Find the volume of the solid generated by the revolution of the curve y=2x when rotated about
x- axis from x=2 to x=4.
Answers
Answer:
The volume of the solid generated by [Math Processing Error], [Math Processing Error] revolved about the x-axis is [Math Processing Error].
Explanation:
Graph of the two curves
Graph of the two curves
Revolving the area between these two curves about the x-axis, we end up with something that looks sort of like a cone... a hollow cone, with a curved inside. Now, imagine for a second taking a cross section parallel to the y-z plane, cutting the cone down the middle. The cross-section is going to look like a washer, with an inner radius equal to the height of [Math Processing Error] wherever we are in the solid, and the outer radius equal to [Math Processing Error] at that same [Math Processing Error].
For reference, the area of a washer is equal to [Math Processing Error], if [Math Processing Error] is the outer radius and [Math Processing Error] is the inner radius. This fact should be immediately obvious - we're just subtracting the area of a circle from the area of another circle.
Now, to find the volume of the solid, we need to sum (integrate) the area of each of cross-sectional washer in the solid. This procedure is commonly called the method of washers .
So, we'll use the method of washers to find the volume of this solid.
The general formula is
[Math Processing Error]
where [Math Processing Error] is a function giving the outer radius of the washer at any x, and [Math Processing Error] is a function giving the inner radius of the washer. Note that the integrands here represent the areas of circles - and together, they give the area of each infinitesimal washer.
In our case, [Math Processing Error] is equal to [Math Processing Error] and [Math Processing Error] is equal to [Math Processing Error].
[Math Processing Error]
Next, we need to worry about the limits of the integrals. Since we're only concerned with rotating the region between [Math Processing Error] and [Math Processing Error], the limits should involve the locations where [Math Processing Error] and [Math Processing Error] intersect. Intersection occurs at [Math Processing Error] and [Math Processing Error].
[Math Processing Error]
The difficult part - setting up the appropriate integral with correct limits - is done, so I won't walk through evaluating the integral. It evaluates to [Math Processing Error], but if you don't trust me you can evaluate it yourself as an exercise.