Question

Some more tricky calc: Find the volume of a cube with side length \(a\) by evaluating the appropriate triple integral in spherical coordinates.

Answer 1

I had what I thought to be the proper setup, but Mathematica tells me otherwise... The closest I've gotten was something like \(\dfrac{a^3\pi}{4}\), which is clearly wrong.

Answer 2

well, volume is just the integral of area which is the integral of circumference...

Answer 3

volume of a cube using spherical coordinates??

Answer 4

How could I not have seen it before? lol

Answer 5

And yes, spherical. I was looking for a challenge :D

Answer 6

I understand how you would calculate \(\rm \rho\) but as for \(\phi\) and \(\theta\)....

Answer 7

so maybe we can use that, if you put the perimeter of a cube into terms for rho, then integrate?

Answer 8

but you have to subtract the area of the excess

Answer 9

I don't think the derivative relationship holds for cubes...

Answer 10

just use the jacobian

Answer 11

you have to wrap your sphere around the cube. just represent theta, phi and rho as functions of x,y,z,

Answer 12

Pic attached.

In the image, I'm setting \(a=2\) and using a cube centered at the origin. Finding this volume, then multiplying by ... 48? should do it.

From what I can tell, both \(\theta\) and \(\phi\) can be fixed in the range \(\left[0,\dfrac{\pi}{4}\right]\), while \(\rho\) would range from \(0\) at the origin to the uppermost face. This face is described by \(z=\dfrac{a}{2}\), which should give \(\rho=\dfrac{a}{2}\sec\phi\), no?

Answer 13

well, since it is a square we know the radius of the sphere

Answer 14

\(r=a\sqrt2/2\)

Answer 15

sorry, rho*

Answer 16

so then if we just derive the volume using these, we can derive it and see how it works

Answer 17

maybe?

Answer 18

So you're suggesting I find the volume of one of those dome-shaped regions, call it \(V_\text{dome}\), then subtract \(6V_{\text{dome}}\) from the sphere's volume?

Answer 19

yep

Answer 20

that's how I'd go about it

Answer 21

I'm curious if that'll yield something different from the result I get with my initial attempt, using the integral
\[\large48\int_0^{\pi/4}\int_0^{\pi/4}\int_0^{a/2~\sec\phi}\rho^2\sin\phi~d\rho~d\phi~d\theta\]

Answer 22

is that the jacobian? I forget

Answer 23

Using this dome-approach, I'd have
\[\large V_{\text{dome}}=\int_0^{2\pi}\int_0^{\pi/4}\int_{a/2~\sec\phi}^{a/\sqrt2}\rho^2\sin\phi~d\rho~d\phi~d\theta\]
It sure is the Jacobian.

Answer 24

Solve for theta before solving for phi? If that changes anything.

Answer 25

yea, I'd do theta before phi or rho

Answer 26

or separate it all, that too :)

Answer 27

well all that you can

Answer 28

Usually the order for integration in spherical coordinates is \(d\rho d\theta d\phi\)

Answer 29

how did you get that sec of phi?

Answer 30

Yeah I was wondering that too since \(\rho^2 = x^2 +y^2 +z^2\)

Answer 31

I don't see why the position of \(\theta\) would matter...

As for the \(\sec\phi\) question, we have \(z=\rho\cos\phi\), and when \(z=\dfrac{a}{2}\), then you get \(\rho=\dfrac{a}{2}\sec\phi\).

Answer 32

Oh I get it. because \(\frac{1}{\cos(\theta)} = \sec\theta\)

Answer 33

Well it's not over 1 but I get it. lol.

Answer 34

ah ok

Answer 35

now why is z a/2?

Answer 36

The cube has side length \(a\), but the way I have it plotted, the cube is centered at the origin. In the image, \(a=2\). I've cut away all the tetrahedral fat to isolate this one piece.

Answer 37

oh oh oh ok, I did it for general a

Answer 38

I can also see the relation of it being a/2 from her drawing earlier.

Answer 39

ok, nope, thought I understood, but I don't. Can you elaborate some more?

Answer 40

why shouldn't z=a?

Answer 41

omg yes...

Answer 42

wow... ok so I see how you got your integral now, and I agree, are you multiplying it by 6 and subtracting?