Finding Volume of the solid, Integral

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Homework Help Overview

The problem involves finding the volume of a solid formed by revolving a circular disk defined by the inequality x² + y² ≤ a² around the line x = a, where a > 0. Participants are exploring the appropriate method for integration and the implications of the setup.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning whether to integrate with respect to x or y, considering the method of washers versus cylindrical shells. There is also discussion about the significance of the line x = a in the context of the problem.

Discussion Status

Some participants have suggested drawing sketches to better understand the problem. There are attempts to clarify the correct setup for the volume integral, but confusion remains regarding the representation of the volume elements and the limits of integration.

Contextual Notes

Participants note that there may be missing information regarding the interval of integration and the correct formulation of the volume element. There is an emphasis on visualizing the problem to aid understanding.

lovemake1
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Homework Statement



Question Reads: A circular disk x ^2 + y^2 <= a ^ 2 , a > 0 is revolved about the line x = a.
Find the volume of the resulting solid.


Homework Equations


v = integral(a, b) (2pi)y [F(y) - G(y)] dy


The Attempt at a Solution



Im currently confused, should i take intergral with respect to x or y?
and what does this x = a tell me ?
does it just mean integral from (0, a)
 
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lovemake1 said:

Homework Statement



Question Reads: A circular disk x ^2 + y^2 <= a ^ 2 , a > 0 is revolved about the line x = a.
Find the volume of the resulting solid.


Homework Equations


v = integral(a, b) (2pi)y [F(y) - G(y)] dy


The Attempt at a Solution



Im currently confused, should i take intergral with respect to x or y?
It depends on whether you use washers or cylindrical shells for your typical volume elements.

I'm guessing that you haven't drawn a sketch of the disk, and one of the solid of revolution. If that's the case, draw them. It's harder to get a handle on these kinds of problems if you don't have a good sense of what the region being revolved and the resulting solid look like.
lovemake1 said:
and what does this x = a tell me ?
does it just mean integral from (0, a)
x = a is the vertical line that the disk (the circle and its interior) is revolved around.
 
ok so since this is a shell method.
i would have to represent in y integral.sqrt(y^2 - a ^2) = x
V= integral(0, a) 2pix(sqrt(y^2 - a ^2))dx

is that correct?
 
lovemake1 said:
ok so since this is a shell method.
i would have to represent in y integral.
I don't understand what you mean. If you mean an integral with dy, then no.
lovemake1 said:
sqrt(y^2 - a ^2) = x
There's no reason to solve for x, but there's a very good reason to solve for y. In any case, your equation above is incorrect. If you square both sides, you get y2 - a2 = x2, or y2 - x2 = a2. That's not what you started with.
lovemake1 said:
V= integral(0, a) 2pix(sqrt(y^2 - a ^2))dx

is that correct?
No. Did you draw a picture? If you had, you would see that the interval of integration is not [0, a].
In my drawing, this is the formula for the typical volume element. What is the interval over which [itex]\Delta x[/itex] ranges?
[tex]\Delta V = 2\pi (a - x)2y\Delta x[/tex]
 

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