# What is the volume of the solid generated when S is revolved about the line #y=3# where S is the region enclosed by the graphs of #y=2x# and #y=2x^2# and x is between [0,1]?

First let us determine where the two functions intersect

So the interval over which we will integrate is

I am going to use the method of washers to find the volume

Outer radius is

Inner radius is

Integral for volume is

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To find the volume of the solid generated when region S is revolved about the line y=3, you can use the disk method.

First, find the points of intersection between the two curves y=2x and y=2x^2: 2x = 2x^2 x = 0 and x = 1

Now, integrate the area of the disks formed by revolving the region S around the line y=3 from x=0 to x=1: [V = \int_{0}^{1} \pi [(\text{outer radius})^2 - (\text{inner radius})^2] dx]

Outer radius is the distance from the line of revolution (y=3) to the outer curve (y=2x), which is (3 - 2x). Inner radius is the distance from the line of revolution to the inner curve (y=2x^2), which is (3 - 2x^2).

[V = \pi \int_{0}^{1} [(3 - 2x)^2 - (3 - 2x^2)^2] dx]

Calculate the integral to find the volume.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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