Disc Method: x- or y-Axis - Calculus 1

Let $f(x)$ be a piecewise function defined as $f(x) = \begin{cases} x & 0 \le x \le 2 \\ 4-x & 2 < x \le 4 \end{cases}$ . Let $R$ be the region bounded by the graph of $f(x)$ and the x-axis. What is the volume of the solid generated by revolving $R$ about the x-axis?
Answer: $\frac{16\pi}{3}$ When you encounter a piecewise function in a volume of revolution problem, you need to set up separate integrals for each piece and add them together. To find the volume when revolving region $R$ about the x-axis, use the disk method: $V = \pi \int_a^b [f(x)]^2 \, dx$ Since $f(x)$ has two pieces, you'll need two integrals: - For $0 \le x \le 2$ : $f(x) = x$ , so $[f(x)]^2 = x^2$ - For $2 < x \le 4$ : $f(x) = 4-x$ , so $[f(x)]^2 = (4-x)^2$ The total volume is: $V = \pi \int_0^2 x^2 \, dx + \pi \int_2^4 (4-x)^2 \, dx$ For the first integral: $\int_0^2 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}$ For the second integral: $\int_2^4 (4-x)^2 \, dx = \int_2^4 (16-8x+x^2) \, dx = \left[16x-4x^2+\frac{x^3}{3}\right]_2^4 = \frac{8}{3}$ Therefore: $V = \pi \cdot \frac{8}{3} + \pi \cdot \frac{8}{3} = \frac{16\pi}{3}$ , which is answer C. Answer A ( $4\pi$ ) likely comes from forgetting to square the function in the disk formula. Answer B ( $\frac{8\pi}{3}$ ) represents calculating only one of the two integrals. Answer D ( $16\pi$ ) results from forgetting the $\pi$ factor or miscalculating the integrals. Study tip: Always break piecewise functions into separate integrals at their boundary points, and remember that the disk method requires $[f(x)]^2$ , not just $f(x)$ .
The integral $V = \pi \int_{0}^{\pi/4} \tan^2(x) dx$ represents the volume of a solid of revolution. Which of the following describes the solid?
Answer: The solid formed by revolving the region bounded by $y = \tan(x)$ , the x-axis, and $x = \pi/4$ about the x-axis. When you encounter an integral with the form $\pi \int_a^b [f(x)]^2 dx$ , this immediately signals the disk method for finding volumes of solids of revolution about the x-axis. The key insight is recognizing that the integrand represents the square of the radius function. In this problem, we have $V = \pi \int_{0}^{\pi/4} \tan^2(x) dx$ . This means we're rotating a region about the x-axis, where the radius of each circular cross-section is $\tan(x)$ . Since we integrate from $x = 0$ to $x = \pi/4$ , and the radius is $\tan(x)$ , the original region must be bounded by $y = \tan(x)$ , the x-axis (where $y = 0$ ), and the vertical line $x = \pi/4$ . This matches option D perfectly. Option A is incorrect because if we rotated $y = \sec(x)$ about the x-axis, the integral would contain $\sec^2(x)$ , not $\tan^2(x)$ . Option B suggests the function itself is $y = \tan^2(x)$ , but then rotating this about the x-axis would give us $\pi \int_0^{\pi/4} [\tan^2(x)]^2 dx = \pi \int_0^{\pi/4} \tan^4(x) dx$ . Option C describes rotation about the y-axis, which would require the washer method with a completely different integral setup involving $dy$ and different bounds. Remember: when you see $\pi \int_a^b [f(x)]^2 dx$ , think "disk method about the x-axis" where $f(x)$ is the radius function. The square in the integrand is your clue that $f(x)$ itself (not $[f(x)]^2$ ) defines the boundary of the original region.
Let $R$ be the region bounded by the graph of $y = \ln(x)$ , the x-axis, and the line $x = e^2$ . What is the volume of the solid generated by revolving $R$ about the x-axis?
Answer: $2\pi(e^2-1)$ The volume is found using the disc method. The region starts where $y = \ln(x)$ intersects the x-axis, which is at $x=1$ . The bounds of integration are from $x=1$ to $x=e^2$ . The volume $V$ is given by the integral $V = \pi \int_{1}^{e^2} (\ln(x))^2 dx$ . This integral requires integration by parts twice. Let $u = (\ln x)^2$ and $dv = dx$ . Then $du = \frac{2\ln x}{x} dx$ and $v = x$ . The integral becomes $x(\ln x)^2 - \int x \frac{2\ln x}{x} dx = x(\ln x)^2 - 2\int \ln x dx$ . The integral of $\ln x$ is $x\ln x - x$ . So the antiderivative is $x(\ln x)^2 - 2(x\ln x - x)$ . Evaluating from $1$ to $e^2$ : $V = \pi [ (e^2(\ln e^2)^2 - 2e^2\ln e^2 + 2e^2) - (1(\ln 1)^2 - 2(1)\ln 1 + 2(1)) ] = \pi [ (e^2(2)^2 - 2e^2(2) + 2e^2) - (0 - 0 + 2) ] = \pi[4e^2 - 4e^2 + 2e^2 - 2] = \pi(2e^2-2) = 2\pi(e^2-1)$ .
Let $R$ be the region bounded by the graph of $y=\sqrt{x-1}$ , the x-axis, and the line $x=5$ . What is the volume of the solid generated by revolving $R$ about the x-axis?
Answer: $8\pi$ When you encounter a problem about revolving a region around an axis, you're dealing with the disk/washer method for finding volumes of revolution. The key is identifying the radius function and setting up the correct integral. For this region bounded by $y = \sqrt{x-1}$ , the x-axis, and $x = 5$ , you need to first determine the bounds of integration. Since $y = \sqrt{x-1}$ intersects the x-axis when $y = 0$ , we have $0 = \sqrt{x-1}$ , so $x = 1$ . The region extends from $x = 1$ to $x = 5$ . When revolving around the x-axis, each cross-section perpendicular to the x-axis forms a disk with radius $r(x) = \sqrt{x-1}$ . The volume formula is: $V = \pi \int_1^5 [\sqrt{x-1}]^2 \, dx = \pi \int_1^5 (x-1) \, dx$ Evaluating: $V = \pi \left[\frac{(x-1)^2}{2}\right]_1^5 = \pi \left[\frac{16}{2} - 0\right] = 8\pi$ Answer C is correct. Answer A ( $12\pi$ ) likely comes from incorrectly using $\int_1^5 \sqrt{x-1} \, dx$ instead of squaring the radius function. Answer B ( $\frac{15\pi}{2}$ ) might result from computational errors in the integration bounds. Answer D ( $\frac{16\pi}{3}$ ) could stem from incorrectly treating this as a cubic integration or using wrong integration techniques. Remember: for volumes of revolution using the disk method, always square the radius function. The formula is $V = \pi \int_a^b [r(x)]^2 \, dx$ , not $V = \pi \int_a^b r(x) \, dx$ .
Consider the region $R$ bounded by $y=x^p$ , the x-axis, and the line $x=1$ , where $p$ is a positive constant. When $R$ is revolved about the x-axis, the resulting volume is $\pi/5$ . What is the value of $p$ ?
Answer: $2$ This problem tests your ability to set up and solve volume integrals using the disk method for solids of revolution. When you see a region being revolved around an axis, immediately think about which method applies and how to set up the integral. To find the volume when region $R$ is revolved about the x-axis, you'll use the disk method: $V = \pi \int_0^1 [f(x)]^2 \, dx$ . Here, $f(x) = x^p$ , so the volume integral becomes: $V = \pi \int_0^1 (x^p)^2 \, dx = \pi \int_0^1 x^{2p} \, dx$ Evaluating this integral: $V = \pi \left[ \frac{x^{2p+1}}{2p+1} \right]_0^1 = \pi \cdot \frac{1^{2p+1} - 0^{2p+1}}{2p+1} = \frac{\pi}{2p+1}$ Since we're told the volume equals $\pi/5$ , you can set up the equation: $\frac{\pi}{2p+1} = \frac{\pi}{5}$ Solving: $2p+1 = 5$ , so $p = 2$ . Looking at the wrong answers: Choice (A) $p = 4$ would give volume $\frac{\pi}{9}$ , which is too small. Choice (C) $p = \frac{1}{2}$ would give volume $\frac{\pi}{2}$ , which is too large. Choice (D) $p = \frac{5}{2}$ would give volume $\frac{\pi}{6}$ , also too small. Study tip: When solving volume problems with the disk method, always remember the formula involves squaring the function, so $V = \pi \int [f(x)]^2 dx$ . Set up your integral carefully, then use the given volume to solve for unknown parameters.
A region $R$ is enclosed by the graph of $y = \sec(x)$ , the x-axis, the y-axis, and the line $x=\pi/3$ . What is the volume of the solid generated by revolving $R$ about the x-axis?
Answer: $\pi\sqrt{3}$ When you encounter a volume problem involving revolution around the x-axis, you need to use the disk method: $V = \pi \int_a^b [f(x)]^2 \, dx$ . Here, the region is bounded by $y = \sec(x)$ , the axes, and $x = \pi/3$ , so you're integrating from $x = 0$ to $x = \pi/3$ . Setting up the integral: $V = \pi \int_0^{\pi/3} [\sec(x)]^2 \, dx = \pi \int_0^{\pi/3} \sec^2(x) \, dx$ The key insight is recognizing that $\sec^2(x)$ is the derivative of $\tan(x)$ . Therefore: $V = \pi[\tan(x)]_0^{\pi/3} = \pi[\tan(\pi/3) - \tan(0)] = \pi[\sqrt{3} - 0] = \pi\sqrt{3}$ This confirms answer D is correct. Answer A likely comes from incorrectly trying to integrate $\sec(x)$ instead of $\sec^2(x)$ , leading to a complex expression involving $\ln|\sec(x) + \tan(x)|$ . Answer B represents the antiderivative of $\sec(x)$ evaluated at the bounds, which would be relevant if you were finding arc length or area under the curve, not volume. Answer C suggests the student might have forgotten to square the function or made an error in evaluating $\tan(\pi/3)$ . Remember: for disk method problems, always square the function before integrating. Also, memorize that the antiderivative of $\sec^2(x)$ is $\tan(x)$ – this integral appears frequently in calculus problems involving trigonometric functions.
The region $R$ is bounded by the graph of $y = x^3$ , the y-axis, and the line $y=8$ . What is the volume of the solid formed by revolving $R$ about the y-axis?
Answer: $\frac{96\pi}{5}$ For revolution about the y-axis, we integrate with respect to $y$ . The radius of each disc is $x$ . We must express $x$ as a function of $y$ : from $y=x^3$ , we get $x = y^{1/3}$ . The region is bounded by the y-axis ( $x=0$ ) and extends to the curve, with $y$ ranging from $0$ to $8$ . The volume is $V = \pi \int_{0}^{8} (\text{radius})^2 dy = \pi \int_{0}^{8} (y^{1/3})^2 dy = \pi \int_{0}^{8} y^{2/3} dy = \pi [\frac{3}{5}y^{5/3}]_0^8 = \frac{3\pi}{5} (8^{5/3} - 0) = \frac{3\pi}{5} ( (8^{1/3})^5 ) = \frac{3\pi}{5} (2^5) = \frac{3\pi}{5}(32) = \frac{96\pi}{5}$ .
A solid is generated by revolving the region bounded by the upper semi-circle $y = \sqrt{4-x^2}$ and the x-axis about the x-axis. A hole of radius 1 is then drilled through the center of the solid along the x-axis. Which integral represents the volume of the remaining object?
Answer: $\pi \int_{-\sqrt{3}}^{\sqrt{3}} (3-x^2) dx$ When finding the volume of a solid with a cylindrical hole drilled through it, you need to subtract the volume of the hole from the original solid using the washer method. The original semicircle $y = \sqrt{4-x^2}$ has radius 2, so when revolved around the x-axis, it creates a sphere. After drilling a hole of radius 1 along the x-axis, you have washers with outer radius $R(x) = \sqrt{4-x^2}$ and inner radius $r(x) = 1$ . The washer method formula is $V = \pi \int_a^b [R(x)^2 - r(x)^2] dx$ . Substituting: $V = \pi \int_a^b [(4-x^2) - 1^2] dx = \pi \int_a^b (3-x^2) dx$ . The key insight is determining the limits of integration. The hole only exists where the original solid exists, but we must also consider where the hole actually removes material. Since the hole has radius 1 and the semicircle has radius $\sqrt{4-x^2}$ , the hole only affects the region where $\sqrt{4-x^2} \geq 1$ , which means $4-x^2 \geq 1$ , so $x^2 \leq 3$ , giving us $x \in [-\sqrt{3}, \sqrt{3}]$ . Answer B is correct: $\pi \int_{-\sqrt{3}}^{\sqrt{3}} (3-x^2) dx$ . Answer A uses the wrong integration limits (the full sphere diameter). Answer C incorrectly applies $(\sqrt{4-x^2}-1)^2$ as if using cylindrical shells. Answer D uses the correct limits but forgets to subtract the hole's area, giving just the original sphere's volume over the reduced interval. Study tip: Always identify where holes actually remove material—integration limits should reflect the intersection of the original solid and the drilled region.
The region in the first quadrant bounded by $y=kx^2$ , the y-axis, and the line $y=4$ is revolved about the y-axis to generate a solid. If the volume of the solid is $8\pi$ , what is the value of the positive constant $k$ ?
Answer: $1$ When you encounter a volume problem involving revolution about an axis, you need to set up the integral using either the disk/washer method or cylindrical shells. Since we're revolving about the y-axis and have a function in terms of x, the washer method works well here. The region is bounded by $y = kx^2$ , the y-axis, and $y = 4$ . When revolved about the y-axis, this creates washers with outer radius determined by solving $y = kx^2$ for x: $x = \sqrt{\frac{y}{k}}$ . The inner radius is 0 (since one boundary is the y-axis). Using the washer method, the volume is: $V = \pi \int_0^4 \left(\sqrt{\frac{y}{k}}\right)^2 dy = \pi \int_0^4 \frac{y}{k} dy = \frac{\pi}{k} \int_0^4 y \, dy$ Evaluating: $V = \frac{\pi}{k} \cdot \frac{y^2}{2}\Big|_0^4 = \frac{\pi}{k} \cdot \frac{16}{2} = \frac{8\pi}{k}$ Since we're told the volume equals $8\pi$ , we have: $\frac{8\pi}{k} = 8\pi$ Solving for k: $k = 1$ Choice A ( $\frac{4}{9}$ ) would give a volume of $18\pi$ , which is too large. Choice B ( $\frac{1}{2}$ ) would yield $16\pi$ , also too large. Choice D ( $\frac{16}{25}$ ) would produce $12.5\pi$ , still incorrect. Remember: when setting up revolution problems, always identify which variable you're integrating with respect to, and make sure your radius function is expressed in terms of that variable. Double-check your setup by verifying the bounds make geometric sense.
Let $R$ be the region in the first quadrant bounded by the graph of $y = 4-x^2$ , the x-axis, and the y-axis. What is the volume of the solid generated when $R$ is revolved about the y-axis?
Answer: $8\pi$ To revolve around the y-axis, we use the disc method with integration with respect to $y$ . The radius of a disc is $x$ . We must express $x$ in terms of $y$ : $y = 4-x^2 \implies x^2 = 4-y \implies x = \sqrt{4-y}$ (since we are in the first quadrant). The region is bounded by the y-axis ( $x=0$ ) and the x-axis ( $y=0$ ). The maximum $y$ -value is at $x=0$ , which is $y=4$ . So, the bounds for $y$ are from $0$ to $4$ . The volume is $V = \pi \int_{0}^{4} (\text{radius})^2 dy = \pi \int_{0}^{4} (\sqrt{4-y})^2 dy = \pi \int_{0}^{4} (4-y) dy = \pi [4y - \frac{y^2}{2}]_0^4 = \pi [(16 - \frac{16}{2}) - (0)] = 8\pi$ .
The region in the second quadrant bounded by the parabola $y = 9 - x^2$ and the coordinate axes is revolved about the y-axis. What is the volume of the resulting solid?
Answer: $\frac{81\pi}{2}$ When you encounter a volume of revolution problem, you need to identify the region, the axis of rotation, and choose the appropriate method. Here we're revolving around the y-axis, so the disk/washer method with horizontal slices is most efficient. First, find where the parabola $y = 9 - x^2$ intersects the axes in the second quadrant. When $x = 0,y = 9$ . When $y = 0,x^2 = 9$ , so $x = -3$ (taking the negative value for the second quadrant). The region is bounded by $x = -3,x = 0,y = 0$ , and the parabola. Using the disk method with horizontal slices, each disk at height $y$ has radius equal to the distance from the y-axis to the parabola. From $y = 9 - x^2$ , we get $x = -\sqrt{9-y}$ (negative since we're in the second quadrant). The radius is $|\sqrt{9-y}| = \sqrt{9-y}$ . The volume is: $V = \pi \int_0^9 (\sqrt{9-y})^2 \, dy = \pi \int_0^9 (9-y) \, dy = \pi[9y - \frac{y^2}{2}]_0^9 = \pi(81 - \frac{81}{2}) = \frac{81\pi}{2}$ Answer B is correct. Answer A, $\frac{162\pi}{5}$ , likely comes from incorrectly using the shell method or making algebraic errors. Answer C, $18\pi$ , suggests forgetting to square the radius. Answer D, $\frac{243\pi}{5}$ , might result from integration mistakes or wrong bounds. Strategy tip: For y-axis revolutions, horizontal slices with the disk method often work best. Always express the radius in terms of your integration variable and double-check your algebra when squaring expressions under the integral.

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