The velocity graph of a car accelerating from rest to a ...



[pic]

[pic]

[pic]

(a) y = x and y = sqrt(x) ( sqrt(x) – x = 0 ( sqrt(x) [1 – sqrt(x)] = 0 ( x = {0, 1}

When x = 0, y = 0 and when x = 1, y = 1

( 0 and 1 are the limits of integration.

When the portion enclosed between the curves y = sqrt(x) and y = x is revolved about the line x = 2, it forms a shape whose volume can be approximated as the sum of the volumes of cylinders with their radii lying ( the line x = 2.

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(b) The inner radius of the small disc shown in the figure above is r = 2 – x = 2 – y, and the outer radius is

R = 2 – x = 2 – y^2. The thickness of the disc is dy.

Volume = Integral over [0, 1] of [((R^2 – r^2)]dy

= ( [(2 – y^2)^2 – (2 – y)^2] dy over [0, 1]

= ( [4 – 4y^2 + y^4 – 4 + 4y – y^2] dy over [0, 1]

= ( [y^4 – 5y^2 + 4y] dy over [0, 1]

= ( [(1/5) y^5 – (5/3) y^3 + 2y^2] over [0, 1]

= ( [(1/5) – (5/3) + 2]

= 8/15.

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