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Published byMilton Robertson Modified over 9 years ago
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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
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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
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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:
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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.
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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
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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
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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)
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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
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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
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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
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Step 3: Set up and Solve Trig identity π/2 units3
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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
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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!
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