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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? L The volume dV of the sheet is: L dV = L W dx This sheet is rolled into a hollow cylinder of height L and radius r W This is called a shell. The width W of the sheet has now become the circumference of a circle of radius r The circumference C of a circle of radius r is r C = 2r

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**The volume dV of the sheet is:**

r dV = L W dx C = 2r Since W has become C, the volume dV of the shell can be written as: L L dV = L 2r dx W 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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**Revolution on the y-axis**

Consider the graph of a function f(x) The area enclosed by this graph with the x-axis from x = a to x = b is shaded. f(x) We will use the shell method to find the volume generated when this area is revolved about the x-axis. L dA L(x) = f(x) dx Take a strip of area dA of width dx parallel to the y-axis x = a x = b 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. = f(x) – 0 = f(x) L(x) = f(x)

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**If this area is revolved about the y-axis, a shell is generated.**

What is the radius of this shell? Radius of the shell is the distance of dA from the axis of revolution. f(x) L dA The distance of dA from the y-axis is x L(x) = f(x) The radius of the shell is: r(x) = x dx x = a x = b 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**

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

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**The volume dV of the shell generated is: f(y)**

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

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**1. Take dA parallel to the axis**

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 L If dA is vertical, L(x) = g(x) – f(x) = Function above – function below If dA is horizontal, L(y) = g(y) – f(y) = right function – left function 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

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**Draw the graphs of y = x2, y = 0, x = 4 and shade the area enclosed.**

Use the shell method to find the volume of revolution of the area enclosed by y = x2, y = 0 and x = 4 about the y axis. Draw the graphs of y = x2, y = 0, x = 4 and shade the area enclosed. y = x2 To revolve this area about the y-axis, take dA parallel to it. The length of dA is: L(x) = x2 – 0 = x2 x = 4 When dA is revolved about the y-axis, the radius of the shell generated is the distance of dA from y-axis. dA x2 y = 0 dx r(x) = x x The volume dV of the shell is: dx x dV = 2 r(x)L(x) dx L(x) = x2 = 2 x · x2 dx = 2 x3 dx

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**The position of dA can change from x = 0 to x = 4**

dV= 2 x3 dx The position of dA can change from x = 0 to x = 4 y = x2 The volume V generated by revolving the area enclosed about the y-axis is: x = 4 dA x2 y = 0 dx x dx x L(x) = x2

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**Take dA of width dy parallel to the x-axis.**

Use the shell method to find the area of revolution of the area enclosed by y = x2, y = 0 and x = 4 about the x axis. Take dA of width dy parallel to the x-axis. Express the functions to the left and right of dA as functions of y y = x2 The function on the left is: y = x2 x = 4 Solving for x gives: dy dA y The function on the right is: x = 4 y = 0 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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**The position of dA varies from y = 0 to y = 16**

r(y) = y To find the limits, we find the points of intersection of x = y and x = 4 y = 16 y = x2 y = 4 gives: y = 16 The position of dA varies from y = 0 to y = 16 x = 4 dy The volume of revolution is: dA y y = 0

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y = 16 y = x2 x = 4 dy dA y y = 0

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

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x = 0 y = 2x y = 4 dx 4 – 2x x x = 2

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Find the volume generated by revolving the area enclosed by y = x2 and y = 4x – x2 about the line x = 2 x = 2 Draw the graphs of y = x2, y = 4x – x2 and x = 2 and shade the area enclosed. 2 y = x2 Find the points of intersection of y = x2 and y = 4x – x2. x 4x – x2 = x2 y = 4x – x2 2 – x 4x – x2 – x2 = 0 2x(2 – x) = 0 x = 0, 2 dx Take dA of width dx parallel to x = 2 x = 0 x = 2 The length of dA is: L(x) = 4x – x2 – x2 = 4x –2x2 The distance of dA from x = 2 is: 2 – x r(x) = 2 – x

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L(x) = 4x – 2x2 r(x) = 2 – x x = 2 The volume of revolution V is: 2 y = x2 x y = 4x – x2 2 – x dx x = 0 x = 2

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x = 2 2 y = x2 x y = 4x – x2 2 – x dx x = 0 x = 2

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**Draw dA of width dy parallel to the x-axis. x = 2 – y**

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 – y dA dy Draw dA of width dy parallel to the x-axis. x = 2 – y 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 4 x = 4 y = 2 – x Solving for x gives: x = 2 – y L(y) = 4 – (2 – y) L(y) = 2 + y dA is below the x-axis. The distance of dA from the x-axis is: – y r(y) = – y

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**To find the limits, solve 2 – y = 4**

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

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

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4/30/2015 Perkins AP Calculus AB Day 4 Section 7.2.

4/30/2015 Perkins AP Calculus AB Day 4 Section 7.2.

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