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Improper Integral Volume Around x-axis

  • Thread starter Slimsta
  • Start date
  • #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 dont have a and b..
 

Answers and Replies

  • #2
33,505
5,191
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
179
2

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 dont 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
33,505
5,191
rs1n, Good point. Slimsta doesn't have the integrand set up correctly to get the volume.
 
  • #5
190
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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
33,505
5,191
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
190
0
i just dont 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 doesnt look right
 
  • #8
33,505
5,191
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
190
0
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
33,505
5,191
No. What's the volume of a disk of radius r and thickness dx?
 
  • #11
190
0
oh my bad. pi*r^2
so pi*[tex] \int _0^{\infty } [(1/(0.6 x+1.7))^2 - 0 ]dx [/tex]
 
  • #12
33,505
5,191
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
190
0
sweet. then i get 3.079!
solid!
 
  • #14
33,505
5,191
No, it's not 0. You're not evaluating your antiderivative correctly.
 

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