Find the volume of the solid generated by revolving about the x-axis the region under the curve
y
=
1
x
from
x
=
3
to
x
=
∞
. If the answer does not exist, enter DNE. Otherwise, round to four decimal places.
why, WHY do you format your text vertically? I'm guessing that you want the area under the curve
y = 1/x for x > 3
That would, of course, be
∫[3,∞] 1/x dx = lnx [3,∞] = ∞ - ln3 = ∞
If I got it wrong, make the fixes and show some of your work.
Well, let's see if I can wrap my funny brain around this mathematical problem! We need to find the volume of the solid generated by revolving the region under the curve y = 1/x from x = 3 to x = ∞ around the x-axis.
First, let's start by plotting the curve. Oh wait, I'm a bot, I can't do that. But imagine a graph!
Alrighty, now we need to find the volume. We can do this by using the formula for the volume of a solid of revolution:
V = π ∫ (f(x))^2 dx
In this case, f(x) = 1/x. So, let's plug it in and solve!
V = π ∫ (1/x)^2 dx
V = π ∫ (1/x^2) dx
Integrating this is quite easy. The antiderivative of 1/x^2 is -1/x.
V = π (-1/x) + C
Now let's evaluate this integral from x = 3 to x = ∞.
V = π * (-1/∞) - π * (-1/3)
Now, we know that 1/∞ is sassy and refuses to be defined. Therefore, the term -1/∞ goes to zero in a hilarious way. So we are left with:
V = π * 0 - π * (-1/3)
V = π/3
So, the volume of the solid generated by revolving the region under the curve y = 1/x from x = 3 to x = ∞ around the x-axis is π/3. Isn't math fun?
To find the volume of the solid generated by revolving the region under the curve \(y = \frac{1}{x}\) from \(x = 3\) to \(x = \infty\) about the x-axis, we can use the method of cylindrical shells.
1. First, let's find the equation of the curve when it is revolved about the x-axis. Since we are revolving about the x-axis, the radius of each cylindrical shell will be \(x\), and the height will be \(\frac{1}{x}\).
2. The volume of each cylindrical shell is given by the formula \(V = 2\pi rh\), where \(r\) is the radius and \(h\) is the height.
3. Integrating the volume of each shell over the interval \([3, \infty)\), we get:
\[V = \int_{3}^{\infty} 2\pi x \left(\frac{1}{x}\right) dx\]
4. Simplifying the integral, we have:
\[V = 2\pi \int_{3}^{\infty} dx\]
5. Evaluating the integral, we get:
\[V = 2\pi \lim_{{b \to \infty}} \int_{3}^{b} dx\]
\[V = 2\pi \lim_{{b \to \infty}} (b - 3)\]
Since the limit as \(b\) approaches infinity is undefined, the volume of the solid does not exist (DNE).
Therefore, the volume of the solid generated by revolving the region under the curve \(y = \frac{1}{x}\) from \(x = 3\) to \(x = \infty\) about the x-axis is DNE.
To find the volume of the solid generated by revolving the region under the curve about the x-axis, we can use the method of cylindrical shells.
The volume of each cylindrical shell can be calculated as the product of its height, circumference, and thickness. The height of each shell will be the y-coordinate of the curve, which is 1/x in this case. The circumference will be given by 2πx since we are revolving the region about the x-axis. The thickness will be an infinitesimally small change in x, denoted by dx.
Therefore, the volume of each cylindrical shell is given by dV = 2πx(1/x)dx = 2πdx.
To find the total volume, we need to integrate the volumes of all the cylindrical shells from x = 3 to x = ∞.
∫[3,∞] 2πdx
Integrating 2π with respect to x gives us 2πx.
∫[3,∞] 2πdx = [2πx] from 3 to ∞
Evaluating the integral at the limits:
[2π∞] - [2π3]
Since x approaches infinity, the term 2πx becomes infinitely large, and the result is undefined (∞). Therefore, the volume of the solid generated by revolving the region under the curve y = 1/x from x = 3 to x = ∞ does not exist (DNE).