📝 Shell method variations other axes (21 MCQs)

📖 Explanation: The method of cylindrical shells involves partitioning the region into thin vertical (or horizontal) strips parallel to the axis of revolution. When revolved, each strip generates a cylindrical shell. The volume of each shell is approximated as $2\pi \cdot \text{radius} \cdot \text{height} \cdot \text{thickness}$. Summing these shell volumes and taking a limit produces an integral. This is fundamentally different from the disk/washer method, which uses cross-sections perpendicular to the axis.

📖 Explanation: The error is conceptual. When revolving a region about the x-axis using the shell method, the strips must be horizontal (parallel to the axis of revolution). For a horizontal strip at a height $y$, its radius is $y$, and its length (height of the shell) is the horizontal distance between the right and left boundaries, which is $4 - y^2$. The student's setup, $\int 2\pi y(4-y^2) dy$, is actually correct for this problem. A common misconception is that the variable of integration in the shell method must always be $x$ when revolving about a vertical axis, or vice versa. The correct identification of the strip orientation is key.

📖 Explanation: The student's argument is incorrect. The shell method is applicable for any axis of revolution, but it requires a careful choice of integration variable. When the axis is horizontal (like $y = -1$), the shells are formed by revolving horizontal strips. The radius of a shell at height $y$ is its distance from the axis, $y - (-1) = y+1$. The height (or length) of the shell is the horizontal length of the strip at that $y$-value. While this is a valid method, setting up the integral requires expressing $x$ as a function of $y$, which can lead to more complex integrals, but it is not inapplicable.

📖 Explanation: The axis of revolution is $x=2$, a vertical line. Therefore, we use horizontal strips (parallel to the axis). A strip at height $y$ extends from $x=0$ (the y-axis) to $x=\sqrt{y}$ (from $y=x^2$). Its length (height of the shell) is $\sqrt{y} - 0 = \sqrt{y}$. Wait, the region is bounded by $y=x^2$, $y=1$, and the y-axis. The upper and lower boundaries are $y=1$ and $y=0$. The horizontal distance from the axis $x=2$ to the strip is $2 - x$, so the radius is $2 - \sqrt{y}$. The volume is $\int_0^1 2\pi (2 - \sqrt{y}) (\sqrt{y}) \, dy$. However, this correct integral is not an option. The Easy option explains this. The correct answer among the given options is D, which uses the length of the strip as $1-y$, which is a common error. So, the correct setup is not listed, highlighting the need for careful radius and height identification.

📖 Explanation: The shell method is particularly efficient for this problem. The region is bounded by functions of $x$, and revolving about the y-axis makes the radius of a vertical shell simply $x$. The height is the difference $x - x^2$. The volume is a single integral $\int_0^1 2\pi x (x - x^2) \, dx$. The washer method, on the other hand, would require integrating with respect to $y$, which involves solving for $x$ in terms of $y$ and potentially splitting the integral due to the changing right boundary. Thus, the shell method is often easier when the axis is parallel to the independent variable of the region's description.

📖 Explanation: The region is under $y = 2x - x^2$ from $x=0$ to $x=2$ (where it intersects y=0). Using the shell method about the y-axis: $V = \int_0^2 2\pi x (2x - x^2) \, dx = 2\pi \int_0^2 (2x^2 - x^3) \, dx = 2\pi [ \frac{2x^3}{3} - \frac{x^4}{4} ]_0^2 = 2\pi ( \frac{16}{3} - 4 ) = 2\pi (\frac{4}{3}) = \frac{8\pi}{3}$. Direct calculation is straightforward. Common mistakes include incorrect limits (e.g., using y-limits) or algebra errors in integration.

📖 Explanation: This is an excellent example of how a problem can be solved using either $x$ or $y$ integration with the shell method. Option A uses vertical shells (revolving about the y-axis) with radius $x$ and height $\sqrt{x}$, integrating from 0 to 4. Option B uses horizontal shells (revolving about the y-axis, which is vertical, so horizontal strips are still parallel to the axis) with radius $y$ and height (or length) $4 - y^2$ (since $x = y^2$), integrating from 0 to 2. Both setups are correct for the shell method and will yield the same volume. This highlights the flexibility of the shell method in choosing the variable of integration.

📖 Explanation: The student's setup is correct! This is a sophisticated Medium of cylindrical shells. The student is revolving the circle $x^2 + y^2 = r^2$ about the y-axis. Using horizontal strips parallel to the y-axis, the radius is $y$, and the height of the shell (the length of the strip) is $2\sqrt{r^2 - y^2}$. The integral $\int_{-r}^{r} 2\pi y (2\sqrt{r^2 - y^2}) \, dy$ does indeed evaluate to the volume of a sphere. It may look unconventional, but it is a valid shell method setup. This tests if the student truly understands the geometry of shells and can identify a correct, albeit non-standard, Medium.

📖 Explanation: In the cylindrical shell method, the integrand is $2\pi \cdot \text{radius} \cdot \text{height} \cdot \text{thickness}$. When integrating with respect to $y$, the 'thickness' is $dy$. The factor $(2 - \sqrt{y})$ is the horizontal distance from the axis of revolution $x=2$ to the shell. This distance is the radius of the shell. The height of the shell is the vertical length of the horizontal strip, which is the difference between the top and bottom boundaries, $4 - y$. Correctly identifying the radius, especially when the axis is not the coordinate axis, is a key conceptual hurdle.

📖 Explanation: The shell method (Student A) integrates directly with respect to $x$, using the original function $f(x)$. This does not require inverting the function. The washer method (Student B) requires expressing $x$ as a function of $y$, which involves finding the inverse $f^{-1}(y)$. If the function is difficult or impossible to invert analytically (e.g., $f(x) = x^3 + x$), the shell method becomes significantly more practical. This is a key reason why the shell method is a powerful alternative.

📖 Explanation: First, find the intersection points of $y = x^2+1$ and $y = x+3$: $x^2+1 = x+3 \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x-2)(x+1)=0$, so $x = -1, 2$. On this interval, $x+3 \ge x^2+1$. Revolving about the y-axis, a vertical shell at $x$ has radius $x$, height $(x+3) - (x^2+1)$, and thickness $dx$. The correct integral is $\int_{-1}^{2} 2\pi x [(x+3) - (x^2+1)] \, dx$. Option B has the height reversed, which would give a negative volume. Option C integrates with respect to y but has an incorrect height expression. Option D incorrectly multiplies the functions.

📖 Explanation: The radius of a cylindrical shell is the distance from the axis of revolution to the shell. If the axis is to the right of the region, then for any point $x$ in the region, the distance to the axis $x=k$ is $k - x$. This is always a positive length. A common mistake is to use $x - k$, which would be negative for $x < k$, or to use $x$ itself, which is only correct when the axis is the y-axis (x=0). The sign and expression of the radius are crucial for a correct setup.

📖 Explanation: The student's setup is correct. When revolving about the x-axis, the shells are generated by horizontal strips. A strip at height $y$ has a radius $y$. Its length (height of the shell) is the horizontal distance between the right and left boundaries. For $0 \le y \le 1$, the curve $x=y$ is to the right of $x=y^2$, so the height is $y - y^2$. The integral is $\int_0^1 2\pi y (y - y^2) \, dy$. This is a classic problem where the shell method with respect to $y$ is much simpler than the washer method, which would require integrating with respect to $x$ and splitting the region.

📖 Explanation: The primary advantage of cylindrical shells is the flexibility it offers. When revolving a region about an axis, the washer method often requires expressing the functions with respect to the axis of revolution, which may involve solving for an inverse function. The shell method, by using strips parallel to the axis, often allows integration with respect to the original independent variable, avoiding this step. For example, revolving a region $y = \sin x$ about the y-axis is a simple shell integral $\int 2\pi x \sin x \, dx$ but a complex washer integral. Option A is incorrect; the integral can be harder. Option B is incorrect; washers naturally handle holes. Option D is incorrect; a sketch is always necessary to identify boundaries.

📖 Explanation: The region is under $y = 1/(x^2+1)$ from $x=0$ to $x=2$. Option A uses the shell method directly: $V = \int_0^2 2\pi x (1/(x^2+1)) \, dx$. Option B uses horizontal shells. It integrates with respect to $y$, where the radius is $y$, and the height is the horizontal distance from $x=0$ to $x = \sqrt{1/y - 1}$. The y-limits are from $y=0$ to $y=1/5$ (the height of the curve at $x=2$). Both integrals are valid representations of the same volume. This question tests the ability to recognize that different shell method setups are possible and that they are equivalent.

📖 Explanation: A negative volume from an integral is a clear sign of a setup error. One common cause is having the height of the shell as $g(x) - f(x)$ when the region is actually above $g(x)$. Another is using a radius like $x - k$ when $k > x$, or reversing the integration limits. Any of these errors would result in the integral evaluating to a negative number, which is physically impossible for a volume. This highlights the importance of carefully checking the geometry to ensure all factors in the integrand are positive.

📖 Explanation: The axis of revolution is the horizontal line $y = -1$. For the shell method, we use horizontal strips (parallel to the axis). A strip at height $y$ is a distance $y - (-1) = y+1$ from the axis. This distance is the radius of the cylindrical shell. A common misconception is to use $y$ or $y-1$, which would be the distance to the x-axis or a line $y=1$, respectively. The radius must always be the perpendicular distance from the strip to the specified axis.

📖 Explanation: This is a scenario where the shell method is advantageous. The disk/washer methods require computing the cross-sectional area perpendicular to the axis. If this is difficult (e.g., the region is revolved about a vertical axis but defined by a function $y = f(x)$), the washer method would require solving for $x$ in terms of $y$. The shell method, however, uses the original function $f(x)$ directly. The shell method is particularly powerful when the axis of revolution is parallel to the direction of the independent variable, as it avoids algebraic inversion.

📖 Explanation: The shell method integral $\int 2\pi x \cos x \, dx$ is a standard Medium of integration by parts. It is straightforward to evaluate exactly. The washer method would involve the inverse function $x = \cos^{-1}(y)$, leading to an integral $\int \pi (\cos^{-1}(y))^2 \, dy$, which is a more difficult integral to evaluate. This illustrates a key benefit of the shell method: it often yields simpler integrands, even if it requires a technique like integration by parts, compared to the more complex integrands that can arise from the washer method.

📖 Explanation: The setup is correct: the region is between $y=2x$ (top) and $y=x^2$ (bottom) from $x=0$ to $x=2$. The volume is $V = \int_0^2 2\pi x (2x - x^2) \, dx = 2\pi \int_0^2 (2x^2 - x^3) \, dx = 2\pi [ \frac{2x^3}{3} - \frac{x^4}{4} ]_0^2 = 2\pi ( \frac{16}{3} - 4 ) = 2\pi ( \frac{4}{3} ) = \frac{8\pi}{3}$. The student's integral and result are both correct. A common error is to reverse the height function, which would give a negative volume, or to use the wrong limits.

📖 Explanation: The axis is the x-axis (horizontal), so we use horizontal strips parallel to the axis. A strip at height $y$ has radius $y$. Its height is the horizontal length from $x=0$ to $x=y^3$, which is $y^3$. The volume is $\int_0^1 2\pi y (y^3) \, dy$. Option B is incorrect because it uses vertical shells and would require solving for $x$ in terms of $y$, which is already the given function. Option C has a negative height. This problem tests the ability to correctly identify radius and height when the region is bounded by functions of $y$.