1. Drill: Give a formula for the area of the plane region in terms of the single variable x.
A = x2
s2 + s2 = x2
2s2 = x2
s2 = x2/2
A = x2/2
A = ½ πx2
A = ½ π(x/2)2
A = ½ π(x2/4)
A = (1/8) πx2
(½ x)2 + h2 = x2
¼ x2 + h2 = x2
h2 = ¾ x2
A square with side length x
A square with diagonals of length x
A semicircle with radius x
A semicircle with diameter x
An equilateral triangle with side lengths of x

2. Day 1: p. 406-407: 1-17 (odd) Day 2: p. 406-407: 2-18 (even)in-class
7.3: Volumes
Day 1: p : 1-17 (odd)
Day 2:
p : 2-18 (even)in-class

3. Volume of a Solid
The volume of a solid of a known integrable cross section area A(x) from x = a to x = b is the integral of A from a to b:

4. How to Find the Volume by the Method of Slicing
Sketch the solid and a typical cross section.
Find a formula for A(x).
Find the limits of integration.
Integrate A(x) to find the volume.

5. SQUARE CROSS SECTIONS Ex: A Square-Based Pyramid
A pyramid 3 m higher has congruent triangular sides and a square base that is 3 m on each side. Each cross section of the pyramid parallel to the base is a square. Find the volume of the pyramid.
Step 1: Sketch the pyramid with its vertex at the origin. Note that the altitude is along the x-axis with 0 < x < 3
Sketch a typical cross section. 3 3

6. Step 2: Find a formula for A(x), the area of the cross section.
In this case, the cross section is a square, with side x, meaning A(x) = x2
Step 3: Find the limits of integration.
The squares go from 0 to 3
Step 4: Integrate to find the volume.
27/3 -0/3 = 9 m2

7. Circular Cross Sections Ex: A Solid of Revolution
The region between the graph f(x) = 2 + xcosx and the x-axis over the interval [-2, 2] is resolved about the x-axis to generate a solid. Find the volume of the solid. (Note x-scale is .5)

8. Set up a definite integral Evaluate using the calculator:
If you were to revolve the figure, the cross sections would would be circular.
The area of a circle is πr2 .
The radius of each circle will be the equation that has been given: 2 + xcosx (Note, the given function will ALWAYS be the radius when revolving an equation around the x-axis.)
A(x) = π(2+xcosx)2
Set up a definite integral
Evaluate using the calculator:
fnInt(π(2+xcosx)2, x, -2, 2)be in RADIAN mode!
The volume is units3

9. Washer Cross Section
The region in the first quadrant enclosed by the y-axis, and the graphs y = cosx and y = sinx is revolved about the x-axis to form a solid. Find its volume.
Step 1: Graph.
x-scale: π/2
y-scale: .5

10. Step 1: Determine the limits of integration.
When the figure is revolved around the x-axis, it will generate a solid with a cone shaped cavity in its center, AKA, as washer
The area of a washer can be found by subtracting the inner area of the outer area.
Step 1: Determine the limits of integration.
x = 0 (y-axis is right hand boundary)
x=π/4 (where cosx = sinx)
Step 2: Determine your outer radius and inner radius:
Outer: cosx
Inner: sinx

11. Step 3: Set up and Solve
Trig identity
π/2 units3

12. Drill
The region bounded by the curve y = x2 + 1 and the line y = -x + 3 is revolved about the x-axis to generate a solid Find the volume of the solid. (See the “washer” example from notes.)
23.4π units3

13. THAT’S IT….YOU DID IT! No more new stuff!
Now it’s time to get serious…..
Today: practice yesterday’s lesson, p : 2-18(even)
We will have practice 7.1 through 7.3 tomorrow and have a quiz Friday.
We will start reviewing for the AP exam next week. Please study the formulas in the back of the AP Test Prep book!