Drill: Give a formula for the area of the plane region in terms of the single variable x.

• 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

• 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 x22

3xh

4

3

2

3

2

1 2xxxA

7.3: Volumes

Day 1: p. 406-407: 1-17 (odd)Day 2:

p. 406-407: 2-18 (even)in-class

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:

b

a

dxxA )(

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.

SQUARE CROSS SECTIONSEx: 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

• 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.

3

0

2dxx3

33

0

x 27/3 -0/3 = 9 m2

Circular Cross SectionsEx: 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)

• 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 52.43 units3

2

2

2)cos2( dxxx

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

• 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

Step 3: Set up and Solve

4/

0

22 )sin(cos

dxxx 4/

0

22 )sin(cos

dxxx

4/

0

)2(cos

dxx

Trig identity

2

2sin4/

0

x

2

)0(2sin

2

)4/(2sin

2

0

2

1

2

0sin

22

sin

π/2 units3

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.)

1

2

222 )1()3( dxxx

1

2

242 )12(96 dxxxxx

1

2

242 )1296( dxxxxx

1

2

42 )86( dxxxx

23.4π units3

THAT’S IT….YOU DID IT!

• No more new stuff!• Now it’s time to get serious…..• Today: practice yesterday’s lesson, p. 406-407:

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!