- #1
DarK1
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x = 2y - y2
x = y
x = y
(Integration) Find volume using disk method when rotated about the X-axis
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- Thread starter DarK1
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- Disk Integration Method Volume
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Discussion Overview
The discussion revolves around finding the volume of a solid of revolution using the disk method when the region bounded by the curves \(x = 2y - y^2\) and \(x = y\) is rotated about the x-axis. The conversation includes aspects of integration, volume calculation, and the application of different methods such as the washer and shell methods.
Discussion Character
- Technical explanation
- Mathematical reasoning
- Homework-related
Main Points Raised
- One participant suggests sketching the region to be revolved and asks for clarification on the axis of rotation.
- Another participant identifies the axis of rotation as the x-axis and prompts for the outer radius, inner radius, and limits of integration.
- A further reply discusses the washer method, providing formulas for the volume of an arbitrary washer and the specific outer and inner radii based on the given curves.
- Participants derive the inner radius from the parabolic curve and express it in terms of \(y\) as \(r^2=(1-\sqrt{1-x})^2\).
- There is a calculation of the limits of integration by equating the two curves, leading to the conclusion that \(x\) ranges from 0 to 1.
- One participant computes the volume using the washer method and confirms the result with the shell method, both yielding \(\frac{\pi}{6}\). However, the correctness of these calculations is not established as a consensus.
Areas of Agreement / Disagreement
Participants provide various approaches and calculations, but there is no explicit consensus on the correctness of the methods or results presented. The discussion remains open with multiple viewpoints on the integration process.
Contextual Notes
Some assumptions regarding the curves and their intersections may not be fully explored, and the discussion does not resolve potential discrepancies in the calculations or methods used.
- #2
MarkFL
Gold Member
MHB
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Re: finding VOLUME using integration (DISK METHOD)
The first thing I would do is sketch the region to be revolved:
View attachment 7885
Now, what is the axis of rotation?
The first thing I would do is sketch the region to be revolved:
View attachment 7885
Now, what is the axis of rotation?
Attachments
- #3
MarkFL
Gold Member
MHB
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I didn't realize you had posted the same problem twice, and in the second one I saw, gave the axis of rotation. I have since merged the two threads.
Okay, so we have the axis of rotation as the $x$-axis. Can you identify the following:
Okay, so we have the axis of rotation as the $x$-axis. Can you identify the following:
- The outer radius
- The inner radius
- The limits of integration
- #4
MarkFL
Gold Member
MHB
- 13,284
- 12
To follow up, the washer method gives us:
$$V=\pi\int_{x_1}^{x^2} R^2-r^2\,dx$$
The volume of an arbitrary washer is:
$$dV=\pi\left(R^2-r^2\right)\,dx$$
The outer radius is the line $y=x$, and so:
$$R^2=x^2$$
The inner radius is the parabolic curve, but we are given this curve in the form $x(y)$, and we need it in the form $y(x)$. So, we may write it in the form:
$$y^2-2y+x=0$$
We see the axis of symmetry is $y=-\dfrac{-2}{2(1)}=1$, and so we want the lower half of the curve.
Apply the quadratic formula, and discard the larger root:
$$y=\frac{2-2\sqrt{1-x}}{2}=1-\sqrt{1-x}$$
Thus:
$$r^2=(1-\sqrt{1-x})^2=1-2\sqrt{1-x}+1-x=2-2\sqrt{1-x}-x$$
Hence:
$$dV=\pi\left(x^2+x+2\sqrt{1-x}-2\right)\,dx$$
To determine the limits of integration, we equate the two curves:
$$x=1-\sqrt{1-x}$$
Let $u=1-x$, and we have:
$$u=\sqrt{u}\implies u\in\{0,1\}\implies x\in\{0,1\}$$
Hence, adding the slices by integration, we obtain:
$$V=\pi\int_0^1 x^2+x+2\sqrt{1-x}-2\,dx=\frac{\pi}{6}$$
Let's check our result using the shell method:
$$V=2\pi\int_0^1 y\left(\left(2y-y^2\right)-y\right)\,dy=2\pi\int_0^1 y^2-y^3\,dy=\frac{\pi}{6}\quad\checkmark$$
$$V=\pi\int_{x_1}^{x^2} R^2-r^2\,dx$$
The volume of an arbitrary washer is:
$$dV=\pi\left(R^2-r^2\right)\,dx$$
The outer radius is the line $y=x$, and so:
$$R^2=x^2$$
The inner radius is the parabolic curve, but we are given this curve in the form $x(y)$, and we need it in the form $y(x)$. So, we may write it in the form:
$$y^2-2y+x=0$$
We see the axis of symmetry is $y=-\dfrac{-2}{2(1)}=1$, and so we want the lower half of the curve.
Apply the quadratic formula, and discard the larger root:
$$y=\frac{2-2\sqrt{1-x}}{2}=1-\sqrt{1-x}$$
Thus:
$$r^2=(1-\sqrt{1-x})^2=1-2\sqrt{1-x}+1-x=2-2\sqrt{1-x}-x$$
Hence:
$$dV=\pi\left(x^2+x+2\sqrt{1-x}-2\right)\,dx$$
To determine the limits of integration, we equate the two curves:
$$x=1-\sqrt{1-x}$$
Let $u=1-x$, and we have:
$$u=\sqrt{u}\implies u\in\{0,1\}\implies x\in\{0,1\}$$
Hence, adding the slices by integration, we obtain:
$$V=\pi\int_0^1 x^2+x+2\sqrt{1-x}-2\,dx=\frac{\pi}{6}$$
Let's check our result using the shell method:
$$V=2\pi\int_0^1 y\left(\left(2y-y^2\right)-y\right)\,dy=2\pi\int_0^1 y^2-y^3\,dy=\frac{\pi}{6}\quad\checkmark$$
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