7.3 Day One: 7.3 Day One: Volumes by SlicingVolumes by Slicing

•Volumes by slicing can be found by adding up each slice of the solid as the thickness of the slices gets smaller and smaller, in other words the number of slices goes to infinity.

•To find the volume of each slice of the pyramid you would find the area of each square then multiply the area by the thickness of the slice. The thickness would be dx because we are slicing with respect to the x-axis. Next you would add the volumes of each slice together to find the total volume.•This would be a Riemann sum with the limit as n the number of slices going to infinity.

n

k

b

a

kkx

dxxAxxAV1

*

0max

)()(lim

Volume of Known Cross-SectionsConsider the humble cylinder- we can think of a cylinder as a “stack of circles” whose Volume = (height )( ). We can think of this as the area of a cross-section [a circle] times the height of the cylinder.

2r

Here is an example using a cylinder

Find the volume of the cylinder using the formula and slicing with respect to the x-axis.

A = r2 A = 22 = 4

6

2

4 dx

16824)2(4)6(44 62 x

• Now use the formula you learned in Geometry to find the area

• V= r2 h=(22)(4) = 16

You can also work problems like these with respect to the y-axis.

Method of Slicing:

1

Find a formula for V(x).(Note that I used V(x) instead of A(x).)

Sketch the solid and a typical cross section.

2

3 Find the limits of integration.

4 Integrate V(x) to find volume.

Find the volume of an object whose base is the relation and when the cross sections are squares.

122 yx

Here’s what it looks like with a few squares:

With a lot more squares:

Area of a cross section = area of a squareIf , then with a domain of [-1,1]. Since our square goes from one curve to the other we can let the side of our cross section be

122 yx 21 xy

212 xs

dxsV

1

1

2

dxxV

1

1

22 )12(

)1,( 2xx

)1,( 2xx

What would be the volume if the cross sections weresemi-circles instead of squares?

)1,( 2xx

)1,( 2xx

Radius of cross section is 21 x

Area of a cross section is

2

2r

1

1

2

2dxrV

1

1

221

2dxxV

Let’s change our cross section to equilateral triangles!

Area of a cross section = Area of a triangle

Base of triangle =

Height of triangle =

Area of a cross section =

212 x

213 x

bh2

1

1

1-

2

1dxbhV

1

1

22 )13)(12(2

1dxxxV

What if the base region is between two curves?Volume = Area of base * heightFind the volume of the solid when the base is the region bounded by y = x and and whose cross sections perpendicular to the x-axis are squares.

2xy

Area of a cross section = 2s

2xxs

1

0

22 )( dxxxV

Cavalieri’s Theorem:

Two solids with equal altitudes and identical parallel cross sections have the same volume.

Identical Cross Sections