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Consider the solid obtained by rotating the region bounded by the curves about the x-axis. Find the volume (V) of this solid using appropriate mathematical methods.

User Munjal
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Final answer:

To find the volume of a solid of revolution, one would typically use the disk or washer methods with the disk method formula V = πr²h. Dimensional analysis helps to confirm the consistency of a formula like V = πr²h for a cylinder's volume or V = 4/3πr³ for a sphere's volume.

Step-by-step explanation:

To calculate the volume of a solid of revolution, we can use the disk method or the washer method, depending on the shape of the region being rotated. The disk method involves taking a cross-sectional area perpendicular to the axis of rotation and integrating this area along the axis. In particular, if we are rotating about the x-axis, the formula for the cross-sectional area, A, is A = πr2, where r is the radius of the disk, which corresponds to the y-value of the function at a specific x.

The general formula for the volume using the disk method is V = ∫abA dx, where [a, b] is the interval on the x-axis over which the region is being rotated.

Consider a cylinder example: the volume V of a cylinder (which is a type of solid of revolution) is the area of the circular base A times the height h, V = Ah. If we are given the radius of the base r and the height h of the cylinder, we apply the formula V = πr2h to find the volume.

Dimensional analysis confirms whether our formulas make sense. For instance, the formula for the volume of a cylinder, V = πr2h, passes dimensional analysis since both sides of the equation represent a volume. The unit for volume is length cubed (L3), and since the radius and height are both lengths, the formula is dimensionally consistent.

To illustrate the application of these principles, consider finding the volume of a sphere. Referring back to the check your understanding section, we can determine using dimensional analysis that the volume of a sphere of radius r is given by the formula V = ⅔πr3, as opposed to its surface area, which is 4πr2.

User Dgor
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