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Volume of Revolution, Shell Method A flat sheet of plastic is of length L, width W and thickness dx What is the volume of this sheet? The volume dV of the sheet is: dV = L W dx L This sheet is rolled into a hollow cylinder of height L and radius r L r The width W of the sheet has now become the circumference of a circle of radius r r The circumference C of a circle of radius r is C = 2 r This is called a shell.W

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W The volume dV of the sheet is: dV = L W dx L Since W has become C, the volume dV of the shell can be written as: L r C = 2 r dV = L 2 r dx If L and r are functions of x then dV is: dV = 2 L(x)r(x) dx If L and r are functions of y then dV is: dV = 2 L(y)r(y) dy

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Consider the graph of a function f(x) L The area enclosed by this graph with the x-axis from x = a to x = b is shaded. x = ax = b f(x) We will use the shell method to find the volume generated when this area is revolved about the x-axis. Take a strip of area dA of width dx parallel to the y-axis Take note: For shell method, dA is taken parallel to the axis of revolution. What is the length of this area strip? Length of the strip = upper function – lower function. dA dx = f(x) – 0= f(x) L(x) = f(x) Revolution on the y-axis

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If this area is revolved about the y-axis, a shell is generated. L x = ax = b f(x) What is the radius of this shell? Radius of the shell is the distance of dA from the axis of revolution. The distance of dA from the y-axis is x The radius of the shell is: dA dx r(x) = x The length of the shell is:L(x) = f(x) The volume dV of the shell is: dV = 2 L(x)r(x) dx The volume V generated by revolving the area from x = a to x = b is:

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Revolution on the x-axis L f(y) In order to revolve this area about the x-axis, dA of width dy is taken parallel to the x-axis To find the length of dA, the functions must of y dy f(y) is on the right of dA and g(y) on its left. L(y) = f(y) – g(y) The length of dA is When dA is revolved about the x-axis, a shell is generated g(y) L(y) = f(y) – g(y) The radius r(y) of the shell is the distance of dA from the axis of revolution. The distance of dA from the x-axis is y y y r(y) = y dA

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L f(y) dA dy L(y) = f(y) – g(y) r(y) = y L(y) = f(y) – g(y) The volume dV of the shell generated is: y dV = 2 r(y)L(y) dy The volume V generated by revolving the area from y = c to y = d is: y = c y = d

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L To use shell method to obtain the volume of revolution, we use the following steps 1. Take dA parallel to the axis 2. Obtain the length of dA If dA is vertical, L(x) = g(x) – f(x) If dA is horizontal, L(y) = g(y) – f(y) 3. Obtain the radius of revolution Radius r(x) or r(y) is the distance of dA from the axis of revolution. 3. Use the appropriate formula to obtain V = Function above – function below = right function – left function

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Draw the graphs of y = x 2, y = 0, x = 4 and shade the area enclosed. y = 0 To revolve this area about the y-axis, take dA parallel to it. x = 4 dx The length of dA is:L(x) = x 2 – 0 Use the shell method to find the volume of revolution of the area enclosed by y = x 2, y = 0 and x = 4 about the y axis. dA y = x 2 = x 2 x2 x2 When dA is revolved about the y-axis, the radius of the shell generated is the distance of dA from y-axis. x r(x) = x L(x) = x 2 xdxThe volume dV of the shell is: dV = 2 r(x)L(x) dx = 2 x · x 2 dx = 2 x 3 dx

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y = 0 x = 4 dx dA y = x 2 x 2 x L(x) = x 2 xdx dV= 2 x 3 dx The position of dA can change from x = 0 to x = 4 The volume V generated by revolving the area enclosed about the y-axis is:

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y = 0 x = 4 dy Use the shell method to find the area of revolution of the area enclosed by y = x 2, y = 0 and x = 4 about the x axis. dA y = x 2 y Take dA of width dy parallel to the x-axis. Express the functions to the left and right of dA as functions of y The function on the left is:y = x 2 Solving for x gives: The function on the right is:x = 4 g(y) = 4 The length of dA is: L(y) = g(y) – f(y) The radius of revolution is the distance of dA from the x-axis r(y) = y

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y = 0 x = 4 dy dA y = x 2 y To find the limits, we find the points of intersection of x = y and x = 4 y = 4 gives: y = 16 The position of dA varies from y = 0 to y = 16 The volume of revolution is: y = 16 r(y) = y

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y = 0 x = 4 dy dA y = x 2 y y = 16

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x = 0 Use the shell method to find the volume of revolution of the area enclosed by y = 2x, x = 0 and y = 4 about the y axis. Take dA of width dx parallel to the y-axis dx The length of dA is: y = 4 Draw the graphs of y = 2x, x = 0, y = 4 and shade the area enclosed. y = 2x L(x) = 4 – 2x4 – 2x r(x) = x x The volume of revolution V is: Obtain limits by solving 2x = 4x = 2

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x = 0 dx y = 4 y = 2x 4 – 2x x x = 2

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Find the volume generated by revolving the area enclosed by y = x 2 and y = 4x – x 2 about the line x = 2 Draw the graphs of y = x 2, y = 4x – x 2 and x = 2 and shade the area enclosed. y = x 2 y = 4x – x 2 x = 2 Take dA of width dx parallel to x = 2 Find the points of intersection of y = x 2 and y = 4x – x 2. 4x – x 2 = x 2 4x – x 2 – x 2 = 0 2x(2 – x) = 0x = 0, 2 x = 0x = 2 dx The length of dA is: L(x) = 4x – x 2 – x 2 The distance of dA from x = 2 is: x 2 2 – x r(x) = 2 – x = 4x –2x 2

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y = x 2 y = 4x – x 2 x = 2 x = 0x = 2 dx L(x) = 4x – 2x 2 The volume of revolution V is: x 2 2 – x r(x) = 2 – x

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y = x 2 y = 4x – x 2 x = 2 x = 0x = 2 dx x 2 2 – x

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Use the shell method to find the volume of the solid generated by revolving the region bounded by y = 2 – x, y = 0 and x = 4 about the x-axis. y = 2 – x Draw the graphs of y = 2 – x, y = 0 and x = 4 and shade the area enclosed. y = 0 x = 4 Draw dA of width dy parallel to the x-axis. dA dy In order to find the length of dA, we need to write the functions on the left and right of dA as functions of y y = 2 – xSolving for x gives: x = 2 – y L(y) = 4 – (2 – y) 4 dA is below the x-axis. The distance of dA from the x-axis is:– y r(y) = – y L(y) = 2 + y

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y = 2x y = 0 x = 4 dA dy 2 – y 4 – y r(y) = – yL(y) = 2 + y The volume of revolution V is: To find the limits, solve 2 – y = 4 y = -2

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y = 2x y = 0 x = 4 dA dy 2 – y 4 – y y = -2

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