Volume questions (cylindrical) (1 Viewer)

[googleplex]

By taking cylindrical shells find the volume enclosed in the region enclosed by
(x-1)^2 + y^2 = 1

I ended up getting like the integral of x(2x-x^2)^1/2

I couldn't get it.

Is my integration shice or is there a trick to these kinds of questions, like 2 semi-circles or something.

if you could show me how to do it it would be much appreciated

Thx

 

[CrashOveride]

What are you rotating it around?

 

[googleplex]

oh shit sorry,
y axis

 

[CrashOveride]

Ok i'll write up the solution for ya

 

[CrashOveride]

Actually this probably lends itself beter to the washer method. I end up getting a lot of integrating to do and using a substitution x = 1 +sin& and having to do another substitution after that and i fall out getting V = -2pi². Hmm maybe some error somewhere...whats the answer ?

Also, there isnt really any quicker way for doing this one using shell method anyone ?

PS. hmm i could eliminate the negative i got if i took my height 'y' as neg. in the first place. How do u know whether to take the pos. or neg. 'y' anyone ?

 

[googleplex]

yeh answer is 2pi^2

how'd u get it?

 

[CrashOveride]

if someone could tell me the reason for choosing the particular 'y' value (pos or neg) i'd be happy to explain the rest

 

[CM_Tutor]

This is problem produces an integral that turns up quite often. EVERYONE should take note of this integral, and what to do with it!!!

OK, applying the cylindrical shells method, you should get that the Volume, V, is given by:

V = 4π * ∫ (0 --> 2) x * sqrt(1 - (x - 1)²) dx

Googleplex has simplified this to 4π * ∫ (0 --> 2) x * sqrt(2x - x²) dx, which is another form in which this integral often appears.

In any event, it is the 'factorised' 1 - (x - 1)² form that is easiest to work with. Perform a u = x - 1 substitution, and you should get:

V = 4π * ∫ (-1 --> 1) (u + 1) * sqrt(1 - u²) du

As CrashOveride has found, this can be evaluated after a lot of quite involved substitution / integration. However (and here's the trick) it can also be found very simply, as follows:

V = 4π * ∫ (-1 --> 1) (u + 1) * sqrt(1 - u²) du
= 4π * ∫ (-1 --> 1) u * sqrt(1 - u²) du + 4π * ∫ (-1 --> 1) sqrt(1 - u²) du

The first part involves ∫ (-1 --> 1) u * sqrt(1 - u²) du = 0, as u * sqrt(1 - u²) is odd, and the second part is a semicircle of radius 1.

So, V = 4π * ∫ (-1 --> 1) u * sqrt(1 - u²) du + 4π * ∫ (-1 --> 1) sqrt(1 - u²) du = 4π * 0 + 4π * π * (1)² / 2

So, V = 4π * π / 2 = 2π² cu units, as required.

 

[Supra]

how do u test/prove that u * sqrt(1 - u2) is odd
do u have to do that by f(a)=-f(-a)

 

[Calculon]

Originally posted by Supra
how do u test/prove that u * sqrt(1 - u2) is odd
do u have to do that by f(a)=-f(-a)

Its not hard to do it that way:

f(a)=a(sqrt(1-a²)

f(-a)=-a(sqrt(1-(-a)²)
=-a(sqrt(1-a²)
=-f(a)

 

[CrashOveride]

Could i just ask how do we know wheter to take the height of the cylindrical shell, "y", +ve or -ve ?

 

[:: ck ::]

how can a height be negative.......

 

[CrashOveride]

oops

 

[CrashOveride]

The base of a solid is the circle x^2 + y^2 = 8x and every plane section perpendicular to the x axis is a rectnagle whose height is one third of the distance of the plane of the section from the origin. Show that volume of the solid is 64pi / 3

I get like the answer plus a tiny bit o.o

 

[CM_Tutor]

Originally posted by CrashOveride
Could i just ask how do we know wheter to take the height of the cylindrical shell, "y", +ve or -ve ?

If you think about the diagram, the 'top' of the cylinder is at y = sqrt(1 - (x - 1)²), and the 'bottom' is at
y = -sqrt(1 - (x - 1)²).

So, the height is actually sqrt(1 - (x - 1)²) - -sqrt(1 - (x - 1)²) = sqrt(1 - (x - 1)²) + sqrt(1 - (x - 1)²)
= 2 * sqrt(1 - (x - 1)²).

 

[CrashOveride]

Thanks CM

 

[CrashOveride]

Also for something like the area bounded between y=2x and y = 4x - x^2 is rotated around the X-axis (using washers) ....when im finding the "inner radius" i find that x = 2 +/- sqrt(4-y)
Now because this inner radius would be the smaller of the two x values for a given y value, i would be inclined to use the x=2-sqret(4-y) as it wud be smaller than the other plus sqrt value....my friend said u can just test since 0,0 lies on the curve...and u get the one with the negative....but isnt that just a bit random? if u tested 4,0 ud have to go postive value etc. so we make the decision based on logic ?

 

[CM_Tutor]

Originally posted by CrashOveride
The base of a solid is the circle x^2 + y^2 = 8x and every plane section perpendicular to the x axis is a rectnagle whose height is one third of the distance of the plane of the section from the origin. Show that volume of the solid is 64pi / 3

I get like the answer plus a tiny bit o.o

Taking slices of the solid of volume δV and thickness δx perpendicular to the x-axis a distance x along the x-axis from the origin.

It follows that δV = BHδx, where B is the base of the rectangular slice and H is its height.

Since, for each slice, the "height is one third of the distance of the plane of the section from the origin", H = x / 3.
By a similar method to that above, B = 2 * sqrt(16 - (x - 4)²)

So, δV = (2x / 3) * sqrt(16 - (x - 4)²) * δx

Summing the volumes of the individual slices, the volume of the solid is:

V = lim_{δx--->0}Σ (x = 0 ---> x = 8) δV
= ∫ (0 --> 8) (2x / 3) * sqrt(16 - (x - 4)²) dx

Let u = x - 4:

So, V = (2 / 3) * ∫ (-4 --> 4) (u + 4) * sqrt(16 - u²) du
= (2 / 3) * ∫ (-4 --> 4) u * sqrt(16 - u²) du + (8 / 3) * ∫ (-4 --> 4) sqrt(16 - u²) du
= (2 / 3) * 0 + (8 / 3) * π * (4)² / 2, noting that u * sqrt(16 - u²) is odd and sqrt(16 - u²) is a semicircle
= 0 + (8π / 3) * (16 / 2)
= 64π / 3 cu units

 

[CM_Tutor]

Originally posted by CrashOveride
Also for something like the area bounded between y=2x and y = 4x - x^2 is rotated around the X-axis (using washers) ....when im finding the "inner radius" i find that x = 2 +/- sqrt(4-y)
Now because this inner radius would be the smaller of the two x values for a given y value, i would be inclined to use the x=2-sqret(4-y) as it wud be smaller than the other plus sqrt value....my friend said u can just test since 0,0 lies on the curve...and u get the one with the negative....but isnt that just a bit random? if u tested 4,0 ud have to go postive value etc. so we make the decision based on logic ?

Assuming you mean rotated about the y-axis, I would show that x = 2 +/- sqrt(4 - y), as you have done.

Then, looking at the diagram, I would note that x < 2, and deduce that x = 2 - sqrt(4 - y), as
x = 2 + sqrt(4 - y) > 2 for all y < 4.

In other words, I would use logic - as you suggest. The logic that (from the diagram) you require the smaller of the x values is also valid.