## Volumes of RevalationsBy Francisco, Mario, joseph, joaquin

### Cross Sectional Partitions

A cross section is simply cutting through an object and getting a shape out of it. We can use a definite Integral to find the volume of a cross-section. Determining the terms is simply taking the cross-sections parallel to the axis. So, if the cross section is parallel to the x-axis our terms are in (x), and vise versa, if itβs parallel to the y-axis our terms are in(y).

Steps to find volume

Identify cross section. It can be circular, semi circular, squares, or any other shape.

Find the function for the area.

Plug in function for A(x) and solve the integral.

### Disk Method

How To Solve: Identify The Bounds. F(X) Is The Function Unless It Is Specified That There Is A Shift. If There Is A Shift, Identify The Area Rotated And It Will Be Plugged In A Place Of F(X). Hint: The Result Will Either Be (F(X)+Shift), (Shift-F(X)), Or (F(X)-Shift). Plug Into The Equation And Solve The Integral.

When rotating about the x-axis, our equation will be in terms of (x), or simply y = f(x).

V = π β« [R(x)2]dx

When rotating about the y-axis, our equation will be in terms of (y), or simply x = g(y).

V = π β« [R(y)2]dy

How to solve:

Identify the bounds.

f(x) is the function unless it is specified that there is the a shift.

If there is a shift, identify the area rotated and it will be plugged in a place of f(x).

Hint: The result will either be (f(x)+shift), (shift-f(x)), or (f(x)-shift).

Plug into the equation and solve the integral.

### Washer Method

Typo: The equation should be Ο instead of 2Ο.

When the solid is rotated about the y-axis, our equation will be in terms of (y). V = β« π ( R2outer- r2inner)dy When the solid is rotated about the x-axis, our equation will be in terms of (x). V = β« Ο (R2outer - r2inner)dx. Find the volume.

Example 1: Find the volume of the region generated by rotating Y= x2, and X = y2 about the x-axis.

### Shell Method

When rotating about the y-axis, our equation must be in terms of (x), so we would use this equation β V =2π β« (radius)(height of shell) dx

When rotating about the x-axis, our equation must be in terms of (y), so we would use this equation β V = 2π β« (radius)(height of shell) dy.

How to Solve:

1. Identify the bounds.

2. The equation should be in terms opposite to the axis that itβs being rotated about.

3.Β The radius will be x or y (depending on the axis of revolution) unless it is specified that itβs shifted. If there is a shift, identify the new radius. Hint: It will be (f(x)+shift),(f(x)-shift), or (shift-f(x)).

4. Height, is going to be the function.

5. Plug into equation and solve integral.

Volume of revolution practice problems

Credits:

Created with images by skeeze - "scuba diver boat leaping" β’ tpsdave - "sea ocean water"