Discs and Washers

I.  Quiz

II  Homework

III.  Volumes of Revolution

Suppose you wanted to make a clay vase.  It is made by shaping the clay into a curve and spinning it along an axis.  If we want to determine how much water it will hold, we can consider the cross sections that are perpendicular to the axis of rotation, and add up all the volumes of the small cross sections.  We have the following definition:

where A(x) is the area the cross section at a point x.

Example:  Disks

Find the volume of the solid that is produced when the region bounded by the curve y = x², y = 0, and x = 2 is revolved around the y-axis.

Solution:

Since we are revolving around the x-axis, we have that the cross section is in the shape of a disk with radius equal to the y coordinate of the point.  Hence A(x) = pr² = p[x²]²

We have

= p/5 x⁵| from 0 to 2 = 32p/5.   

Example:  Washers

Find the volume of the solid formed be revolving the region between the curves y = x² and y = sqrt(x) about the x-axis.

Solution

We draw the picture and revolve a cross section about the x-axis and come up with a washer.  The area of the Washer is equal to the area of the outer disk minus the area of the inner disk.  

A = p(R² - r²)

We have that R is the y coordinate of the top curve (sqrt(x)) and r is the y coordinate of the bottom curve (x²).  We have

 A = p([sqrt(x)]² - [x²]²) = p[x - x⁴]

Hence

Example:  Revolving about the y-axis

Find the volume of the solid that is formed by revolving the curve bounded by y = x and y = sqrt(x) about the y-axis.  This time our cross section is perpendicular to the y-axis.  When we revolve, we get a washer with R equal to the x-coordinate of the y =sqrt(x) curve and r equal to the x-coordinate of the y = x curve.  Hence

A = p((y²)² - (y)²) = p(y⁴ - y²)

We get