Find the volume of the solid which is obtained by revolving the discs

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Homework Help Overview

The discussion revolves around finding the volume of a solid obtained by revolving a disc defined by the equation (x - 1)² + y² ≤ 1, as well as calculating the arclength of the curve y = e^x over the interval [0, ln 2]. The subject area includes calculus, specifically integral calculus related to volumes of revolution and arclength calculations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of cylindrical shells for the volume calculation and discuss the integral setup. There are attempts to clarify the nature of the solid formed by the revolution, with some questioning whether it results in a sphere or a toroidal shape. For the arclength problem, participants suggest integration techniques, including u-substitution and integration by parts, while also discussing the complexity of the integral involved.

Discussion Status

The discussion is active, with various approaches being suggested for both problems. Some participants provide hints and guidance on integration techniques, while others express confusion about the original problem statements and their implications. There is no explicit consensus on the best approach yet, as multiple interpretations and methods are being explored.

Contextual Notes

Participants note that the original problem statement may have been misleading regarding the nature of the solid formed by the revolution. There are also indications of varying levels of familiarity with integration techniques among participants, which may affect the discussion's progression.

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Homework Statement



Find the volume of the solid which is obtained by revolving the discs [tex](x - 1)^2 + y^2 \le 1[/tex]

The Attempt at a Solution



I say using cylinders would be easiest so,
[tex]\int_a^b{2\pi r(x)h(x) dx} = \int_0^2{2\pi x\sqrt{1-(x-1)^2} dx}[/tex]

Any hints on how to get started with this integral?

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Homework Statement



Find the arclength of the curve [itex]y = e^x[/itex] with [itex]0 \le x \le ln 2[/itex]

The Attempt at a Solution


[tex]y' =e^x[/tex]
so I need to solve
[tex]\int_a^b{\sqrt{1+(y')^2}dx} = \int_0^{ln 2}{\sqrt{1+e^{2x}}dx}[/tex]

Any hints?
 
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[tex]\int{\sqrt{1+e^{2x}}[/tex]

this can be written as [tex]{1+e^{2x}}^(1/2)[/tex]

then to integrate you add one to the power...etc

does this help?
 
wait that came up wrong i meant:

this can be written as (1+e^(2x))^(1/2)
 
I can't do what you said as it's a composite function...

Any other ideas?
 
sara_87, when doing exponents longer than one character, uses { } instead of ( ). It seems you just forgot, since you did it the first time with the 2x :D
 
o, cscott, would It help If You had it in this form? (Hopefully you know integration by parts ).

Make a u substitution
[tex]u=e^{2x}[/tex]
[tex]\frac{du}{dx}=2e^{2x}[/tex]
Solve for dx
[tex]dx=\frac {du}{2e^{2x}}=\frac{du}{2u}[/tex]
Now you just need to do the rest.

[tex]\int (u+1)^{\frac{1}{2}} 2u du[/tex]

Have fun :D
 
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question 1)Not that I want to ruin your fun here, but by revolving discs do you mean a sphere? If I draw a graph of a revolving disc unless it's simply rotating about it's axis which gives a circle, wouldn't I end up with a sphere? in other words [tex]\frac{4}{3}\pi r^3[/tex]
 
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Gib Z said:
o, cscott, would It help If You had it in this form? (Hopefully you know integration by parts ).

Make a u substitution
[tex]u=e^{2x}[/tex]
[tex]\frac{du}{dx}=2e^{2x}[/tex]
Solve for dx
[tex]dx=\frac {du}{2e^{2x}}=\frac{du}{2u}[/tex]
Now you just need to do the rest.

[tex]\int (u+1)^{\frac{1}{2}} 2u du[/tex]

Have fun :D

You made a slight mistake; the integral is [tex]\int \frac{(u+1)^{\frac{1}{2}}}{2u} du[/tex]
 
Schrödinger's Dog said:
question 1)Not that I want to ruin your fun here, but by revolving discs do you mean a sphere? If I draw a graph of a revolving disc unless it's simply rotating about it's axis which gives a circle, wouldn't I end up with a sphere? in other words [tex]\frac{4}{3}\pi r^3[/tex]

Sorry, it should say "revolve the disc defined by ___ around the y-axis" which I think calls for cylinders to make it easiest and wouldn't be a sphere.
 
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  • #10
cscott said:
Sorry, it should say "revolve the disc defined by ___ around the y-axis" which I think calls for cylinders to make it easiest and wouldn't be a sphere.

Oh you mean revovle a cyclinder around the y axis. I suggest you copy out the question again exactly as it is on paper, as your original question is a little misleading. I thought you meant spinning a circle through 360 degrees.
 
  • #11
cristo- ty for spotting this, my brain seems to be farting a lot recently...

cscott- either way, I still think integration by parts helps
 
  • #12
The first question I'm assuming your talking about a toroidal. which is found from the curve y=e^x. You may want to look up toroidal volumes.

Hint: you should end up with the second question integral, part 1 if you have done it correctly.

i.e. [tex]\int_a^b{\sqrt{1+(y')^2}dx= \int_0^{ln 2}{\sqrt{1+e^{2x}}dx}[/tex]

The second I can only say was too complicated for the level I'm at although once I saw how it was done I could follow the steps easilly enough(I asked Dr.E.P.A Bailey. A colleague to solve it basically) Otherwise known as cheating :smile:

let [tex]u^2=1+e^{2x}[/tex] and then substitute.

then put in the form of partial fractions.

[tex]\frac {f(u)}{g(u)}[/tex]

where f(u) and g(u) are polynomials.

You should end up with this:-

[tex]\int_0^{ln 2} 1+\frac {1}{2(u-1)}. \frac {-1}{2(u+1)}[/tex]

from here it should be fairly straightforward if you express the answer in the form of natural logs.

Hope that helps.
 
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