Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
x = 4(square root of (3y)), x = 0, y = 3;
about the y-axis
V=?????
x=4√(3y)
x^2 = 48y
To use discs, note that the radius of each disc is x, and the thickness is dy, so
v = ∫[0,3] πx^2 dy
= 48π∫[0,3] y dy
= 216π
To use shells, each shell has radius x, and height 3-y, so
v = ∫[0,12] 2πx(3-y) dx
= 2π∫[0,12] x(3-x^2/48) dx
= 216π
Well, well, well, looks like we've got ourselves a lovely math problem here. Now, since we're talking about rotating stuff, I can't help but picture a twirling circus act. Imagine acrobats spinning around the y-axis! Alright, let's crunch some numbers and find the volume of our spinning extravaganza.
To find the volume V, we'll be using the method of cylindrical shells. It might sound complicated, but don't worry, I'll guide you through it!
First, let's sketch the region bounded by the curves. We've got the parabola x = 4√(3y), which starts at the origin (0,0) and goes up to the point (12,3). Then we have the y-axis as our axis of rotation. Picture it with me!
Now, let's pick a random vertical strip (or "shell") with width Δy. Its height will vary based on which y-value we choose. Alright, take a deep breath, here we go!
The circumference of each shell will be given by 2π(radius), which is simply 2πx. Since x = 4√(3y), we can rewrite it as 2π(4√(3y)).
Now, to find the volume of each shell, we multiply its circumference by its height, which is Δy. So, the volume of a single shell is 2π(4√(3y))Δy.
To find the total volume V, we sum up all these shell volumes from y = 0 to y = 3. We express it as an integral:
V = ∫[0,3] 2π(4√(3y))Δy
Now, let's take the integral of that expression and evaluate it from 0 to 3. Drum roll, please!
V = 2π ∫[0,3] 4√(3y) dy
Evaluating this integral will give us the final answer for the volume V of our rotating region. Time to put on your math hat and crunch those numbers, my friend!
To find the volume V of the solid obtained by rotating the region bounded by the given curves about the y-axis, we can use the method of cylindrical shells.
First, let's sketch the curves x = 4√(3y), x = 0, and y = 3 to better visualize the region.
The intersection points between x = 4√(3y) and y = 3 can be found by setting the equations equal to each other:
4√(3y) = 3
Square both sides:
48y = 9
y = 9/48
y = 3/16
So the region is bounded by x = 4√(3y), x = 0, and y = 3/16.
To find the volume, we integrate the circumference of each cylindrical shell multiplied by its height. The height of each shell is dx, and the radius is given by x.
The region's y-range is from y = 0 to y = 3/16, but it's easier to work with x for integration purposes. So let's express the bounds of integration in terms of x.
To find the x-range, we need to solve for y in terms of x in the equation x = 4√(3y):
x = 4√(3y)
x/4 = √(3y)
(x/4)^2 = 3y
y = (x/4)^2 / 3
Now we can set up the integral:
V = ∫[0, 3/16] 2πx * (dx)
= 2π ∫[0, 3/16] x * (dx)
Evaluating the integral of x * (dx), we get:
V = 2π [x^2/2] [0, 3/16]
= π * (3/16)^2/2
= π * 9/256
= 9π/256
Therefore, the volume V of the solid obtained by rotating the region about the y-axis is 9π/256 cubic units.
To find the volume V of the solid obtained by rotating the region bounded by the curves about the y-axis, we can use the method of cylindrical shells.
First, let's plot the curves x = 4√(3y), x = 0, and y = 3 to visualize the region:
y
^
|
+----|------------+
| | |
| | |
| | |
| | |
x <-------|----|------------|
| | |
+----|------------+
|
|----------------> x
The region bounded by the curves lies between y = 0 and y = 3, and the rotation is around the y-axis.
To calculate the volume using cylindrical shells, we integrate with respect to y. The volume of each cylindrical shell is given by the product of the height of the shell (which is the difference between the corresponding x-coordinates of the curves) and the circumference of the shell (which is the distance around the y-axis at that y-value).
The height of the shell can be found by subtracting the x-coordinate of the curve x = 0 (which is 0) from the x-coordinate of the curve x = 4√(3y). So, the height of the shell is 4√(3y).
The circumference of the shell is given by 2πr, where r is the distance from the y-axis to the curve x = 4√(3y). The distance from the y-axis to the curve x = 4√(3y) is simply the x-coordinate, which is 4√(3y).
Therefore, the volume of an individual shell is given by dV = 2π(4√(3y))(4√(3y))dy.
To find the total volume V, we integrate this expression over the interval from y = 0 to y = 3:
V = ∫[0,3] 2π(4√(3y))(4√(3y))dy
Simplifying this expression, we have:
V = 2π∫[0,3] 48y dy
V = 96π∫[0,3] y dy
V = 96π[(1/2)y^2] from 0 to 3
V = 96π(1/2)(3^2 - 0^2)
V = 96π(9/2)
V = 432π
Therefore, the volume of the solid obtained by rotating the region about the y-axis is V = 432π.