Improper Integral Volume Around x-axis

In summary: Note that your integrand is f(x) = (1/(.6x + 1.7))^2. So your antiderivative should be -1.5ln(0.6x + 1.7). Check your work again.
  • #1
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Homework Statement



By rotating R=[tex]$\{ (x,y)|x\geq0, 0\leq y\leq \frac{1}{0.6 x+1.7}\}$[/tex] about the x-axis we obtain a solid with the volume V = ______

Homework Equations


The Attempt at a Solution


[tex]$\int _0^{\infty }\frac{dx}{0.6 x+1.7}$[/tex] is divergent
but what do i do to get the volume? if don't have a and b..
 
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  • #2
Break the integral up into two integrals. You can split them anywhere; I'm choosing 1.
[tex]\int_a^1 \frac{dx}{0.6x + 1.7} + \int_1^b \frac{dx}{0.6x + 1.7}[/tex]

Both are improper integrals. Take the limit of the first as a --> 0 and the second as b --> infinity.
 
  • #3
Slimsta said:

Homework Statement



By rotating R=[tex]$\{ (x,y)|x\geq0, 0\leq y\leq \frac{1}{0.6 x+1.7}\}$[/tex] about the x-axis we obtain a solid with the volume V = ______

Homework Equations


The Attempt at a Solution


[tex]$\int _0^{\infty }\frac{dx}{0.6 x+1.7}$[/tex] is divergent
but what do i do to get the volume? if don't have a and b..

Is your integral correct? It appears your integral computes the area of R and not the volume of rotating it about the x-axis. Look up Disc Method in your calculus book for a hint on properly setting up the integral.
 
  • #4
rs1n, Good point. Slimsta doesn't have the integrand set up correctly to get the volume.
 
  • #5
Mark44 said:
rs1n, Good point. Slimsta doesn't have the integrand set up correctly to get the volume.

well the question is give like that..
[tex]$f(x)=\frac{1}{0.6 x+1.7}$[/tex]
[tex]$\int _0^t\frac{dx}{0.6 x+1.7}=\left. \frac{\ln (0.6 x+1.7 )}{0.6 }\right| _0^t=$ [/tex] [ln(0.6 t+1.7 )]/0.6 - [ln(1.7 )]/0.6

and taking the limit as t[tex]$\to \infty$[/tex] we conclude that
[tex]$\int _0^{\infty }\frac{dx}{0.6 x+1.7}$[/tex] is divergent .

Therefore the region R[tex]$=\{ (x,y)|x\geq 0, 0\leq y\leq \frac{1}{0.6 x+1.7}\}$[/tex] has infinite area.

By rotating R[tex]$=\{ (x,y)|x\geq0, 0\leq y\leq \frac{1}{0.6 x+1.7}\}$[/tex] about the x-axis we obtain a solid with the volume V =___

so if i use
[tex]
\int_a^1 \frac{dx}{0.6x + 1.7} + \int_1^b \frac{dx}{0.6x + 1.7}
[/tex]
[tex]
\int_0^1 \frac{dx}{0.6x + 1.7} + \int_1^\infty \frac{dx}{0.6x + 1.7}
[/tex]

==> [ln(0.6+1.7 )]/0.6 - ln(1.7 )]/0.6] + [infinity - ln(0.6+1.7 )]/0.6]

how do i get a value for the volume?
 
  • #6
OK, they're doing two things: 1) determining the area of region R, and 2) determining the volume of the solid of revolution.

If the graph of f(x) = 1/(.6x + 1.7) is revolved around the x-axis, what integral represents the volume of the solid?

Hint: This ain't it.
[tex]\int _0^{\infty }\frac{dx}{0.6 x+1.7}[/tex]

Draw a sketch of the solid. You're probably going to want to use disks to calculate the volume.

Note: My previous advice for splitting the integral into two integrals was incorrect. For some reason I was thinking that the integrand was undefined at 0 - it isn't. Your technique of integrating from 0 to t is the right way to go. Sorry for the errant advice.
 
  • #7
i just don't understand what R [tex] $=\{ (x,y)|x\geq0, 0\leq y\leq \frac{1}{0.6 x+1.7}\}$ [/tex] means.
my guess would be
[tex] \int _x^y \frac{dx}{0.6 x+1.7} [/tex] x= 0, y=0.6 ?
but that doesn't look right
 
  • #8
That definition just means that R is the set of all points (x, y) for which the x value is >= 0, and the y value is between the x-axis and the curve y = 1/(.6x + 1.7). So R is the region in the first quadrant above the x-axis and below that curve.

What you need to do is to revolve that region around the x-axis and find the volume of the resulting solid.

Do you know how to find the volume of a solid of revolution? There are a couple of ways to do it. One way involves disks, and another way involves cylindrical shells. The disk method is probably the way to go here.

The integral you show DOES NOT give you the volume of the solid. Forget that one; you need a different integral.
 
  • #9
V= 2pi[tex] \int _0^{\infty } yt-yb [/tex]
V= 2pi[tex] \int _0^{\infty } [1/(0.6 x+1.7)-0 ] dx [/tex]
isnt that it?
 
  • #10
No. What's the volume of a disk of radius r and thickness dx?
 
  • #11
oh my bad. pi*r^2
so pi*[tex] \int _0^{\infty } [(1/(0.6 x+1.7))^2 - 0 ]dx [/tex]
 
  • #12
Right, and that one you can integrate as you did with the other integral, integrating from 0 to t, say, and then taking the limit as t --> infinity.
 
  • #13
sweet. then i get 3.079!
solid!
 
  • #14
No, it's not 0. You're not evaluating your antiderivative correctly.
 

1. What is an improper integral volume around x-axis?

An improper integral volume around x-axis is the calculation of the volume of a solid where one or more of the boundaries of the integral are infinite. This means that the limits of integration for the variable of interest (in this case, x) extend to infinity.

2. How is an improper integral volume around x-axis different from a regular integral?

An improper integral volume around x-axis differs from a regular integral in that it involves integrating over an infinite interval. This introduces additional challenges in calculating the integral and may require special techniques or adjustments to the standard methods of integration.

3. What are some real-world applications of improper integral volume around x-axis?

Improper integral volume around x-axis has several real-world applications, such as calculating the volume of a solid with infinite length, such as a skyscraper, or finding the volume of a continuously changing 3D shape, such as a fluid in motion.

4. How do you solve an improper integral volume around x-axis?

To solve an improper integral volume around x-axis, you first need to identify the boundaries of the integral and determine if they extend to infinity. Then, you can use techniques such as the limit comparison test, the comparison test, or the integral test to determine if the integral converges or diverges. Based on the result, you can then use appropriate methods of integration to calculate the volume.

5. What is the significance of calculating improper integral volume around x-axis?

Calculating improper integral volume around x-axis is important in many fields of science and engineering, such as physics, chemistry, and fluid mechanics. It allows us to accurately determine the volume of complex and continuously changing 3D shapes, which is essential in understanding and predicting various natural phenomena and designing structures and systems.

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