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A k = area of k th rectangle, f(c k ) – g(c k ) = height, x k = width. 6.1 Area between two curves

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Figure 4.23: When the formula for a bounding curve changes, the area integral changes to match. (Example 5) Find the area of the region between the curves

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A Section 6.2 Figure 5 Approximating the volume of a sphere with radius 1 (a) Using 5 disks, V (b) Using 10 disks, V (c) Using 20 disks, V

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Figure 5.6: The region (a) and solid (b) in Example Volumes – Solid of revolution y = f(x) is rotated about x-axis on [a,b]. Find the volume of the solid generated. A cross-sectional slice is a circle and a slice is a disk.

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Figure 5.6: The region (a) and solid (b) in Example 4. Volumes – Solid of revolution is rotated about the x-axis on [0, 4] Find the volume of the solid generated.

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Find the volume of the solid generated by revolving a region between the y-axis and the curve x = 2/y from y = 1 to y = 4. Volumes by disk-y axis rotation

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Find the volume of the solid generated by revolving a region between the y-axis and the curve x = 2/y from y = 1 to 4.

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Figure 5.10: The cross sections of the solid of revolution generated here are washers, not disks, so the integral A(x) dx leads to a slightly different formula. baba Washers If the region revolved does not border on or cross the axis of revolution, the solid has a hole in it. The cross sections perpendicular to the axis are washers. V = Outside Volume – Inside Volume

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. The region bounded by the curve y = x 2 +1 and the line y = -x + 3 is revolved about the x-axis to generate a solid. Find the volume of the solid of revolution.

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The inner and outer radii of the washer swept out by one slice. Outer radius R = - x + 3 and the inner radius r = x 2 +1

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Find the limits of integration by finding the x- coordinates of the points of intersection. x 2 + 1= - x + 3 x 2 + x –2=0 ( x+ 2 )(x – 1) = 0 x = -2 x = 1

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Outer radius R = - x + 3 and the inner radius r = x 2 +1 Calculation of volume

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The region bounded by the parabola y = x 2 and the line y = 2x in the first quadrant is revolved about the y-axis to generate a solid. Find the volume of the solid. Drawing indicates a dy integration so solve each equation for x as a function of y Set = to find y limits of integration y = 0 and y = 4 are limits y-axis rotation

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The washer swept out by one slice perpendicular to the y-axis.

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The region bounded by the parabola y = x 2 and the line y = 2x in the first quadrant is revolved about the y-axis to generate a solid. Find the volume of the solid. calculation

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Figure 5.17: Cutting the solid into thin cylindrical slices, working from the inside out. Each slice occurs at some x k between 0 and 3 and has thickness x. (Example 1) 6. 3 Cylindrical Shells Used to find volume of a solid of revolution by summing volumes of thin cylindrical shells or sleeves or tree rings.

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) Imagine cutting and unrolling a cylindrical shell to get a (nearly) flat rectangular solid. Its volume is approximately V = length height thickness. volume of a shell V shell =2 (radius)(height)(thickness)

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The region enclosed by the x-axis and the parabola y = f(x) = 3x – x 2 is revolved about the y – axis. Find the volume of the solid of revolution. V shell =2 (radius)(height)(thickness) problem

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The shell swept out by the kth rectangle. Notice this axis or revolution is parallel to the red rectangle drawn.

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The region bounded by the curve y = /x,, the x –axis and the line x = 4 is revolved about the y-axis to generate a solid. Find the volume of the solid. problem

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The region, shell dimensions, and interval of integration in

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The shell swept out by the rectangle in.

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Summary-Volumes-which method is best Axis of rotation x-axis shell parallel dx dy disk perpendicular dx dy y-axis

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Lengths of Plane curves Find the length of the arc formed by u = 1 + 4x du = 4dx du/4 = dx

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Follow the link to the slide. Then click on the figure to play the animation. A Figure Figure Figure 6.3.7

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Section 6.3 Figures 3, 4 Volumes by Cylindrical Shells

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Section 1 / Figure 1 Computer-generated picture of the solid in Example 9 A

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