Let R denote the region in the plane consisting of all points (x,y) where x ≥ 1 and 0 ≤ y ≤ 2/x. Let S denote the solid formed by rotating R about the x-axis. Observe that S is an unbounded region; that is, it extends indefinitely in the direction of the positive x-axis. Nevertheless, its volume is finite and can be expressed in terms of the improper integral
I just have to set up the integral from 1 to infinity but I don't understand how I can do it without a function
you have a function: y=2/x
volume = Integral pi*r^2 dx
where r = y = 2/x
pi*Int(4/x^2 dx) [1,oo]
now take it from there. piece of cake.
Not done is inevitable This post reoanstes with me once I see how something can be done, and prove to myself that I can do it, I too may often abandon it.I've a few thoughts, beyond the obvious focus and finish : enjoy your doodling, abandon earlier, and share on wikis. In detail:Firstly, I would consider this more kindly a proof-of-concept execution is like an artist's sketch, and while it may be a start of a finished work, it may also be a carefully worked doodle: there is certainly fun, and some value, in working something out even if nothing comes of it, as you note. So give yourself a break!Secondly, one issue is that your time is limited and that these half-finished projects are a serious drain. To reduce the time drain, abandon projects earlier! Simply, be satisfied or distracted earlier, without spending such time filling in details that will never be completed to your satisfaction. Cut your losses!Thirdly, may I encourage you to contribute more to wikis? Your (very nice) explanation of the Kelly criterion would make a wonderful (section of) a Wikibook (, say), and one of the reasons wikis work so well is because they combine different spurts of effort some-one else can pick up and build from your abandoned work. Help us build together!More fundamentally, you're noting a conflict between what you find emotionally fulfilling (your id: learning, figuring things out), and what you find rationally fulfilling (your superego: accomplishing, benefiting others). This emotional conflict, not poor task-management, is the root issue, and might be reconciled or managed.Lastly, might I counsel some Japanese philosophy, specifically the aesthetic of ? All that exists is imperfect, impermanent, and incomplete, and this shouldn't be a source of sadness rather one may feel a warm melancholy at the inevitable flaws and loose ends of one's works.Thanks again for your thoughtful and provocative post.
To set up the integral for calculating the volume of the solid formed by rotating the given region R about the x-axis, we can use the method of cylindrical shells.
First, visualize the region R in the coordinate plane. It consists of all points (x, y) satisfying the conditions: x ≥ 1 and 0 ≤ y ≤ 2/x. As x increases, the region extends indefinitely towards the positive x-axis but remains bounded in the y-direction.
To find the volume of the solid, we divide it into infinitesimally thin cylindrical shells aligned parallel to the y-axis. The height of each shell will be Δy, and its radius (distance from the y-axis) will be the corresponding value of x.
To determine the range of y-values across the region R, rearrange the inequality 0 ≤ y ≤ 2/x as x ≥ 2/y. This tells us that y can take any positive value up to a maximum y-value of 2, as x can be greater than or equal to 1. Therefore, the integral for the volume will be from y = 0 to y = 2. Since we are integrating with respect to y, we need to rewrite our integral limits in terms of y.
Next, consider a shell at a particular y-value. The corresponding x-value can be found by rearranging the inequality 0 ≤ y ≤ 2/x as x ≥ 2/y. So, the radius of the cylindrical shell, R(y), is 2/y. The height of the cylindrical shell, h(y), can be expressed as the difference in x-values at that y, which is x - 1. Therefore, the height is x - 1.
The volume of each cylindrical shell, dV, is given by the formula: dV = 2πR(y)h(y)dy. Substituting the values we found, we get dV = 2π(2/y)(x - 1)dy.
Now, we integrate this expression with respect to y over the limits of integration 0 to 2:
V = ∫[0 to 2] 2π(2/y)(x - 1)dy
To proceed further, we need to express x in terms of y. From the given inequality x ≥ 2/y, we can solve for x:
x ≥ 2/y
xy ≥ 2
y ≤ 2/x
Since we already have the upper and lower limits for y, we can replace the inequality in our integral:
V = ∫[0 to 2] 2π(2/y)(x - 1)dy = ∫[0 to 2] 2π(2/y)(y - 1)dy
Simplifying and factoring out constants:
V = 4π ∫[0 to 2] (y - 1)/y dy
Now we can evaluate this integral from 0 to 2 to find the volume of the solid formed by rotating R about the x-axis.
Hope this helps! Let me know if you have any further questions.