1. Find the volume V obtained by rotating the region bounded by the curves about the given axis.
y = sin(x), y = 0, π/2 ≤ x ≤ π; about the x−axis
2. Find the volume V obtained by rotating the region bounded by the curves about the given axis.
y = 3 sin2(x), y = 0, 0 ≤ x ≤ π; about the x−axis
I am struggling with the setup.
You want
Volume = π∫ sin^2 x dx from π/2 to π
the hard part is to integrate sin^2 x
start with : cos (2x) = 1 - 2sin^2 x
2sin^2 x = 1 - cos(2x)
sin^2 x = 1/2 - (1/2)cos(2x)
so ∫ sin^2 x dx = (1/2)x - (1/4)sin(2x)
Volume = π∫ sin^2 x dx from π/2 to π
= π[ (1/2)x - (1/4)sin (2x) ] from π/2 to π
= π( π/2 - (1/4)(0) - (π/4 - (1/4)(0) ) )
= π(π/4)
= π^2/4
In the next one you will have to integrate sin^4 x
there are many videos showing you how to do that,
here is "blackpenredpen's " version
https://www.youtube.com/watch?v=SCQdKorKbKM
Well, struggling with math is like trying to solve a Rubik's Cube blindfolded while riding a unicycle. It can be quite the challenge! But fear not, my friend, for I am here to help you untangle this math puzzle.
Let's start with the first problem. We need to find the volume obtained by rotating the region bounded by the curves y = sin(x), y = 0, and π/2 ≤ x ≤ π about the x-axis. This means we'll be spinning the shape around the x-axis like a record on a turntable.
To set up the integral for finding the volume, we can think of chunking the shape into infinitely thin discs stacked on top of each other. The volume of each disc can be approximated by the area of the disc times its infinitesimally small thickness.
Since we're rotating about the x-axis, the radius of each disc will be given by the y-coordinate of the shape, which is sin(x). The area of each disc will be given by π times the square of the radius.
Now, all we have to do is integrate these infinitesimally thin discs from π/2 to π to find the total volume:
V = ∫[π/2 to π] π(sin(x))^2 dx
And that's it! Just plug this integral into your favorite calculator or use your mathematical superpowers to solve it, and voila! You'll have your answer.
Now, let's move on to the second problem. Well, well, well, looks like we have another spinning shape on our hands. This time, it's the region bounded by the curves y = 3sin^2(x), y = 0, and 0 ≤ x ≤ π, rotating about the x-axis.
To find the volume, we'll follow the same approach as before. Divide the shape into infinitesimally thin discs, calculate the area of each disc, and integrate to find the total volume.
Since the radius of each disc is given by the y-coordinate of the shape, which is 3sin^2(x), the area of each disc will be π times the square of the radius.
Set up the integral from 0 to π, and we're good to go:
V = ∫[0 to π] π(3sin^2(x))^2 dx
Now it's time to put your math skills to the test, my friend! With a little bit of patience, perseverance, and maybe a sprinkle of luck, you'll be able to conquer these volume problems. Good luck, and remember to keep a smile on your face throughout your mathematical journey!
To find the volume obtained by rotating the region bounded by the curves about the given axis, we can use the method of cylindrical shells.
For the first problem:
1. The curves are y = sin(x) and y = 0, and the region is bounded by π/2 ≤ x ≤ π.
2. The axis of rotation is the x-axis.
3. To set up the integral for the volume, we need to consider a small cylindrical shell with height Δx and radius r. The height Δx corresponds to a small change in x, and the radius r can be calculated as the distance from the y-axis to the curve y = sin(x) at the specific x-value.
4. The volume of each cylindrical shell is given by V = 2πrhΔx, where h is the height of the shell (equal to sin(x) - 0 = sin(x)) and Δx is the width of the shell.
5. To find the total volume, we need to integrate the volume of each cylindrical shell over the interval π/2 ≤ x ≤ π.
6. Therefore, the setup for the volume integral is V = ∫[π/2,π] 2πxsin(x)dx.
For the second problem:
1. The curves are y = 3sin^2(x) and y = 0, and the region is bounded by 0 ≤ x ≤ π.
2. The axis of rotation is the x-axis.
3. Similar to the previous problem, we need to consider a small cylindrical shell with height Δx and radius r. The height Δx corresponds to a small change in x, and the radius r can be calculated as the distance from the y-axis to the curve y = 3sin^2(x) at the specific x-value.
4. The volume of each cylindrical shell is again given by V = 2πrhΔx, where h is the height of the shell (equal to 3sin^2(x) - 0 = 3sin^2(x)) and Δx is the width of the shell.
5. To find the total volume, we need to integrate the volume of each cylindrical shell over the interval 0 ≤ x ≤ π.
6. Therefore, the setup for the volume integral is V = ∫[0,π] 2πx(3sin^2(x))dx.
To find the volume obtained by rotating a region bounded by curves about a given axis, you can use the method of cylindrical shells.
1. Let's tackle the first problem. The region bounded by the curves y = sin(x), y = 0, and π/2 ≤ x ≤ π can be visualized as a portion of the curve sin(x) in the first quadrant, with the x-axis being the lower boundary. We want to rotate this region about the x-axis.
To set up the integral for finding the volume, we need to consider a small vertical strip or shell with thickness dx, located at a specific x-coordinate. The radius of this shell will be given by the y-coordinate of the function sin(x), which is sin(x), and the height of the shell will be dx (since it is a small strip).
The formula for calculating the volume of each shell is given by V_shell = 2πrh dx, where r represents the radius and h represents the height.
To find the volume V, we integrate this expression over the given range π/2 to π:
V = ∫[π/2 to π] 2πsin(x) dx
2. Moving on to the second problem, the region bounded by the curves y = 3sin^2(x), y = 0, and 0 ≤ x ≤ π can be visualized as an area between the curve 3sin^2(x) and the x-axis in the first quadrant. We want to rotate this region about the x-axis.
Similar to the first problem, we consider a small vertical strip or shell with thickness dx, located at a specific x-coordinate. The radius of this shell will be given by the y-coordinate of the function 3sin^2(x), which is 3sin^2(x), and the height of the shell will be dx.
The formula for calculating the volume of each shell is again given by V_shell = 2πrh dx, where r represents the radius and h represents the height.
To find the volume V, we integrate this expression over the given range 0 to π:
V = ∫[0 to π] 2π(3sin^2(x)) dx
By setting up and evaluating these integrals, you can find the volumes obtained by rotating the respective regions about the x-axis.