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

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 = x 2 s 2 + s 2 = x 2

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

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  1. 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 x2

  2. 7.3: Volumes Day 1: p. 406-407: 1-17 (odd) Day 2: p. 406-407: 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 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

  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 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)

  8. 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

  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. 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. 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!

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