SUNY Geneseo Department of Mathematics
Wednesday, December 9
Math 221 02
Fall 2020
Prof. Doug Baldwin
In problem set 11, question 3, is substitution a legitimate way to do the proof? Yes.
Based on “Volume and the Slicing Method” in section 6.2 of our textbook and this discussion of volume by slices.
A pyramid with a right-triangle base is lying with one side on the x axis so that it’s point is at the origin and its base is at x = 2. The two legs of the right triangle that form the base are each 1 unit long, and the pyramid tapers linearly to its point. Find its volume.
The basic idea in all volume-by-slices problems is to think of the shape whose volume you want as being carved into thin slices, slices so thin that they have sides approximately parallel to the (x, in this case) axis — in that case the volume of a slice is just its thickness, Δx, times its area. Denote the area as a(x), to emphasize that it typically varies with x. Now the volume of the shape is just the sum of the volumes of all the slices, which is a Riemann sum — i.e., the volume of the shape is an integral of areas of slices:
So the first step in this case is to work out what the area of a slice is. Each slice is a triangle, so use the formula for the area of a triangle. The description of the end of the pyramid indicates that the triangle’s base and height are equal, and since the sides taper linearly to the point, that relationship holds through the whole triangle. We also noticed that the sum of the base and height equals the x coordinate of each triangle, a relationship that also holds over the whole triangle because of its linear sides. Then we can solve for the length of the base and height. (We could also get the base and height by noticing that each is half of x at the end of the pyramid, yet another relationship that holds throughout the pyramid because of linearity.) However you find the base and height in terms of x, you can use it to come up with a formula for area in terms of x:
Finally, plug the area formula into the integral we came up with, and evaluate:
The process we followed, and which generally works for all volumes by slices, is...
Similar to the above shape, except that instead of the sides tapering linearly to the point, they follow parabolic curves. In particular, at each x value, the sides of the triangular cross-section have length 2√x.
Finding the areas of the slices is easier than in the first example, since this problem gives the relationship between x and the base and height of each triangular slice:
Once you know the area formula, plug it into the integral for volume by slices and evaluate:
A frustum is a pyramid with its top cut off, i.e., a tapered shape with a rectangular cross section that doesn’t come to a point. Imagine a frustum lying on the x axis and extending from x = 1 to x = 3; at any given x coordinate between those bounds, the width of the frustum is x, and the height is x/2. What is the frustum’s volume?
Here the shape doesn’t have triangular cross sections, but rather rectangular ones. So use the formula for area of a rectangle to find the areas of the slices...
...and then integrate the areas over the range of x values occupied by the frustum:
How does this idea apply to question 4 in problem set 11?
Looking at the picture in the question, think of the length, h, of the shape as extending along the x axis, from x = 0 to x = h. The cross-sectional area is then A, which is constant. Plug that area and the range of x values into the integral formula for volume.
Finding volumes produced by rotating a curve around the x axis.
Please read “Solids of Revolution” and “The Disk Method” in section 6.2 of the textbook by class time Friday.
Please also contribute to a this discussion of volumes of revolution by class time Friday.