Volumes of Solids
Cross Sections
The following 3D demo represents the integral:
\int_{0}^{10}(\sqrt{x})^2dx
This integral is the volume of the solid with square cross sections that have a width and height of
\sqrt{x}
from x = 0 to x = 10.These cross sections are the squares that are shown in the demonstration. By using an integral to find the accumulation of the areas of these square cross sections, we can get the volume of the solid.
Disk Method
The following 3D demo represents the area under the function:
f(x) = -\frac{1}{2}x + 2
in Quadrant 1 revolved around the x-axis.The result is a solid cone. The volume of this cone can be found with the integral:
\pi\int_{0}^{4}(-\frac{1}{2}x^2+2)^2dx
This integral represents the accumulation of all the areas of the circular cross sections that make up the cone, which is equal to the volume of the cone.
Washer Method
Similar to the cone demo, this 3D demo represents the area inside the functions:
f(x) = -\frac{1}{2}x + 2
and y = 1
in Quadrant 1 revolved around the x-axis.This results in a cone with a cylinder in the center cut out. We can treat the volume of this solid like the accumulation of the area of all the "washers" that make it up.
The radius of the inside of the "washer" is 1. The radius of the whole "washer" is equal to f(x). The area of the "washer" at any given point can be calculated by subtracting the area of the inside circle from the area of the entire circle. This means that the volume of this solid is equal to:
\pi\int_{0}^{2}(-\frac{1}{2}x^2+2)^2 - (1)^2dx