Finding Volume of the solid, Integral

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To find the volume of the solid formed by revolving a circular disk defined by x^2 + y^2 <= a^2 around the line x = a, it is essential to determine whether to integrate with respect to x or y, which depends on the chosen method (washers or cylindrical shells). The line x = a indicates the axis of revolution, and visualizing the disk and resulting solid is crucial for understanding the problem. The discussion emphasizes that the interval of integration is not simply [0, a], and the correct setup for the volume element must be established. The participants clarify that the volume formula should reflect the correct bounds and relationships derived from the geometry of the problem. Drawing a sketch is recommended to better grasp the dimensions and limits involved in the integration process.
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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 \Delta x ranges?
\Delta V = 2\pi (a - x)2y\Delta x
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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