Shell Method - Calculus 2

1. A solid is formed by revolving the region bounded by the curve $y=x-x^4$ and the x-axis about the y-axis. Setting up the volume integral using the washer method would be difficult because it requires solving for $x$ in terms of $y$. Which integral correctly sets up the volume using the more convenient shell method?
   Answer: $\int_0^1 2\pi x(x-x^4) \,dx$ To use the shell method for revolution about the y-axis (a vertical axis), we integrate with respect to $x$. The intersection points of $y=x-x^4$ and $y=0$ are $x(1-x^3)=0$, so $x=0$ and $x=1$. These are the limits of integration. The radius of a cylindrical shell is the distance from the y-axis, so $r(x)=x$. The height of the shell is the function value, $h(x) = x-x^4$. The volume is $V = \int_0^1 2\pi r(x)h(x) \,dx = \int_0^1 2\pi x(x-x^4) \,dx$.
2. A region is bounded by $y = 2x$, $y = 6$, and $x = 0$. When revolved around the $y$-axis, both the shell method and disk/washer method can be used. If the shell method gives volume $V_s$ and requires evaluating $\int_0^3 f(x) dx$, what is $f(x)$?
   Answer: $f(x) = 2\pi x(6 - 2x)$ When using the shell method to find volumes of revolution around the y-axis, you're essentially summing up cylindrical shells with radius $x$, height determined by the function, and thickness $dx$. The general formula is $V = 2\pi \int x \cdot (\text{height}) \, dx$. First, let's understand this region. The boundaries are $y = 2x$, $y = 6$, and $x = 0$. These intersect at $(0,0)$, $(3,6)$, and $(0,6)$, forming a triangle. When you slice this region with vertical strips (for the shell method), each strip at position $x$ extends from $y = 2x$ up to $y = 6$, giving a height of $6 - 2x$. For the shell method around the y-axis, the volume is $V_s = 2\pi \int_0^3 x(6 - 2x) \, dx$. This means $f(x) = 2\pi x(6 - 2x)$, which is answer C. Let's examine why the other options are wrong. Option A, $f(x) = x(6 - 2x)$, gives you just the area element without the $2\pi$ factor required for the shell method volume formula. Option B, $f(x) = \pi x^2(6 - 2x)$, incorrectly uses $\pi$ instead of $2\pi$ and has an extra factor of $x$, suggesting confusion with the disk method. Option D, $f(x) = 2\pi x^2(6 - 2x)$, has the correct $2\pi$ but includes an extra $x$ factor. Remember: shell method integrands always have the form $2\pi \times \text{radius} \times \text{height}$, where radius is the distance from the axis of rotation.
3. Let R be the region bounded by $y=e^{x^2}$, $y=0$, $x=0$, and $x=1$. When revolving R about the y-axis, the shell method is preferable. If the volume is approximated by $n$ cylindrical shells of equal thickness, what is the volume of the $i$-th shell (where $x_i$ is the right endpoint of the $i$-th subinterval)?
   Answer: $V_i = \frac{2\pi}{n} x_i e^{x_i^2}$ The interval $[0,1]$ is divided into $n$ subintervals of thickness $\Delta x = 1/n$. The right endpoint of the $i$-th interval is $x_i = i/n$. For the shell method, the volume of the $i$-th shell is approximated by $2\pi r_i h_i \Delta x$. Here, the radius is $r_i = x_i$ and the height is $h_i = e^{x_i^2}$. The thickness is $\Delta x = 1/n$. So, the volume of the $i$-th shell is $V_i = 2\pi x_i e^{x_i^2} \Delta x = 2\pi (i/n) e^{(i/n)^2} (1/n) = \frac{2\pi i}{n^2} e^{(i/n)^2}$. Option B uses $x_i$ instead of $i/n$, which is a more general representation: $V_i = 2\pi x_i e^{x_i^2} (1/n) = \frac{2\pi}{n} x_i e^{x_i^2}$. Let's re-examine C. It's $\frac{2\pi i}{n^2} e^{(i/n)^2}$. This matches my derivation. Let's check B again. $\frac{2\pi}{n} x_i e^{x_i^2} = \frac{2\pi}{n} (i/n) e^{(i/n)^2} = \frac{2\pi i}{n^2} e^{(i/n)^2}$. So B and C are equivalent if we substitute $x_i=i/n$. B is a slightly better form. I will choose B. The key is $2\pi r h \Delta x$.
4. The integral $\int_0^2 2\pi(4-y)(y/2) dy$ represents the volume of a solid of revolution. If the same volume is to be calculated using the washer method, what would be the integral?
   Answer: $\int_0^1 \pi(4^2 - (4-2x)^2) dx$ From the shell integral (wrt $y$), we identify the components. Axis of revolution: horizontal, from radius $r(y)=4-y$, so the axis is $y=4$. Height of shell: $h(y)=y/2$, so $x=y/2$ is the curve. The region is a triangle bounded by $x=y/2$ (or $y=2x$), $x=0$ (from $h(y)$ at $y=0$) and $y=2$ (upper limit). When $y=2$, $x=1$. The region is bounded by $y=2x$, $y=2$ and $x=0$. For the washer method (wrt $x$ for horizontal axis), the integration is from $x=0$ to $x=1$. The outer radius is the distance from $y=4$ to the bottom curve $y=2x$, so $R(x)=4-2x$. The inner radius is the distance from $y=4$ to the top curve $y=2$, so $r(x)=4-2=2$. The integral is $V = \int_0^1 \pi((4-2x)^2 - 2^2)dx$. This does not match C. Let's re-read the shell integral. $h(y)=y/2$. Limits $y=0$ to $y=2$. Region is bounded by $x=y/2$, $x=0$, $y=0$, $y=2$. Axis $y=4$. Washer method: $x$ from 0 to 1. Outer radius from $y=4$ to $y=0$, $R=4$. Inner radius from $y=4$ to $y=2x$, $r=4-2x$. So $V=\int_0^1 \pi(4^2-(4-2x)^2)dx$. This matches C.
5. The region bounded by $y = x^2 - 1$ and $y=3$ is revolved about the y-axis. Which integral gives the volume of the resulting solid?
   Answer: $\int_{-1}^3 \pi (\sqrt{y+1})^2 \,dy$ This problem can be solved by either the shell or washer method. The washer method is more direct. Revolving around the y-axis means integrating with respect to $y$. The region spans from $y=-1$ to $y=3$. The curve is $x^2=y+1$, so $x=\pm\sqrt{y+1}$. The radius of a washer at height $y$ is $R(y)=\sqrt{y+1}$. The volume is $V = \int_{-1}^3 \pi R(y)^2 \,dy = \int_{-1}^3 \pi (\sqrt{y+1})^2 \,dy$. Using the shell method requires care because the radius must be positive; the integral is $\int_{-2}^2 2\pi |x|(3-(x^2-1)) \,dx = \int_{-2}^2 2\pi |x|(4-x^2) \,dx$. Note that $\int_{-2}^2 2\pi x(4-x^2) \,dx = 0$ because the integrand is an odd function over a symmetric interval.
6. The region bounded by $x = y^2$, $x=0$, and $y=2$ is revolved around the line $y=3$. Which integral gives the volume of the resulting solid using the shell method?
   Answer: $V = \int_0^2 2\pi (3-y)(y^2) \,dy$ For revolution about a horizontal axis $y=3$, the shell method requires integration with respect to $y$. The shells are stacked from $y=0$ to $y=2$. The radius $r(y)$ of a shell at height $y$ is its distance from the axis of revolution, so $r(y) = 3-y$. The height of the shell (which is its length in the x-direction) is given by $h(y) = x = y^2$. The volume is therefore $V = \int_0^2 2\pi r(y)h(y) \,dy = \int_0^2 2\pi (3-y)(y^2) \,dy$.
7. A solid is generated by rotating the region R bounded by $y=x^2$ and $y=\sqrt{x}$ about the line $x=-1$. What is the volume of the solid?
   Answer: $29\pi/30$ We use the shell method. The curves intersect when $x^2 = \sqrt{x}$, giving $x = 0$ and $x = 1$. For $0 \leq x \leq 1$, we have $\sqrt{x} \geq x^2$. The radius of each shell is $r(x) = x - (-1) = x + 1$ and the height is $h(x) = \sqrt{x} - x^2$. The volume is $V = \int_0^1 2\pi(x+1)(\sqrt{x}-x^2)dx = 2\pi \int_0^1 (x^{3/2} - x^3 + x^{1/2} - x^2)dx = 2\pi[\frac{2}{5}x^{5/2} - \frac{x^4}{4} + \frac{2}{3}x^{3/2} - \frac{x^3}{3}]_0^1 = 2\pi(\frac{2}{5} - \frac{1}{4} + \frac{2}{3} - \frac{1}{3}) = 2\pi(\frac{2}{5} - \frac{1}{4} + \frac{1}{3}) = 2\pi \cdot \frac{24 - 15 + 20}{60} = \frac{29\pi}{30}$.
8. Let R be the region bounded by $y=1/x^2$, $y=0$, $x=1$, and $x=2$. Let $V_1$ be the volume when R is revolved about the y-axis and $V_2$ be the volume when R is revolved about the x-axis. Which statement is true?
   Answer: $V_1 = 2\pi \ln 2$ and $V_2 = 7\pi/24$ $V_1$ (revolve about y-axis): Use the shell method. $V_1 = \int_1^2 2\pi x (1/x^2) \,dx = \int_1^2 2\pi (1/x) \,dx = 2\pi [\ln|x|]_1^2 = 2\pi(\ln 2 - \ln 1) = 2\pi \ln 2$. $V_2$ (revolve about x-axis): Use the disk method. $V_2 = \int_1^2 \pi (1/x^2)^2 \,dx = \int_1^2 \pi x^{-4} \,dx = \pi [x^{-3}/(-3)]_1^2 = -\frac{\pi}{3} [1/x^3]_1^2 = -\frac{\pi}{3} (1/8 - 1) = -\frac{\pi}{3} (-7/8) = 7\pi/24$. Thus, $V_1 = 2\pi \ln 2$ and $V_2 = 7\pi/24$.
9. A solid is formed by revolving the region bounded by $y = x^2$, $y = 0$, and $x = 2$ around the line $x = 3$. If we set up the shell method integral as $V = 2\pi \int_a^b R(x) \cdot H(x) dx$, what are the correct values of $R(x)$ and $H(x)$?
   Answer: $R(x) = 3 - x$ and $H(x) = x^2$ When setting up the shell method for a solid of revolution, you need to identify two key components: the radius of each cylindrical shell and the height of each shell. For this problem, you're revolving the region bounded by $y = x^2$, $y = 0$, and $x = 2$ around the vertical line $x = 3$. Since you're revolving around a vertical line, you'll integrate with respect to $x$ from $x = 0$ to $x = 2$. The radius $R(x)$ is the horizontal distance from any point $x$ in your region to the axis of revolution $x = 3$. Since your region spans from $x = 0$ to $x = 2$ (all to the left of $x = 3$), this distance is $3 - x$. The height $H(x)$ is the vertical extent of the region at any given $x$-value, which runs from $y = 0$ up to $y = x^2$, giving us $H(x) = x^2$. Looking at the wrong answers: Choice A incorrectly uses $R(x) = 3 + x$, which would be the distance if you were measuring from $x = -3$ instead of to $x = 3$. Choice B uses $R(x) = x - 3$, which gives negative values since $x < 3$ in our region—radius must be positive. Choice D correctly identifies the radius but mistakenly uses $H(x) = 2 - x^2$, which doesn't represent the height of our bounded region. Study tip: Always sketch the region and axis of revolution. For shell method radius, ask yourself: "How far is this strip from the axis?" The distance formula will guide you to the correct expression.
10. The region bounded by $y=x^2$ and $y=2x$ is revolved about the line $x=3$. The volume of the resulting solid is given by $V = \int_0^2 2\pi (3-x)h(x) \,dx$. What is the correct function for $h(x)$?
    Answer: $h(x) = 2x - x^2$ The problem uses the shell method, as indicated by the form of the integral. The term $h(x)$ represents the height of the cylindrical shells. The region is bounded by $y=2x$ (the upper curve) and $y=x^2$ (the lower curve) on the interval where they intersect, which is $[0, 2]$. The height of a vertical strip at a given $x$ is the difference between the upper function and the lower function. Therefore, $h(x) = 2x - x^2$.
11. A solid is generated by revolving the region bounded by $y=f(x)$ and $y=g(x)$, where $f(x) \ge g(x)$ on the interval $[a,b]$, about the vertical line $x=c$ where $c < a$. Which integral correctly represents the volume of this solid using the shell method?
    Answer: $\int_a^b 2\pi (x-c)(f(x)-g(x)) \,dx$ For the shell method with a vertical axis of revolution, we integrate with respect to $x$. The height of a cylindrical shell at position $x$ is the difference between the upper and lower curves, $h(x) = f(x)-g(x)$. The radius $r(x)$ is the distance from the axis of revolution $x=c$ to the shell's position $x$. Since $c < a \le x$, the distance is $r(x) = x-c$. The volume integral is $V = \int_a^b 2\pi r(x)h(x) \,dx = \int_a^b 2\pi (x-c)(f(x)-g(x)) \,dx$.
12. Consider the region bounded by $y = e^x$, $y = 1$, and $x = \ln(3)$. When this region is revolved around the $y$-axis using the shell method, which integral expression gives the correct volume?
    Answer: $2\pi \int_0^{\ln(3)} x(e^x - 1) dx$ When using the shell method to find volumes of revolution around the y-axis, you're essentially summing up cylindrical shells. Each shell has radius x, height given by the difference between functions, and thickness dx. First, let's visualize this region. The curve $y = e^x$ intersects $y = 1$ when $x = 0$ (since $e^0 = 1$). The vertical line $x = \ln(3)$ gives us our right boundary, where $y = e^{\ln(3)} = 3$. So we have a region from $x = 0$ to $x = \ln(3)$, bounded above by $y = e^x$ and below by $y = 1$. For the shell method around the y-axis, each vertical strip at position x becomes a cylindrical shell with: - Radius: x (distance from y-axis) - Height: $e^x - 1$ (top function minus bottom function) - Volume element: $2\pi x(e^x - 1)dx$ Integrating from $x = 0$ to $x = \ln(3)$ gives us $2\pi \int_0^{\ln(3)} x(e^x - 1) dx$, which is answer D. Answer A uses the washer method setup but incorrectly squares the radius term. Answer B also uses washer method thinking but with the wrong integrand. Answer C has the shell method format but incorrectly subtracts $e^x$ from 1, giving a negative height—remember that the exponential function is always above the horizontal line in this region. Study tip: Always sketch the region first and identify which function is on top. For shell method, the integrand should be radius × (outer function - inner function).
13. The finite region enclosed by $y = x^2 - 4x + 3$ and $y = x - 1$ is revolved around the line $y = -2$. Using the shell method, what integral setup would be most efficient?
    Answer: $2\pi \int_{-2}^0 (y + 2)[x_{right}(y) - x_{left}(y)] dy$ The curves intersect where $x^2 - 4x + 3 = x - 1$, giving $x^2 - 5x + 4 = 0$, so $x = 1$ or $x = 4$. Between these points, $x - 1 > x^2 - 4x + 3$. Since we're revolving around the horizontal line $y = -2$, horizontal shells are more natural. For horizontal shells, the radius is $y - (-2) = y + 2$, and we integrate with respect to $y$ over the range of $y$-values in the region. The $y$-values range from the minimum of the parabola (at $x = 2$: $y = 4 - 8 + 3 = -1$) to the intersection point values. Choice A uses vertical shells with wrong radius function. Choice C uses the wrong radius (should be distance to axis $y = -2$, not the parabola value plus 2). Choice D uses the line value plus 2 as radius, which is incorrect.
14. A region in the first quadrant is bounded by $x = y^2$ and $x = 8 - y^2$. When revolved around the line $x = -1$, the volume using the shell method requires integration with respect to which variable and over what interval?
    Answer: Integration with respect to $y$ over $[0, 2]$ The curves intersect where $y^2 = 8 - y^2$, so $2y^2 = 8$ and $y = 2$ (in the first quadrant). For shell method around $x = -1$, we can use horizontal shells. Each shell has radius $r = y - (-1) = y + 1$ and height $h = (8 - y^2) - y^2 = 8 - 2y^2$. We integrate with respect to $y$ from $0$ to $2$. Choice B would give vertical shells, but that's more complex since we'd need to solve for $y$ in terms of $x$. Choice C uses wrong limits (should be 2, not 4). Choice D also uses wrong variable and limits.
15. Let R be the region in the first quadrant bounded by $y=4-x^2$. Let $V_x$ be the volume when R is revolved about the x-axis, and $V_y$ be the volume when R is revolved about the y-axis. Which method is most efficient for calculating $V_y$, and what is the ratio $V_y/V_x$?
    Answer: Shell method is most efficient; ratio is $15/32$. $V_x$ (revolve about x-axis): Use disk method. Limits are $x=0$ to $x=2$. $V_x = \int_0^2 \pi (4-x^2)^2 \,dx = \pi \int_0^2 (16-8x^2+x^4) \,dx = \pi[16x - \frac{8}{3}x^3 + \frac{1}{5}x^5]_0^2 = \pi(32 - 64/3 + 32/5) = 256\pi/15$. $V_y$ (revolve about y-axis): Shell method is efficient. Limits $x=0$ to $x=2$. Radius $r(x)=x$, height $h(x)=4-x^2$. $V_y = \int_0^2 2\pi x(4-x^2) \,dx = 2\pi[2x^2 - x^4/4]_0^2 = 2\pi(8-4) = 8\pi$. The washer method for $V_y$ is also easy: $V_y = \int_0^4 \pi (\sqrt{4-y})^2 \,dy = \pi[4y-y^2/2]_0^4 = \pi(16-8)=8\pi$. So both methods are efficient. The ratio is $V_y/V_x = (8\pi) / (256\pi/15) = 8 \cdot 15 / 256 = 120/256 = 15/32$. Since both methods are efficient for $V_y$, we check the options. Both C and D have the correct ratio. But the shell method setup for $V_y$ is arguably more direct from the function's given form $y=f(x)$.
16. The region bounded by $y=\sin(x)$ and the x-axis from $x=0$ to $x=\pi$ is revolved about the line $x=-1$. What is the volume of the resulting solid?
    Answer: $2\pi(\pi+2)$ We use the shell method with respect to $x$. The axis of revolution is $x=-1$. The radius of a shell at $x$ is the distance from $x$ to $-1$, which is $r(x) = x - (-1) = x+1$. The height of the shell is $h(x) = \sin(x)$. The limits are from $0$ to $\pi$. The volume is $V = \int_0^\pi 2\pi(x+1)\sin(x) \,dx = 2\pi \int_0^\pi (x\sin(x) + \sin(x)) \,dx$. Using integration by parts, $\int x\sin(x) \,dx = -x\cos(x) + \sin(x)$. The integral becomes $2\pi [-x\cos(x) + \sin(x) - \cos(x)]_0^\pi = 2\pi[(-\pi(-1) + 0 - (-1)) - (0+0-1)] = 2\pi[(\pi+1) - (-1)] = 2\pi(\pi+2)$.
17. Consider the solid formed by revolving the region bounded by $y=x^2$ and $y=4$ about the line $x=3$. Let $V_S$ be the integral for the volume using the shell method and $V_W$ be the integral for the volume using the washer method. Which pair of integrals is correct?
    Answer: $V_S = \int_{-2}^2 2\pi (3-x)(4-x^2) dx$ and $V_W = \int_0^4 \pi ((3+\sqrt{y})^2 - (3-\sqrt{y})^2) dy$ Shell Method (integrate wrt $x$): Axis is $x=3$. The region spans $x \in [-2, 2]$. Radius $r(x)=3-x$. Height $h(x)=4-x^2$. So, $V_S = \int_{-2}^2 2\pi (3-x)(4-x^2) dx$. Washer Method (integrate wrt $y$): Axis is $x=3$. The region spans $y \in [0, 4]$. The right boundary is $x=\sqrt{y}$ and left is $x=-\sqrt{y}$. The outer radius is the distance from $x=3$ to the far curve $x=-\sqrt{y}$, so $R(y) = 3-(-\sqrt{y}) = 3+\sqrt{y}$. The inner radius is the distance from $x=3$ to the near curve $x=\sqrt{y}$, so $r(y) = 3-\sqrt{y}$. Thus, $V_W = \int_0^4 \pi (R(y)^2 - r(y)^2) dy = \int_0^4 \pi ((3+\sqrt{y})^2 - (3-\sqrt{y})^2) dy$.
18. In the formula for the volume of a solid of revolution using the shell method, $V = \int_a^b 2\pi r(x) h(x) \,dx$, what does the expression $2\pi r(x) h(x) \Delta x$ represent geometrically?
    Answer: The volume of a thin cylindrical shell with radius $r(x)$, height $h(x)$, and thickness $\Delta x$. The shell method approximates the volume by summing the volumes of infinitesimally thin cylindrical shells. The term $2\pi r(x)$ is the circumference of the shell, $h(x)$ is its height, and $\Delta x$ is its thickness. The product of circumference and height, $2\pi r(x) h(x)$, gives the lateral surface area of the cylinder. Multiplying this area by the thickness $\Delta x$ gives the approximate volume of that thin shell.
19. Let R be the region in the first quadrant bounded by the graphs of $y=x^3$, the x-axis, and the line $x=1$. Which integral represents the volume of the solid generated by revolving R about the line $x=2$ using the shell method?
    Answer: $V = \int_0^1 2\pi (2-x)(x^3) \,dx$ Using the shell method for a vertical axis of revolution, the integral is with respect to $x$. The region is bounded by $x=0$ and $x=1$. For a shell at a given $x$, the height $h(x)$ is given by the function $y=x^3$, so $h(x)=x^3$. The radius $r(x)$ is the distance from the axis of revolution $x=2$ to the shell at $x$, which is $r(x)=2-x$. The volume of a shell is $2\pi r(x)h(x)dx$. Therefore, the total volume is given by the integral $V = \int_0^1 2\pi (2-x)(x^3) \,dx$.
20. Which integral represents the volume of the solid generated by revolving the region bounded by $y=x^2$ and $y=x$ about the line $y=2$, using the shell method?
    Answer: $\int_0^1 2\pi(2-y)(\sqrt{y}-y) \,dy$ To use the shell method for revolution about a horizontal line ($y=2$), we must integrate with respect to $y$. First, express $x$ in terms of $y$: $x=y$ and $x=\sqrt{y}$. The region is bounded by $y=0$ and $y=1$. For a given $y$, the right boundary is $x=\sqrt{y}$ and the left is $x=y$. The height of a horizontal shell is $h(y) = \sqrt{y}-y$. The radius is the distance from the axis $y=2$ to the shell, so $r(y) = 2-y$. The volume is $V = \int_0^1 2\pi r(y)h(y) \,dy = \int_0^1 2\pi(2-y)(\sqrt{y}-y) \,dy$.

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