Probably a typo - the volume is (4/3)##\pi r^3##.Feodalherren said:
The book is using disks that are oriented vertically (the x-axis intersects each disk perpendicularly). The volume of such a disk is its area times its thickness, dx.Feodalherren said:Homework Equations
calculus, integration
The Attempt at a Solution
I have the solution in the book but it's confusing me, I'll attach a picture.
So I get lost where it starts talking about a cross-sectional area all of a sudden multiplying by ∏. What's going on here?!
The formula for finding the volume of a sphere using calculus and integration is V = ∫πr^2dx, where r is the radius of the sphere and dx is the small element of volume.
The process for solving for the volume of a sphere using calculus and integration involves setting up the integral using the formula V = ∫πr^2dx, determining the limits of integration, and then evaluating the integral using integration techniques such as substitution or u-substitution.
The limits of integration for finding the volume of a sphere using calculus and integration are -r to r, as this represents the range of the radius of the sphere.
Yes, the volume of a sphere can also be found using the formula V = (4/3)πr^3 which is derived from geometry and does not require calculus or integration.
The volume of a sphere found using calculus and integration is more accurate as it takes into account the infinitesimal elements of the sphere, while the formula V = (4/3)πr^3 is an approximation and may have a larger margin of error.