Finding Volume of the solid, Integral

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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)
 

Answers and Replies

  • #2
34,886
6,626

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.
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.
 
  • #3
150
1
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?
 
  • #4
34,886
6,626
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.
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.
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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