = \sin x$ and $y = 0$ around the horizontal line $y = -2$. Since this axis is below your region, you'll use the disk method. The key insight is determining the radius function: it's the distance from any point on your curve to the axis of rotation. For any $x$ value, the upper boundary is $y = \sin x$ and the lower boundary is $y = 0$. The distance from $y = \sin x$ to the line $y = -2$ is $\sin x - (-2) = \sin x + 2$. The distance from $y = 0$ to $y = -2$ is $0 - (-2) = 2$. Since you're rotating the entire region between these curves, you get a washer with outer radius $\sin x + 2$ and inner radius $2$. Wait—actually, since $\sin x \geq 0$ on $[0,\pi]$, the region is solid (no hole), so this becomes a simple disk with radius $\sin x + 2$. The volume is $\pi \int_0^\pi (\sin x + 2)^2 \, dx$, which is choice C. Choice A incorrectly subtracts the squared radius from 4. Choice B treats this like a washer method with wrong radii. Choice D ignores the $\sin x$ boundary entirely. Study tip: Always sketch the region and axis of rotation first. The radius is always the distance from your curve to the axis—add or subtract coordinates accordingly based on their relative positions.
3. The region enclosed by $y=\cos(x)$, $y=1$, $x=0$, and $x=\pi/2$ is revolved about the line $y=1$. Which of the following integrals gives the volume of the generated solid?
   Answer: $\pi \int_{0}^{\pi/2} (1 - \cos(x))^2 dx$ The axis of revolution is horizontal ($y=1$), so we integrate with respect to $x$. The bounds are given as $x=0$ to $x=\pi/2$. Since the region is bounded by $y=1$ and revolved about $y=1$, this is a disc method problem. The radius of a disc at $x$ is the distance from the axis of revolution $y=1$ to the curve $y=\cos(x)$. This distance is $R(x) = 1 - \cos(x)$, since $1 \ge \cos(x)$ on the interval. The volume is given by the formula $V = \pi \int_{a}^{b} [R(x)]^2 dx$, which in this case is $V = \pi \int_{0}^{\pi/2} (1 - \cos(x))^2 dx$.
4. Consider the region bounded by $x = y^2 - 1$ and $x = 3$ for $y \in [-2, 2]$. When revolved around the line $x = -3$, the volume using the disc method is:
   Answer: $\pi \int_{-2}^2 [(6)^2 - (y^2 + 2)^2] \, dy$ The region extends from $x = y^2 - 1$ to $x = 3$ for $y \in [-2, 2]$. When revolving around $x = -3$, each horizontal slice creates a washer. The outer radius is the distance from $x = -3$ to $x = 3$, which is $3 - (-3) = 6$. The inner radius is the distance from $x = -3$ to $x = y^2 - 1$, which is $(y^2 - 1) - (-3) = y^2 + 2$. Volume is $\pi \int_{-2}^2 [6^2 - (y^2 + 2)^2] \, dy$. Choice A uses only the inner radius. Choice B has the correct structure but writes $y^2 - 1 + 3$ instead of simplifying to $y^2 + 2$. Choice D uses the wrong variable of integration.
5. The region bounded by $y = \tan(x)$, $x = \pi/4$, and the x-axis is revolved around the line $x = \pi/4$. Which of the following integrals represents the volume of the resulting solid?
   Answer: $\pi \int_{0}^{1} (\frac{\pi}{4} - \arctan(y))^2 dy$ The axis of revolution is vertical ($x = \pi/4$), so we must integrate with respect to $y$. The function must be expressed as $x$ in terms of $y$: $x = \arctan(y)$. The bounds for $y$ correspond to $x=0$ and $x=\pi/4$. When $x=0$, $y=\tan(0)=0$. When $x=\pi/4$, $y=\tan(\pi/4)=1$. So, $y$ ranges from 0 to 1. The axis of revolution $x=\pi/4$ is a boundary of the region. The radius of a disc at height $y$ is $R(y) = \frac{\pi}{4} - x = \frac{\pi}{4} - \arctan(y)$. The volume is $V = \pi \int_{0}^{1} [R(y)]^2 dy = \pi \int_{0}^{1} (\frac{\pi}{4} - \arctan(y))^2 dy$.
6. The region between $y = x^2$ and $y = 2x$ in the first quadrant is revolved around the line $y = 4$. The correct integral for the volume is:
   Answer: $\pi \int_0^2 [(4 - x^2)^2 - (4 - 2x)^2] \, dx$ The curves intersect where $x^2 = 2x$, so $x = 0$ or $x = 2$. For $x \in [0, 2]$, we have $2x \geq x^2$, so the region is between $y = x^2$ (bottom) and $y = 2x$ (top). When revolving around $y = 4$, each vertical slice creates a washer. The outer radius is $4 - x^2$ (distance from $y = 4$ to bottom curve) and inner radius is $4 - 2x$ (distance to top curve). Volume is $\pi \int_0^2 [(4 - x^2)^2 - (4 - 2x)^2] \, dx$. Choice B reverses the radii. Choices C and D attempt integration with respect to $y$ but have errors in identifying the correct boundaries and radii.
7. The region bounded by $y = \sec(x)$, $y=\sqrt{2}$, $x=-\pi/4$, and $x=\pi/4$ is revolved about the line $y=\sqrt{2}$. Which integral represents the volume?
   Answer: $\pi \int_{-\pi/4}^{\pi/4} (\sqrt{2} - \sec(x))^2 dx$ The axis of revolution is horizontal ($y=\sqrt{2}$), so we integrate with respect to $x$. The bounds are given as $x=-\pi/4$ to $x=\pi/4$. On this interval, $1 \le \sec(x) \le \sqrt{2}$, so the curve $y=\sec(x)$ is below the axis of revolution $y=\sqrt{2}$. The axis of revolution is a boundary of the region. The radius is $R(x) = \sqrt{2} - \sec(x)$. The volume is given by $V = \pi \int_{-\pi/4}^{\pi/4} (\sqrt{2} - \sec(x))^2 dx$. Choice C is mathematically equivalent because of the square, but the radius is defined as a positive distance, making A the better conceptual choice.
8. The region enclosed by $x = y^2 - 4$ and $x = 2y$ is revolved around the line $x = 6$. What is the correct setup for finding the volume using the disc method?
   Answer: $\pi \int_{-2}^4 [(6 - y^2 + 4)^2 - (6 - 2y)^2] \, dy$ First, find intersection points: $y^2 - 4 = 2y$ gives $y^2 - 2y - 4 = 0$, so $y = -2, 4$. For $y \in [-2, 4]$, we have $2y > y^2 - 4$, so the region extends from $x = y^2 - 4$ to $x = 2y$. When revolving around $x = 6$, we use the washer method (not pure disc method) since there's a hole. The outer radius is $6 - (y^2 - 4) = 10 - y^2$ and inner radius is $6 - 2y$. Volume is $\pi \int_{-2}^4 [(10 - y^2)^2 - (6 - 2y)^2] \, dy$. Note that $10 - y^2 = 6 - y^2 + 4$. Choices A and C use only one radius. Choice D reverses the radii.
9. The region R is bounded by $y = x^2$, the x-axis, and the line $x = a$ for $a > 0$. R is revolved about the line $x = a$. If the volume of the resulting solid is $\frac{8\pi}{3}$, what is the value of $a$?
   Answer: $2$ The axis of revolution is vertical ($x=a$), so we integrate with respect to $y$. We rewrite the function as $x=\sqrt{y}$. The y-bounds are from $y=0$ to $y=a^2$. The axis $x=a$ is a boundary of the region. The radius is $R(y) = a - \sqrt{y}$. The volume is $V = \pi \int_{0}^{a^2} (a - \sqrt{y})^2 dy = \pi \int_{0}^{a^2} (a^2 - 2ay^{1/2} + y) dy$. Integrating gives $V = \pi [a^2y - 2a(\frac{2}{3})y^{3/2} + \frac{1}{2}y^2]_{0}^{a^2} = \pi [a^2(a^2) - \frac{4a}{3}(a^2)^{3/2} + \frac{1}{2}(a^2)^2] = \pi [a^4 - \frac{4a}{3}(a^3) + \frac{1}{2}a^4] = \pi [a^4 - \frac{4}{3}a^4 + \frac{1}{2}a^4] = \pi a^4 (1 - \frac{4}{3} + \frac{1}{2}) = \pi a^4 (\frac{6-8+3}{6}) = \frac{\pi a^4}{6}$. We are given $V=\frac{8\pi}{3}$. So, $\frac{\pi a^4}{6} = \frac{8\pi}{3} \implies a^4 = \frac{8 \cdot 6}{3} = 16$. Thus, $a = 2$ (since $a>0$).
10. The region bounded by $y=\ln(x)$, $y=1$, and $x=1$ is revolved about the line $x=1$. What is the volume of the solid?
    Answer: $\pi (\frac{1}{2}e^2 - 2e + \frac{5}{2})$ The axis of revolution is vertical ($x=1$), so we integrate with respect to $y$. We express the curve as $x = e^y$. The region is bounded by $x=1$ and $x=e^y$. The y-bounds are determined by the horizontal lines bounding the region. The curve $y=\ln(x)$ passes through $(1,0)$, so the lower bound is $y=0$. The upper bound is given as $y=1$. The axis $x=1$ is a boundary of the region. The radius is $R(y) = e^y - 1$. The volume is $V = \pi \int_{0}^{1} (e^y - 1)^2 dy = \pi \int_{0}^{1} (e^{2y} - 2e^y + 1) dy$. Evaluating the integral gives $V = \pi [\frac{1}{2}e^{2y} - 2e^y + y]_{0}^{1} = \pi [(\frac{1}{2}e^2 - 2e + 1) - (\frac{1}{2}e^0 - 2e^0 + 0)] = \pi [(\frac{1}{2}e^2 - 2e + 1) - (\frac{1}{2} - 2)] = \pi [\frac{1}{2}e^2 - 2e + 1 + \frac{3}{2}] = \pi [\frac{1}{2}e^2 - 2e + \frac{5}{2}]$.
11. The region bounded by $y=1$, $y=1/\sqrt{x}$, and $x=4$ is revolved about the line $y=1$. What is the volume of the solid?
    Answer: $\pi (\ln(4) - 1)$ When you encounter a solid of revolution problem, you need to identify the region being revolved and choose the appropriate method. Since we're revolving around a horizontal line (y = 1), the washer method is ideal. First, visualize the region. The curve $y = \frac{1}{\sqrt{x}}$ intersects $y = 1$ when $\frac{1}{\sqrt{x}} = 1$, so $x = 1$. The region is bounded by these curves from $x = 1$ to $x = 4$. For the washer method revolving around $y = 1$, you need the outer radius $R(x)$ and inner radius $r(x)$. Since $\frac{1}{\sqrt{x}} > 1$ when $x < 1$ and $\frac{1}{\sqrt{x}} < 1$ when $x > 1$, in our interval $[1,4]$, the curve $y = \frac{1}{\sqrt{x}}$ lies below $y = 1$. Therefore: - Outer radius: $R(x) = 1 - 1 = 0$ (distance from axis $y = 1$ to line $y = 1$) - Inner radius: $r(x) = 1 - \frac{1}{\sqrt{x}}$ (distance from axis to curve) Wait - this setup is incorrect. The outer radius should be $R(x) = 1 - \frac{1}{\sqrt{x}}$ and inner radius is 0, giving us: $V = \pi \int_1^4 \left(1 - \frac{1}{\sqrt{x}}\right)^2 dx = \pi \int_1^4 \left(1 - \frac{2}{\sqrt{x}} + \frac{1}{x}\right) dx$ $= \pi \left[x - 4\sqrt{x} + \ln|x|\right]_1^4 = \pi[(4 - 8 + \ln 4) - (1 - 4 + 0)] = \pi(\ln 4 - 1)$ This confirms answer D. Answer A uses $\ln 4$ but with wrong arithmetic. Answer B substitutes $\ln 2$ incorrectly. Answer C has the wrong sign relationship. Strategy tip: Always sketch the region first and carefully identify which function is farther from the axis of revolution - this determines your outer radius.
12. The region bounded by $y=x^2-2$ and $y=-2$ from $x=0$ to $x=1$ is revolved about the line $y=-2$. Find the volume of the solid.
    Answer: $\pi/5$ When you encounter a volume of revolution problem, start by visualizing the region and identifying the axis of rotation. Here, you're rotating the region between $y = x^2 - 2$ and $y = -2$ around the line $y = -2$. Since the axis of rotation is $y = -2$, which is also the lower boundary of your region, this becomes a disk method problem. The radius of each disk at position $x$ is the distance from the axis $y = -2$ to the upper curve $y = x^2 - 2$. This radius is: $r(x) = (x^2 - 2) - (-2) = x^2$. Using the disk method formula: $V = \pi \int_0^1 [r(x)]^2 \, dx = \pi \int_0^1 (x^2)^2 \, dx = \pi \int_0^1 x^4 \, dx$ Evaluating: $V = \pi \left[\frac{x^5}{5}\right]_0^1 = \pi \cdot \frac{1}{5} = \frac{\pi}{5}$ The incorrect answers stem from common mistakes: A) $\frac{2\pi}{3}$ likely comes from incorrectly using the washer method or miscalculating the integral. B) $\frac{2\pi}{5}$ suggests you found the correct integral but doubled the result, possibly by confusion about the setup. C) $\frac{\pi}{3}$ indicates an integration error, perhaps evaluating $\int x^4 dx$ as $\frac{x^4}{3}$ instead of $\frac{x^5}{5}$. The correct answer is D) $\frac{\pi}{5}$. Strategy tip: Always identify whether the axis of rotation forms a boundary of your region. If it does, use the disk method; if not, you'll likely need the washer method.
13. Let R be the region enclosed by the graphs of $y = x^2$, $y = 4$, and $x = 0$. The region R is revolved about the line $y = 4$. What is the volume of the solid generated?
    Answer: $\frac{256\pi}{15}$ The region is revolved around a horizontal line, so we integrate with respect to $x$. The intersection of $y=x^2$ and $y=4$ is at $x=2$. The bounds of integration are from $x=0$ to $x=2$. The axis of revolution $y=4$ is a boundary of the region, so we use the disc method. The radius of a disc at a given $x$ is $R(x) = 4 - x^2$. The volume is given by the integral $V = \pi \int_{0}^{2} (R(x))^2 dx = \pi \int_{0}^{2} (4 - x^2)^2 dx$. Expanding the integrand gives $V = \pi \int_{0}^{2} (16 - 8x^2 + x^4) dx$. Evaluating the integral gives $V = \pi [16x - \frac{8}{3}x^3 + \frac{1}{5}x^5]_{0}^{2} = \pi (32 - \frac{64}{3} + \frac{32}{5}) = \pi (\frac{480 - 320 + 96}{15}) = \frac{256\pi}{15}$.
14. The region enclosed by the parabola $x = y^2+1$ and the line $x=2$ is revolved about the line $x=2$. What is the volume of the resulting solid?
    Answer: $\frac{16\pi}{15}$ The axis of revolution is vertical ($x=2$), so we integrate with respect to $y$. First, find the bounds of integration by setting the x-values equal: $y^2+1 = 2 \implies y^2 = 1 \implies y = \pm 1$. The axis of revolution is a boundary of the region, so we use the disc method. The radius of a disc at height $y$ is the horizontal distance from the axis to the curve: $R(y) = 2 - (y^2+1) = 1-y^2$. The volume is given by $V = \pi \int_{-1}^{1} (1-y^2)^2 dy$. By symmetry, this is $2\pi \int_{0}^{1} (1-2y^2+y^4) dy = 2\pi [y - \frac{2}{3}y^3 + \frac{1}{5}y^5]_{0}^{1} = 2\pi (1 - \frac{2}{3} + \frac{1}{5}) = 2\pi (\frac{15-10+3}{15}) = 2\pi (\frac{8}{15}) = \frac{16\pi}{15}$.
15. A region is enclosed by $x + y = 3$, $x = 0$, and $y = 0$. When this triangular region is revolved around the line $x = 5$, the volume is:
    Answer: $\pi \int_0^3 [(5 - 0)^2 - (5 - (3 - y))^2] \, dy$ The triangular region has vertices at $(0,0)$, $(3,0)$, and $(0,3)$. From the line $x + y = 3$, we get $x = 3 - y$. When revolving around $x = 5$, each horizontal slice at height $y$ creates a washer extending from $x = 0$ to $x = 3 - y$. The outer radius is $5 - 0 = 5$ and the inner radius is $5 - (3 - y) = 2 + y$. Volume is $\pi \int_0^3 [5^2 - (5 - (3 - y))^2] \, dy = \pi \int_0^3 [25 - (2 + y)^2] \, dy$. Choice A uses only inner radius. Choice B has incorrect outer radius calculation. Choice C uses wrong inner radius.
16. The region R is bounded by $y=f(x)$, $y=k$, $x=a$, and $x=b$, where $f(x) \le k$ for $x$ in $[a, b]$. A solid is formed by revolving R about the line $y=k$. Which expression gives the volume of this solid?
    Answer: $\pi \int_{a}^{b} (k - f(x))^2 dx$ The solid is generated by revolving a region about one of its boundaries ($y=k$), so the disc method is applicable. The axis of revolution is horizontal, so the integration is with respect to $x$ from $a$ to $b$. The radius $R(x)$ of a disc at a given $x$ is the distance between the axis $y=k$ and the curve $y=f(x)$. Since $f(x) \le k$, the radius is $R(x) = k - f(x)$. The volume of the solid is given by $V = \pi \int_{a}^{b} [R(x)]^2 dx = \pi \int_{a}^{b} (k - f(x))^2 dx$.
17. The region enclosed by the graphs of $y=1/x$, $y=-1$, $x=1$, and $x=3$ is revolved about the line $y=-1$. Which integral correctly represents the volume of the solid?
    Answer: $\pi \int_{1}^{3} (\frac{1}{x} + 1)^2 dx$ The axis of revolution is the horizontal line $y=-1$, which is a boundary of the specified region. Therefore, the disc method applies. We integrate with respect to $x$ from $x=1$ to $x=3$. The radius $R(x)$ of a disc at a given $x$ is the distance from the axis of revolution $y=-1$ to the other boundary curve, $y=1/x$. Since $1/x > -1$ for $x>0$, the radius is $R(x) = \frac{1}{x} - (-1) = \frac{1}{x} + 1$. The volume formula for the disc method is $V = \pi \int_{a}^{b} [R(x)]^2 dx$. Substituting the radius and bounds gives $V = \pi \int_{1}^{3} (\frac{1}{x} + 1)^2 dx$.
18. The region bounded by $y=2\sqrt{x}$, $y=4$, and $x=0$ is revolved about the line $y=4$. What is the volume of the solid?
    Answer: $\frac{32\pi}{3}$ The axis of revolution is horizontal ($y=4$), so we integrate with respect to $x$. Find the intersection of $y=2\sqrt{x}$ and $y=4$: $4=2\sqrt{x} \implies \sqrt{x}=2 \implies x=4$. The bounds of integration are from $x=0$ to $x=4$. The axis of revolution, $y=4$, is a boundary of the region. The radius of a disc is $R(x) = 4 - 2\sqrt{x}$. The volume is $V = \pi \int_{0}^{4} (4 - 2\sqrt{x})^2 dx = \pi \int_{0}^{4} (16 - 16x^{1/2} + 4x) dx$. Evaluating the integral: $V = \pi [16x - 16(\frac{2}{3})x^{3/2} + 2x^2]_{0}^{4} = \pi [(16(4) - \frac{32}{3}(4^{3/2}) + 2(4^2)) - 0] = \pi [64 - \frac{32}{3}(8) + 32] = \pi [96 - \frac{256}{3}] = \pi [\frac{288 - 256}{3}] = \frac{32\pi}{3}$.
19. Let R be the region bounded by the y-axis, the line $y=x-2$, and the line $y=1$. What is the volume of the solid formed by revolving R about the line $y=1$?
    Answer: $9\pi$ The axis of revolution is horizontal ($y=1$), so we integrate with respect to $x$. The region is bounded by the y-axis ($x=0$), the line $y=1$, and the line $y=x-2$. The intersection of $y=x-2$ and $y=1$ occurs at $1=x-2$, which gives $x=3$. So the bounds of integration are from $x=0$ to $x=3$. The axis of revolution $y=1$ is a boundary of the region. The radius of a disc is the vertical distance from the axis to the other boundary: $R(x) = 1 - (x-2) = 3-x$. The volume is $V = \pi \int_{0}^{3} (3-x)^2 dx$. Evaluating the integral: $V = \pi [-\frac{(3-x)^3}{3}]_{0}^{3} = \pi [0 - (-\frac{3^3}{3})] = \pi [\frac{27}{3}] = 9\pi$.
20. The region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 4$ is revolved around the line $x = -1$. Using the disc method, which integral correctly represents the volume of the resulting solid?
    Answer: $\pi \int_0^2 (y^2 + 1)^2 \, dy$ When revolving around $x = -1$, we need to express the region in terms of $y$. From $y = \sqrt{x}$, we get $x = y^2$. The region extends from $y = 0$ to $y = 2$. The radius from the axis $x = -1$ to any point on the curve is the distance from $x = -1$ to $x = y^2$, which is $y^2 - (-1) = y^2 + 1$. Using the disc method, the volume is $\pi \int_0^2 (y^2 + 1)^2 \, dy$. Choice B uses the wrong variable of integration. Choice C incorrectly applies the washer method. Choice D uses both wrong variable and wrong method.

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