The solid of revolution It is the three-dimensional figure that is generated by rotating a flat surface around the axial axis or axis of revolution. Figure 1 shows an animation of a solid of revolution generated in this way.
Another very easy to visualize example consists of generating a right circular cylinder, rotating a rectangle of height or length h and radius r, around the positive x axis (figure 2). To find its volume there is a well-known formula:
V = area of base x height
Other solids of revolution are the sphere, the right circular cone and various figures, depending on the surface put in rotation and of course, the selected axis..
For example, rotating the semicircle around a line parallel to the diameter gives a solid of hollow revolution.
For the cylinder, the cone, the sphere, both solid and hollow, there are formulas to find the volume, which depends on the radius and the height. But when they are generated by other surfaces, the volume is calculated by definite integrals.
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Solids of revolution can be classified according to the curve that generates them:
It is enough to rotate a semicircle around an axis that will be the diameter of the sphere of radius R. Its volume is:
Vsphere = (4/3) πR3
To obtain a cone of height H and radius R, the surface to be rotated is a right triangle, around the axial axis that passes through one of the legs. Its volume is:
Vcone = (1/3) πHRtwo
Rotating a rectangle around an axial axis that passes through one of the sides, which can be the short side or the long side, we obtain a right circular cylinder of radius R and height H, whose volume is:
Vcylinder = πRtwoH
The torus is shaped like a donut. It is obtained by rotating a circular region around a line in the plane that does not intersect the circle. Its volume is given by:
Vtorus = 2πatwoR
Where a is the radius of the cross section and R is the radius of the torus according to the scheme presented in the figure:
In integral calculus these two methods are frequent:
-Discs and washers
-Shells
When slicing a solid of revolution, the cross section can be a disk, if the solid is solid, or it can be a kind of washer (a disk with a hole in the middle), if it is a hollow solid..
Suppose a planar region is rotated about the horizontal axis. From this flat region we take a small rectangle of width Δx, which is rotated perpendicularly around the axial axis.
The height of the rectangle is between the outermost curve R (x) and the innermost curve r (x). They correspond to the outer radius and inner radius respectively..
By making this rotation a washer of volume ΔV is generated, given by:
ΔV = Full volume - volume of hole (if any)
Remembering that the volume of a right circular cylinder is π. radiotwo x height, we have:
ΔV = π [Rtwo(x) - rtwo(x)] Δx
The solid can be divided into a multitude of small volume portions ΔV. If we add them all together, we will have the full volume.
For this we make the volume ΔV tend to 0, with which Δx also becomes very small, becoming a differential dx.
So we have an integral:
V = ∫tob π [Rtwo(x) - rtwo(x)] dx
In case the solid is solid, then the function r (x) = 0, the slice of the solid that is generated is a disk and the volume remains:
V = ∫tob πRtwo(x) dx
When the axis of revolution is vertical, the above equations take the form:
V = ∫tob π [Rtwo (y) - rtwo (y)] dy y V = ∫tob πRtwo(y) dy
As the name implies, this method consists in assuming that the solid consists of layers of differential thickness. The layer is a thin tube that originates from the rotation of a rectangle parallel to the axis of rotation.
We have the following dimensions:
-The height of the rectangle w
-Its longitude h
-The distance from the center of the rectangle to the axis of rotation p
Knowing that the volume of the layer is exterior volume - interior volume:
π (p + w / 2)twoh - π (p - w / 2)twoh
By developing remarkable products and simplifying, you get:
Layer volume = 2π⋅p⋅w⋅h
Now let's make the height w of the rectangle Δy, as seen in the following figure:
With this the volume ΔV is:
ΔV = 2π p x h x Δy
And making the number of layers n is very large, Δy becomes a differential dy, with which the total volume is the integral:
V = ∫cd 2π p (y) h (y) dy
The procedure described applies similarly when the axis of revolution is vertical:
Find the volume generated by the rotation of the plane region between the curves:
y = xtwo; y = 0; x = 2
Around the y axis.
-The first thing we must do is graph the region that will generate the solid of revolution and indicate the axis of rotation. We have it in the following graph:
-Now we look for the intersections between the curve y = xtwo and the line x = 2. For its part, the line y = 0 is none other than the x axis.
From the graph it is easy to see that the parabola and the line intersect at the point (2,4), which is corroborated by substituting x = 2 in y = xtwo.
-Then one of the methods to calculate the volume is chosen, for example the layer method with vertical axis of revolution:
V = ∫tob 2π p (x) h (x) dx
Important: In the layering method the long side of the rectangle is parallel to the axis of rotation.
The radius of the layer is x
The height of the rectangle is determined by the parabola xtwo.
The integration variable is x, which varies between 0 and 2, with this we have the limits of integration. Substituting the expressions for p (x) and h (x)
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