Find the volume of the solid obtained by rotating the region bounded by the curves y=cos(x), y=0, x=0, and x=π/2 about the line y=1.

radius of cylinder of skin thickness dr (r goes from 0 to 1)

r = 1-cos x
dr = sin x dx
so circumference = 2 π (1-cos x)
height of that cylinderical shell = x
dV = 2 π(1-cos x) x dr
= 2 π (1-cos x)x sin x dx
= 2 π x sin x dx - 2 π x sin x cos x dx

= 2 π [ sin x-x cos x]
- 2 π/8 [ sin 2 x - 2 x cos 2x]
(I used wolfram alpha integral - google it)
evaluate at x = pi/2
= 2 π [ 1]
- 2 π/8 [ 0 + π ]

= 2 π - π^2 /4
now
evaluate at x = 0
= 2 π[ 0 ]
- π/4 [ 0 ] handy
so
2 π - π^2 /4

works for me. I did the integral directly, using

∫[0,1] 2π(1-y) arccos(y) dy

and got the same answer.

I like your technique of avoiding the arccos, though.

Well, well, well...look at you trying to rotate things around like a Cirque du Soleil performer! So, you want to find the volume of the solid obtained by rotating this fancy region about the line y=1? Alright, let's get our clown calculators ready!

To tackle this problem, we'll be using the method of cylindrical shells. We'll be integrating from 0 to π/2, so get your mathematical helmets on!

The radius of each shell can be found by subtracting the function y=1 from y=cos(x), giving us r = cos(x) - 1. And since we're rotating around the line y=1, the height of each shell will just be 1.

So, the volume of each shell will be 2π(r)(h), which simplifies to 2π(cos(x) - 1)(1). Now we integrate this expression from 0 to π/2.

∫[0,π/2] 2π(cos(x) - 1)dx

Time to put on our math wizard hats! Integrating this expression will give us:

2π(sin(x) - x) ∣[0,π/2]

Plug in the upper bound and lower bound, and let's see what kind of mathematical magic we get:

2π(sin(π/2) - π/2) - 2π(sin(0) - 0)

2π(1 - π/2) - 2π(0)

2π - π² - 0

The final result is 2π - π².

So, the volume of the solid obtained by rotating this region about the line y=1 is 2π - π² cubic units. Ta-da!

Now, wasn't that calculation a barrel of laughs? Hope I managed to entertain you while solving that math problem! If you have any more questions or need more laughs, I'm all ears!

To find the volume of the solid obtained by rotating the region bounded by the curves y = cos(x), y = 0, x = 0, and x = π/2 about the line y = 1, we can use the method of cylindrical shells.

First, let's visualize the region bounded by the curves. The region is a portion of the graph of y = cos(x) enclosed by the x-axis and the vertical line at x = π/2.

Next, let's consider a cylindrical shell within this region. The height of the cylindrical shell will be equal to the difference between the y-coordinate of the upper curve (y = cos(x)) and the y-coordinate of the lower curve (y = 0). So the height of the shell will be cos(x) - 0 = cos(x).

The radius of the cylindrical shell will be the distance from the axis of rotation (y = 1) to the curve y = cos(x). Since the axis of rotation is above the curve, the radius will be equal to 1 - cos(x).

To set up the integral that will give us the volume, we need to integrate the circumference of the cylindrical shell multiplied by its height over the interval [0, π/2] (the limits of integration are determined by the x-values of the region).

The circumference of the cylindrical shell can be calculated using the formula 2πr, where r is the radius of the shell. In this case, the circumference is 2π(1 - cos(x)).

So, the integral that will give us the volume is:

V = ∫[0,π/2] 2π(1 - cos(x))(cos(x)) dx

To evaluate this integral, we can use integration techniques such as integration by parts or trigonometric substitution.

lots of algebra here

here is the integral site:
http://www.wolframalpha.com/calculators/integral-calculator/