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**Volumes by Slicing: Disks and Washers**

Objective: To find the volume of figures using the method of disks/washers.

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Volume Recall that the underlying principle for finding the area of a plane region is to divide the region into thin strips, approximate the area of each strip by the area of a rectangle, add the approximations to form a Riemann Sum, and take the limit of the Riemann Sums to produce an integral for the area.

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Volume Under appropriate conditions, the same strategy can be used to find the volume of a solid. The idea is to divide the solid into thin slabs, approximate the volume of each slab, add the approximations to form a Riemann Sum, and take the limit of the Riemann Sums to produce an integral for the volume.

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The Volume Problem Let S be a solid that extends along the x-axis and is bounded on the left and right, respectively, by the planes that are perpendicular to the x-axis at x = a and x = b. Find the volume V of the solid, assuming that its cross-sectional area A(x) is known at each x in the interval [a, b].

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The Volume Problem To solve this problem, we begin by dividing the interval [a, b] into n subintervals, thereby dividing the solid into n slabs. If we assume that the width of the kth subinterval is , then the volume of the kth slab can be approximated by the volume of a right cylinder of width (height) and cross sectional area , where is a point in the kth subinterval.

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The Volume Problem Adding these approximations yields the following Riemann Sum that approximates the volume V: Taking the limit as n increases and the widths of the subintervals approach zero yields the definite integral

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The Volume Formula 7.2.2 Let S be a solid bounded by two parallel planes perpendicular to the x-axis at x = a and x = b. If, for each x in [a, b], the cross-sectional area of S perpendicular to the x-axis is A(x), then the volume of the solid is provided A(x) is integrable.

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The Volume Formula 7.2.2 Let S be a solid bounded by two parallel planes perpendicular to the y-axis at y = c and y = d. If, for each y in [c, d], the cross-sectional area of S perpendicular to the y-axis is A(y), then the volume of the solid is provided A(y) is integrable.

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Solids of Revolution A solid of revolution is a solid that is generated by revolving a plane region about a line that lies in the same plane as the region; the line is called the axis of revolution. Many familiar solids are of this type.

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Volume Problem Let f be continuous and nonnegative on [a, b], and let R be the region that is bounded above by y = f(x), below by the x-axis, and on the sides by the lines x = a and x = b. Find the volume of the solid of revolution that is generated by revolving the region R about the x-axis.

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Volume Problem We can solve this problem by slicing. For this purpose, observe that the cross section of the solid taken perpendicular to the x-axis at the point x is a circular disk of radius f(x). The area of this region is The volume of the solid is

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Method of Disks Because the cross sections are disk shaped, the application of this formula is called the method of disks. The area of this region is The volume of the solid is

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Example 2 Find the volume of the solid that is obtained when the region under the curve over the interval [1, 4] is revolved about the x-axis.

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Example 2 Find the volume of the solid that is obtained when the region under the curve over the interval [1, 4] is revolved about the x-axis.

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Example 3 Derive the formula for the volume of a sphere of radius r.

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Example 3 Derive the formula for the volume of a sphere of radius r.

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Problem Let f and g be continuous and nonnegative on [a, b], and suppose that f(x) > g(x) for all x in the interval [a, b]. Let R be the region that is bounded above by y = f(x), below by y = g(x), and on the sides by the lines x = a and x = b. Find the volume of the solid of revolution that is generated by revolving the region R about the x-axis.

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Volumes By Washers We can solve this problem by slicing. For this purpose, observe that the cross section of the solid taken perpendicular to the x-axis at the point x is the annular or “washer-shaped” region with inner radius g(x) and outer radius f(x). Thus, The area is The volume is

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Example 4 Find the volume of the solid generated when the region between the graphs of the equation and over the interval [0, 2] is revolved about the x-axis.

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Example 4 Find the volume of the solid generated when the region between the graphs of the equation and over the interval [0, 2] is revolved about the x-axis.

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**Perpendicular to the y-axis**

The methods or disks and washers have analogs for regions that are revolved about the y-axis. Using the method of slicing, we should have no problem deducing the following formulas for the volumes of the solids.

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Disks/Washers These formulas are:

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Example 5 Find the volume of the solid generated when the region enclosed by , y = 2, and x = 0 is revolved about the y-axis.

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Example 5 Find the volume of the solid generated when the region enclosed by , y = 2, and x = 0 is revolved about the y-axis.

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Example 6 Find the volume of the solid whose base is the region bounded between the curve and the x-axis from x = 0 to x = 3 whose cross sections taken perpendicular to the x-axis are: a) squares b) semi-circles

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Example 6 Find the volume of the solid whose base is the region bounded between the curve and the x-axis from x = 0 to x = 3 whose cross sections taken perpendicular to the x-axis are: Squares Remember, we will use the method of slicing to find the area of a cross section and then integrate. The side of each square is the value of the function, so the area is the function squared.

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Example 6 Find the volume of the solid whose base is the region bounded between the curve and the x-axis from x = 0 to x = 3 whose cross sections taken perpendicular to the x-axis are: b) semi-circles We need to find the area of a semi-circle to use the method of slicing. The diameter of the circle is the value of the function, so the radius is half of that.

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Example 7 Now, we will try to rotate a figure around a line other than the x or y-axis. We will use the idea of the outer and inner radius to find the correct formula.

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Example 7 Find the volume of the figure generated by rotating the graphs of y = 2x and y = x2 from y = 0 to y = 4 around the line x = 2. r R

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Example 7 Find the volume of the figure generated by rotating the graphs of y = 2x and y = x2 from y = 0 to y = 4 around the line x = 2. The outer radius is The inner radius is The volume is r R

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Example 7 Find the volume of the figure generated by rotating the graphs of y = 2x and y = x2 from y = 0 to y = 4 around the line x = 2. The outer radius is The inner radius is The volume is r R

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Homework Section 6.2 1-25 odd 39

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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.

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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