calculationg volume using integral?

Question

Concern the region bounded by y = x^2. y = 1, and the y-axis. Find the volume of the solid obtained by rotating the region around the y-axis. How do I solve this using integrals?

Answer 1

If you visualize the graph, it's pretty much a bowl shaped solid. There are two methods that are used in rotations - the disk/washer method or the cylindrical shell method. Each one takes your solid and slices it into disks or cylinders, and then sums the volumes of those pieces to find the total volume of the solid. The "slicing" determines which method you use and how your integral is set up.

For the disk/wasker method, we slice the bowl parallel to the x-axis, so that the pieces look like disks. Since we're slicing it horizontally, we have a disk that has a radius of y^(1/2) and a height of dy. The volume of a disk is pi*r^2*h, and since the bowl is bounded by y = 0 and y = 1, we can set up an integral like this:

V = S pi * (y^(1/2))^2 dy, evaluated from 0 to 1 (I'm using S as the integration symbol - I can't figure out how to print that character here)
V = pi S y dy
V = pi (y^2)/2, evaluated from 0 to 1
V = pi/2

Using the cylindrical shell method is similar - we're basically finding the volume of a shell formed by peeling off a layer that is, in this case, 1 - x^2 high and dx thick. The length of our shell is the circumference of the cylinder - 2 * pi * r, where r is our x value. Volume of this shell is simply length * width * height, so here we go:

V = S (2 * pi * x)(1 - x^2) dx, evaluated from 0 to 1 (that's the domain for x in this figure)
V = 2pi S (x - x^3) dx
V = 2pi ((x^2/2) - (x^4)/4), , evaluated from 0 to 1
V = 2pi (1/2 - 1/4)
V = 2pi (1/4)
V = pi/2

You get the same answer either way - one is just easier to get to than the other. It will take a bit of practice to see which method is best for a particular situation.

Answer 2

If I have were given this top your area will strengthen quadratically from 0 to the bottom area A over a top h in the variety of fashion that area at a distance x from the end of the cone = A*(x/h)^2 . Now you pick to combine area*dx from 0 to h to get the quantity?