Consider the solid obtained by rotating the region bounded by the given curves about the line y = 2.

Y = 2+secx -pi/3<= x <= pi/3 , y = 4
Find the volume

Well, well, well, we're talking about rotating curves today! Let's give it a spin, shall we?

So we're given the curves y = 2 + sec(x) and y = 4, and we're rotating the region bounded by these curves about the line y = 2. That's quite a twirl!

To find the volume of this solid, we can use the method of cylindrical shells. Don't worry, there won't be any actual shells involved. It's just a mathematician's way of calculating volumes.

The first step is to find the height of each cylindrical shell. In this case, the height is determined by the difference between y = 4 and y = 2 + sec(x). That's like the distance between the top and bottom of the shell.

Now, we need to determine the radius of each shell. Since we're rotating about the line y = 2, the radius is simply the distance between the line y = 2 and the curve y = 2 + sec(x).

Alright, now let's set up our integral. We'll integrate from x = -pi/3 to x = pi/3 because those are the boundaries of our region.

The integral that will give us the volume is:

V = ∫(2π(radius)(height))dx
= ∫(2π(2 + sec(x) - 2))(dx)
= 2π∫sec(x)dx
= 2πln|sec(x) + tan(x)| + C

Where C is the constant of integration.

And there you have it, the volume of the solid obtained by rotating the region bounded by the curves y = 2 + sec(x) and y = 4 about the line y = 2. It's 2πln|sec(x) + tan(x)| + C. Enjoy your rotating curves, my friend!

To find the volume of the solid obtained by rotating the region bounded by the curves about the line y = 2, we will use the method of cylindrical shells.

First, let's sketch the region bounded by the curves y = 2 + sec(x) and y = 4. This region lies between the x-values -pi/3 and pi/3. The curve y = 2 + sec(x) is an interesting one, as it oscillates between positive and negative values. However, when we rotate this region about the line y = 2, we only need to consider the positive part of the curve, as the negative part will be canceled out.

Next, we will consider a vertical strip within this region. Let's call the width of this strip "dx". The height of the strip will be the difference between the y-coordinates of the top and bottom curves, which is (4 - (2 + sec(x)) = 2 - sec(x).

Now, we will consider a cylindrical shell formed by rotating this strip about the line y = 2. The radius of this shell is the distance between the line y = 2 and the curve y = 2 + sec(x). Since the line y = 2 is at a constant height, the radius of the shell is simply the height of the strip, which is 2 - sec(x).

The volume of this cylindrical shell can be calculated by multiplying its circumference by its height and its thickness. The circumference of the shell is given by 2π(radius) = 2π(2 - sec(x)). The height and thickness of the shell are both dx.

Therefore, the volume of this cylindrical shell is 2π(2 - sec(x)) dx.

To find the total volume, we integrate this expression over the interval -pi/3 to pi/3, which represents the bounds of the region.

Integral from -pi/3 to pi/3 of 2π(2 - sec(x)) dx = 2π * Integral from -pi/3 to pi/3 of (2 - sec(x)) dx

To evaluate this integral, we can use the antiderivative of sec(x), which is ln|sec(x) + tan(x)|. Integrating 2 - sec(x) gives us 2x - ln|sec(x) + tan(x)|.

Now, we can substitute the bounds of the region (-pi/3 and pi/3) into this expression:

2π * [(2(pi/3) - ln|sec(pi/3) + tan(pi/3)|) - (2(-pi/3) - ln|sec(-pi/3) + tan(-pi/3)|)]

Simplifying further:

2π * [(2pi/3 - ln(2 + √3)) - (-2pi/3 - ln(2 - √3))]

Finally:

2π * (4pi/3 + ln(2 - √3) - ln(2 + √3))

This is the volume of the solid obtained by rotating the region bounded by the given curves about the line y = 2.