📝 Shell method volume revolving about y-axis (24 MCQs)

📖 Explanation: The cylindrical shell method requires integrating the surface area of each shell, which is $2\pi \cdot (\text{radius}) \cdot (\text{height})$. Here, the radius is $x$ and the height is $f(x) = \sqrt{x}$, so the integrand is $2\pi x \sqrt{x}$. Options B and C are incorrect because they omit the radius factor or use the wrong height. Option D incorrectly squares the radius component.

📖 Explanation: The cylindrical shell method for revolving about the y-axis requires the region to be defined by $y = f(x)$ with $f(x) \ge 0$ on $[a,b]$ and $0 \le a < b$. This ensures the radius $x$ is nonnegative. The method doesn't require the function to be monotonic. Options C and D are incorrect because they describe different scenarios or conditions.

📖 Explanation: The student's integral correctly sets up the shell method with radius $x$ and height $x^2$, giving $8\pi$. However, when using the disk method about the y-axis, the integral would be $\int_{0}^{4} \pi (\sqrt{y})^2 \, dy = 8\pi$. Option B presents an alternative shell method setup, which is also correct. The question tests whether the student recognizes that the student's claim is correct, even if alternative methods exist. Option C represents the disk method about the x-axis, which is incorrect for this problem.

📖 Explanation: For the shell method, the height of the shell at radius $x$ is the difference between the upper and lower curves, $x - x^2$. The radius is $x$, so the integrand is $2\pi x (x - x^2)$. Option B has the height reversed, giving a negative volume. Option C is the disk/washer method about the x-axis. Option D misses the radius factor $x$, which is essential in the shell method.

📖 Explanation: The cylindrical shell method allows integration with respect to $x$, which is often easier because the functions are already given as $y = f(x)$. The washer method would require solving for $x$ in terms of $y$, which can be difficult or impossible for some functions. Options B, C, and D are not universally true; the choice depends on the specific problem and the complexity of the resulting integrals.

📖 Explanation: The shell method uses the radius $x$ and height $\sin x$, so the integrand is $2\pi x \sin x$. Option B misses the radius $x$, which is crucial. Option C is the disk method about the x-axis. Option D would be the disk method about the y-axis if it were correctly set up, but it's missing the correct bounds and the radius factor.

📖 Explanation: In the shell method, the surface area of a cylindrical shell is $2\pi \cdot (\text{radius}) \cdot (\text{height})$. The student correctly identifies the height as $\sqrt{x}$ but omits the radius $x$. This would lead to an incorrect volume. The correct integrand should be $2\pi x \sqrt{x}$. The bounds are correct for $x$, and the axis is the y-axis, which is correct for this setup.

📖 Explanation: The shell method integrates $2\pi x e^x \, dx$ from 0 to $\ln 2$, which is straightforward. The washer method would require expressing $x$ as $\ln y$ and integrating $\pi (\ln y)^2 \, dy$ from 1 to 2, which is more complex due to the $(\ln y)^2$ term. Thus, shells are simpler here. The disk method about x is not applicable for revolving about the y-axis. This illustrates how the choice of method depends on the geometry.

📖 Explanation: For the shell method, the radius of each shell is the horizontal distance from the y-axis to the shell, which is $x$. To ensure the radius is nonnegative, the region must lie in the domain $x \ge 0$, i.e., entirely to the right of the y-axis. The method does not require the function to be one-to-one (B) or for the region to be bounded by two curves (C). While the axis is the y-axis in this specific case, the method can be adapted to other axes, so D is not a necessary condition for the method itself, but for this specific Medium.

📖 Explanation: For the shell method about the y-axis, the radius is the horizontal distance from the y-axis, which is $x$. The height is the vertical length of the shell, which is given by the function $y = 4 - x^2$. Option B swaps the radius and height. Option C introduces a factor of 2, which is incorrect. Option D is a combination of the wrong radius and a factor of 2. This setup correctly identifies the components needed for the shell integral.

📖 Explanation: While both methods give the same volume, they are not always equally efficient. The choice of method depends on the geometry and the ease of integration. For some functions, expressing $x$ as a function of $y$ (needed for washers about the y-axis) can be difficult or lead to complex integrals. Thus, the shell method may be simpler in those cases. The student's conclusion is too broad. Option B is incorrect because efficiency is about the complexity, not the result. Option C is incorrect because both methods can be used for the same axis. Option D is incorrect because the integrals are different in form, even if they evaluate to the same value.

📖 Explanation: The shell method requires the radius $x$ and the height $f(x) = 1/x$. The integrand is $2\pi x \cdot \frac{1}{x} = 2\pi$. So the integral is $2\pi \int_{1}^{3} dx$. Option A misses the radius $x$, which would incorrectly give the area of a rectangle. Option C is the disk method about the x-axis. Option D incorrectly simplifies the height. This demonstrates that the shell method can sometimes simplify dramatically, but it's important to correctly identify the radius and height.

📖 Explanation: In the cylindrical shell method, the variable of integration corresponds to the axis perpendicular to the shells. When revolving about the y-axis, the shells are formed by vertical strips of thickness $dx$. Thus, $dx$ represents the thickness of each shell. The height is given by $f(x)$, the radius by $x$, and the circumference by $2\pi x$. This is a fundamental concept for setting up shell integrals.

📖 Explanation: The student correctly identifies the radius as $x$ and the height as $\sqrt{x}$. The bounds from $x=1$ to $x=4$ are correct for the given region. Therefore, the integral is correct. Options B and C swap the radius and height, which is a common mistake. Option D suggests using $y$-bounds, which would be the washer method, not the shell method. This question tests whether the student can recognize a correct setup and avoid common errors.

📖 Explanation: The curves intersect at $x=0$ and $x=2$. For $0 \le x \le 2$, the upper curve is $y=2x$ and the lower is $y=x^2$. Using the shell method, the height is $2x - x^2$, and the radius is $x$. The volume is $2\pi \int_{0}^{2} x(2x - x^2) \, dx = 2\pi \int_{0}^{2} (2x^2 - x^3) \, dx = 2\pi \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_{0}^{2} = 2\pi \left( \frac{16}{3} - 4 \right) = 2\pi \cdot \frac{4}{3} = \frac{8\pi}{3}$. Option B is half of this, option C is double, and option D is four times. This requires careful integration and correct identification of the region.

📖 Explanation: When revolving about the y-axis, the washer method would require solving $y = \sin x$ for $x$, which gives $x = \arcsin y$ for $0 \le y \le 1$. However, since $\sin x$ is not one-to-one on $[0,\pi]$, the washer method would need to split the region into two parts: one where $x = \arcsin y$ and another where $x = \pi - \arcsin y$. The shell method integrates $2\pi x \sin x \, dx$ from 0 to $\pi$, which is a single integral. Thus, washers would require splitting, while shells would not. This highlights the advantage of shells in this case.

📖 Explanation: When using the shell method and integrating with respect to $y$, the shells are horizontal. The radius of the shell is $y$, and the height is the horizontal length of the strip, which is the difference between the right and left boundaries: $h(y) - g(y)$. Option B gives a negative height, which is invalid. Option C and D do not represent the length of the strip. This question tests whether the student understands the geometry when the roles of $x$ and $y$ are reversed in the shell method.

📖 Explanation: The region is bounded by $y = x^3$, $y = 8$, and $x = 0$. When using shells about the y-axis, it's easier to integrate with respect to $x$. The radius is $x$, and the height is the vertical distance from the lower boundary $y = x^3$ to the upper boundary $y = 8$, so the height is $8 - x^3$. The bounds for $x$ are from 0 to 2 (since $2^3 = 8$). Thus, the integral is $2\pi \int_{0}^{2} x(8 - x^3) \, dx$. Option A and B are incorrect because they use $y$ as the radius and $y^{1/3}$ as height, which would be the washer method about the y-axis. Option D has the wrong height.

📖 Explanation: The shell method is versatile and can handle regions bounded by two curves. In such cases, the height of the shell is the difference between the upper and lower functions, $f(x) - g(x)$. The student's argument is incorrect because it underestimates the method's applicability. Option B and D are false restrictions. Option C is also false because the method works for both one-curve and two-curve regions. This question addresses a common misconception about the limitations of the shell method.

📖 Explanation: The region is bounded below by the x-axis (y=0) and above by $y = x^2$. For a given $x$, the vertical strip extends from $y=0$ to $y=x^2$, so the height is $x^2 - 0 = x^2$. Option B is incorrect because it represents the vertical line $x=2$, not the height. Option C would be the height if the region were between $y=x^2$ and $y=2$, which is not the case here. Option D is incorrect because it introduces a factor of 2. This question checks the understanding of the geometry of the region.

📖 Explanation: The shell method for revolving about the y-axis uses the radius $x$ and height $\ln x$, so the integrand is $2\pi x \ln x$. The bounds are from $x=1$ to $x=e$. Option B is the washer method about the y-axis, where $x = e^y$ and $y$ goes from 0 to 1. Option C is the disk method about the x-axis. Option D misses the radius $x$. This question tests the ability to choose the correct method and set it up properly.

📖 Explanation: The shell method is advantageous when integrating with respect to the variable that is easier to work with. If the function is naturally given as $x = f(y)$ (i.e., $x$ as a function of $y$), then the washer method about the y-axis might be simpler because it directly uses $x$ as the radius and $f(y)$ as the height. The shell method would require expressing $y$ as a function of $x$, which might be difficult. Options B and C are situations where the shell method is often easier. Option D is exactly when the shell method is preferred. So, the answer is A, as it describes a case where washers might be simpler.