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Find the volume of the solid obtained by rotating the region bounded by the given...
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(Level 1) Associate Educator, Expert, Newton
First the volume of a cylinder is
`pi R^2 h`
If we were to drill a hole through the middle of the cylinder, such that we now had a tube-like thing, or a washer, the volume would be the volume of the original cylinder minus the volume of the hole, or:
`pi R^2 h - pi r^2 h = pi (R^2-r^2) h`
Ok, I don't know how well you can see the various lines in the picture, but here goes:
The black tear drop shape gets rotated around the dotted black line. If you were to slice this shape with vertical slices, you would get a bunch of washers: cylinders with cylindrical holes in them. The formula for the volume of this shape is:
`int_0^1 pi (R^2 - r^2) dx`
where R represents the outer radius of a washer slice and r represents the inner radius of a washer slice.
(The `pi(R^2-r^2)` represents the area of the cross section when you slice it vertically. The dx can be thought of as the "height" of a washer. If you are making vertical slices, your cylinders are lying on their sides, and the heights are the `Delta x` . As you take slices closer and closer together, and let `Delta x -> 0` , this becomes the above integral)
Let's first look at the outer radius:
This is determined by:
The distance from the x axis up to the curve is `x^2`
The distance from the x-axis the the line y=1 is 1
Thus, the distance from the curve to the line y=1 is actually `1-x^2`
So the outer radius is `1-x^2`
(in the first picture, the vertical multicolored line is length 1. The green piece of it is `x^2` . Thus the red piece, which is actually the radius, is `1-x^2` )
Similarly for the inner radius:
The distance from the x-axis to the line y=1 is 1 (represented by the multicolored vertical line).
The distance from the x-axis to the curve is `y=sqrt(x)` (represented by the green line)
Thus, the inner radius (represented by the red line) is `1-sqrt(x)`
So, putting that into the formula for volume using washer slices, we have:
`int_0^1 pi[ (1-x^2)^2 - (1-sqrt(x))^2] dx`
Posted by mlehuzzah on December 29, 2012 at 6:45 AM (Answer #1)
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