Volume with Shell Method for Revolving Curves and Lines

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Homework Help Overview

The problem involves using the shell method to determine the volume of a solid formed by revolving a region bounded by the curve x=2y-y² and the line x=0. The original poster expresses difficulty in visualizing the problem and graphing the curves.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the nature of the curve x=2y-y², questioning how to graph it and where it intersects with x=0. There is an inquiry about the axis of rotation and how to identify the shape of the curve.

Discussion Status

The discussion is ongoing, with participants providing insights into the characteristics of the curve and suggesting methods for visualizing it. There is no explicit consensus yet, but some guidance on graphing techniques and understanding the parabola's orientation has been offered.

Contextual Notes

Participants note the challenge of graphing the curve and the need to clarify the axis of rotation. The original poster's uncertainty about the shape of the curve indicates a potential gap in understanding the relationship between the equations and their graphical representations.

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Homework Statement



Use the shell method to find the volumes of the solid generated by revolving the regions bounded by the curves and lines.

x=2y-y2, x=0

Homework Equations



The shell method is of the format: V = 2 \pi\int x * height dx


The Attempt at a Solution



I cannot picture the problem (not sure exactly how to graph it)
 
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x=2y-y^2 is a parabola and x=0 is a line. What's the problem with graphing them? Where do they intersect? And what what axis are you going to rotate the region around?
 
rotating around the x-axis


how can I tell its a parabola?
 
A usual parabola is given by y = ax^2 + bx + c. In this case, we have x = ay^2 + by + c. The two are much the same really, except that in the second case, the parabola is 'on it's side'.

It is actually two square roots, pasted together though. You can see this by solving the equation for y, resulting in your usual "y = f(x)" graph. For general a, b, c:
y = \frac{-b \pm \sqrt{b^2 - 4ac + 4ax} }{2a}
(Two equations, one for + and one for -!)

If you graph these, we get (using a = 1, b = 2, c = -1 for example):
sqrz8o.jpg

(The two don't meet in the middle exactly because Maple has trouble graphic them there.)If you really can't figure out the shape of a curve, why not simply try to draw a few easy points on paper? Like (0,0) or (0,1), (1,0) etc... You will most likely recognize a familiar shape from that, and you can then go on and analyze the curve equation further.
 
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