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

In summary, cscott was trying to solve an integral involving revolving discs around the y axis and ended up having to do it in a more complicated way than originally planned. Using substitution and partial fractions, he was able to get the answer in the correct form.
  • #1
cscott
782
1

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?

---

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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  • #2
[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?
 
  • #3
wait that came up wrong i meant:

this can be written as (1+e^(2x))^(1/2)
 
  • #4
I can't do what you said as it's a composite function...

Any other ideas?
 
  • #5
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
 
  • #6
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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  • #7
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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  • #8
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]
 
  • #9
Schrodinger'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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1. What is the definition of "volume" in this context?

In this context, volume refers to the amount of space that is occupied by the solid obtained by revolving the discs.

2. How do you find the volume of the solid?

To find the volume of the solid, you will need to use the formula V = π∫ab(f(x))2dx, where a and b are the limits of integration and f(x) is the function that represents the shape of the disc. You will also need to know the radius of the discs and the axis of revolution.

3. Can you provide an example of finding the volume of the solid?

Sure! Let's say we have a disc with a radius of 3 and we are revolving it around the x-axis from x = 0 to x = 4. The function that represents this disc is f(x) = 3, so the volume would be V = π∫04(3)2dx = 36π cubic units.

4. Is there a specific unit for the volume of the solid?

The unit for the volume of the solid will depend on the units used for the radius of the discs. For example, if the radius is given in meters, the volume would be in cubic meters. It is important to keep the units consistent throughout the calculation.

5. What happens if the discs have different radii?

If the discs have different radii, you will need to use the formula V = π∫ab(f(x))2dx for each disc separately and then add the volumes together to get the total volume of the solid.

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