Finding Volume Using Double Integrals: A Question on Cylindrical Coordinates

[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.
$$x^4 + y^4 = \frac12(x^2 + y^2)^2 + \frac12(x^2 - y^2)^2 = \frac12 r^4 + \frac12 r^4(\cos^2\theta - \sin^2\theta)^2 = \frac12 r^4 + \frac12 r^4(\cos(2\theta))^2 \\ = \frac12 r^4 (1 + \cos^2(2\theta))$$

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 !