What axis is this to be revolved around ?whatlifeforme said:Homework Statement
Determine how many integrals are required for disk, washer, and shell method.
Homework Equations
x=3y^2 - 2 and x=y^2 from (-2,0) to (1,1)
The Attempt at a Solution
Since there are no breaks or abnormalities in the graph it appears that 1 integral will solve for each to me. I have no other way of looking at this problem.
how do i know when the disk method requires more than one integral?
SammyS said:What axis is this to be revolved around ?
Have you drawn a sketch of the region that will be rotated?whatlifeforme said:sorry. about the x-axis.
evidently it will take two for disk and washer, and one for shell, but I'm not sure why.
These are all methods for calculating the volume of a solid using multiple integrals. The main difference is the shape of the cross-section used to slice the solid. Disk integrals use circular cross-sections, washer integrals use annular (ring-shaped) cross-sections, and shell integrals use cylindrical cross-sections.
Disk and washer integrals are typically used for solids of revolution, where the shape of the solid can be obtained by rotating a curve around an axis. Shell integrals are used for solids with a cylindrical or prismatic shape. The choice of method depends on the shape of the solid and the cross-sections used to slice it.
The general formula for these integrals is ∫ab π(R(x))2 dx, where R(x) is the radius of the cross-section at a given x-value. The limits of integration, a and b, will depend on the shape and orientation of the solid.
One common mistake is not properly setting up the integral by choosing the correct limits of integration and function to integrate. It is also important to correctly identify the shape and orientation of the solid to determine which method to use. Additionally, make sure to double check the units of measurement to ensure the final answer is in the correct units.
Disk and washer integrals can be thought of as special cases of shell integrals, where the height of the cylindrical cross-section is infinitesimally small. This means that disk and washer integrals can be obtained by integrating a shell integral using the appropriate limits and function. Additionally, the volume calculated using different methods should be the same for the same solid.