SECTION 7.3Volume

VOLUMES OF AN OBJECT WITH A KNOWN CROSS-SECTION

Think of the formula for the volume of a prism: V = Bh.

The base is a cross-section of the prism. Now imagine that the shape has a very tiny thickness, thus forming a solid. We can think of the prism as a stack of many many of these solids.

We can use calculus to find the volume of a solid by writing an expression for the area of a cross-section and integrating that expression over the length of the object.

GENERAL FORMULA

Where A(x) is the area of a cross section.

ANIMATION

http://homework.zendog.org/2012ch7daytwo.pdf

EXAMPLE 1

Find the volume of the solid created on a region whose base is the function For this solid, each cross section perpendicular to the x-axis is a square.

Begin by writing a formula for the area of the square cross section.

EXAMPLE 2

Find the volume created by a solid whose base is the region y = x2 for 0 < x < 3 if the cross sections are

A) semicirclesB) Isosceles right triangles with the leg on the base.

Each cross section is perpendicular to the x-axis.

EXAMPLE 3

Find the volume of an object that has a circular base of radius 2cm and a cross section that is perpendicular to the x-axis and is a right isosceles triangle with the leg on the base.

VOLUME OF REVOLUTION

Imagine if our graph of were rotated around the x-axis to form a solid. Now take a vertical cross section of that solid.

What would the cross section look like?Circle

How would you find the area of that circle?

How would you find the volume of the solid?

DISC METHOD

Take a region R, which consists of the area between a function f(x) and the x- axis over the interval [a, b]. If R is rotated around the x-axis to form a solid, then its cross section will be a circular disc. The volume of the solid produced by the rotation is…

The same procedure can be done for a function of y rotated around the y-axis.

https://www.youtube.com/watch?v=-CPUdbjpnno

EXAMPLE

Find the volume of the solid resulting from revolving the region bounded by the curves and from x = 0 to x = around the y- axis.

WASHER METHOD

Now visualize the region bounded by the curves and y = x/2. If this region were rotated around the x axis….

What would the vertical cross section look like?A washer….a circle with a circular hole cut out of the middle.

How would you find the area of this washer?Area of the outer circle – Area of the hole.

WASHER METHOD CONTINUED

The radius of each circle is equal to the value of their corresponding function.

If for all x in an interval, then the volume of a solid formed by rotating the region bounded by f(x) and g(x) over the x-axis on the interval [a, b] is…

https://www.youtube.com/watch?v=3oAjcLD34kc

EXAMPLE

Let R be the region bounded by the curves and . Find the volume of the solid formed by rotating R around the x-axis.

ROTATING ABOUT A LINE OTHER THAN THE X OR Y AXIS

When rotating around an axis, the value of a function (biggie or smalls) tells you how far the function is from the axis.

When rotating around a line other than an axis, you must write an expression that represents the distance from the curve(s) to that line.

EXAMPLE

Let R be the region bounded by y = 4 – x2 and y = 0. Find the volume of the solids obtained by revolving R about each of the following….

(a) the x axis(b) the line y = -3(c) the line y = 7(d) the line x = 3