Finding Volume Using Double Integrals: A Question on Cylindrical Coordinates

[CAF123]

The fact that a^2 + b^2 = c^2 does not imply that a^4 + b^4 = c^4: for example if a = \sqrt 3, b = 2 and c = \sqrt 7 then a^2 + b^2 = c^2, but a^4 + b^4 = 25 while c^4 = 49.

I don't think we are using the fact that $a^2 + b^2 = c^2,$only that $x^2 + y^2 = r^2$ and making the substitutions $x = a^2, y = b^2, r = c^2$

 

[DryRun]

You are referring to 3 variables here, r, θ, and z.

To find the volume with a double integral rather than a triple integral, you need to have an integrand that's different than 1.

[STRIKE]The obvious integrand, in my opinion, is 16-r⁴, and integrate over r and θ, using the appropriate Jacobian.

Alternatively, you could use an integrand of 2π r and integrate over r and z.[/STRIKE]

Added in Edit:
I've crossed out the baloney !

I apologize for the mess, but honestly, i was unsure at first about the proof, but then a look at SammyS' (usually very trustworthy) post #15 stated that $16-r^4$, which meant that $x^4+y^4$ had to be equal to $r^4$. So, i tried to prove that statement.

Initially, i was going for the $\cos^4 θ +\sin^4 θ= 1$, but then i quickly realized it was wrong, as the expression didn't hold for all values of θ. Further effort seemed to indicate that the Pythagoras' theorem could be the right method. Thanks for crossing out the baloney!

No.
x^4 + y^4 = \frac12(x^2 + y^2)^2 + \frac12(x^2 - y^2)^2<br /> = \frac12 r^4 + \frac12 r^4(\cos^2\theta - \sin^2\theta)^2<br /> = \frac12 r^4 + \frac12 r^4(\cos(2\theta))^2 \\<br /> = \frac12 r^4 (1 + \cos^2(2\theta)) \not\equiv r^4

Or, more directly, x^4 + y^4 = r^4(\cos^4\theta + \sin^4\theta) \not\equiv r^4 but the former expression is easier to integrate with respect to \theta.

Is there any other way of solving this?

 

[CAF123]

@Sharks Where is the fault in your argument?

 

[DryRun]

@Sharks Where is the fault in your argument?

Trying some values to verify that argument... Let the sides of a right-angled triangle be: a = 3, b = 4 and c (hypotenuse) = 5. This set of values works perfectly in the basic Pythagoras' theorem: $(3)^2+(4)^2=(5)^2$ giving $9+16=25$ which is correct.
Now, using that same set of values: $a^2=9, b^2=16$ and $c^2=25$. According to my previous argument, using the Pythagoras' theorem: $(9)^2+(16)^2$ should be equal to $(25)^2$ but, $81+256 \not = 625$. In theory it seemed like it would have worked, but i was clearly wrong.

 

[peripatein]

@peripatein
If you want to use a double integral then that's fine too. The two forms are equivalent.
By using the triple integral you get:$$ \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{16-r^4} r\,dz\,dr\,d\theta$$
By doing the double integral you get: $$ \int_0^{2\pi} \int_0^1 16 - r^4\, r\,dr\,d\theta$$ Because you want a volume, it is implicitly assumed that the surface lies above z = 0, and it's projection will be something in the xy plane (i.e where z = 0)

Rather perplexing. Wishing to verify - may the integration, for obtaining the required volume, still be performed thus:
$$ \int_0^{2\pi} \int_0^1 16 - r^4\, r\,dr\,d\theta$$

?

 

[CAF123]

Trying some values to verify that argument... Let the sides of a right-angled triangle be: a = 3, b = 4 and c (hypotenuse) = 5. This set of values works perfectly in the basic Pythagoras' theorem: $(3)^2+(4)^2=(5)^2$ giving $9+16=25$ which is correct.
Now, using that same set of values: $a^2=9, b^2=16$ and $c^2=25$. According to my previous argument, using the Pythagoras' theorem: $(9)^2+(16)^2$ should be equal to $(25)^2$ but, $81+256 \not = 625$. In theory it seemed like it would have worked, but i was clearly wrong.

That would work as a counterexample, but as you said there seemed to be no mistake in the logic that led you to believe that $x^4 + y^4 = r^4$

 

[CAF123]

Rather perplexing. Wishing to verify - may the integration, for obtaining the required volume, still be performed thus:
$$ \int_0^{2\pi} \int_0^1 16 - r^4\, r\,dr\,d\theta$$

?

Yes, but obviously given the discussion, the fallacy that x⁴ + y⁴ = r⁴ means z in polar coordinates is not the 16 - r⁴.

 

[peripatein]

Hmm. If z is not 16-r^4, how could that double integral yield the requisite volume? I mean, is it not then incorrect?

 

[CAF123]

Hmm. If z is not 16-r^4, how could that double integral yield the requisite volume? I mean, is it not then incorrect?

The set up is right, it is just z is incorrect. So yes, as it stands, it is incorrect.

 

[peripatein]

No.
x^4 + y^4 = \frac12(x^2 + y^2)^2 + \frac12(x^2 - y^2)^2<br /> = \frac12 r^4 + \frac12 r^4(\cos^2\theta - \sin^2\theta)^2<br /> = \frac12 r^4 + \frac12 r^4(\cos(2\theta))^2 \\<br /> = \frac12 r^4 (1 + \cos^2(2\theta))

Should z then not be 1/2r⁴(1+cos²(2θ))?

 

[pasmith]

Should z then not be 1/2r⁴(1+cos²(2θ))?

z=16-(x^4+y^4) = 16 - \frac12 r^4 (1 + \cos^2(2\theta)).

 

[peripatein]

That is what I intended to write, thank you :-)

 

[SammyS]

z=16-(x^4+y^4) = 16 - \frac12 r^4 (1 + \cos^2(2\theta)).

In integrating over θ, it may be helpful to use the identity:

\displaystyle 1 + \cos^2(2\theta)=\frac{1}{2} (3+cos(4 θ))\ \^*​

^* verified by WolframAlpha --

 

[peripatein]

The integration yielded 23.75pi. Would anyone kindly verify?

 

[SammyS]

The integration yielded 23.75pi. Would anyone kindly verify?

Probably not correct ... although, as you all know by now, you really need to carefully check all of SammyS's statements. (LOL)

The top surface lies between z=15 and z=16, so the volume should somewhere between the volume of a cylinder of radius 1 and height 15 and the volume of a cylinder of radius 1 and height 16.

... between 15π and 16π.

 

[peripatein]

You were right, Sammy (for a change ;-)). It ended up yielding (63/4)pi. Thank you!

 

[SammyS]

You were right, Sammy (for a change ;-)). It ended up yielding (63/4)pi. Thank you!

That's correct !