Triple integral - volume

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August 14th 2008, 04:08 AM
asi123

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Triple integral - volume

Hey.
I need to find the volume of this integral.
I'm pretty sure I did the second part fine, but I'm really not sure about the jacobian, can I do that?

And another thing, I know I'm asking a lot of question these past few days, but it's only because I have a big test on Sunday (Headbang)
August 14th 2008, 11:59 AM
ThePerfectHacker

I assume the volume you are trying to find is the set $V=\{ (x,y,z)| (x+y+z)^2+(x-2y-z)^2+(2x-y+z)^2 \leq h^2 \}$ (where $h>0$ ).

The Jacobian theorem says,
$\iiint_R f = \iiint_{\bold{g}^{-1}(R)} (f \circ \bold{g}) \cdot | \det D(\bold{g})|$
Here $\bold{g}:\mathbb{R}^3 \to \mathbb{R}^3$ is a differenciable vector function.
And $\det D (\bold g)$ is the determinant of the Frechet' derivative i.e. the Jacobian.
Of course, $\bold{g}^{-1}(R)$ is the set $\{ (x,y,z)\in \mathbb{R}^3 | \bold{g}(x,y,z) \in R \}$

It is conventient to have $\bold{g}$ be such a function so that $\bold{g}(x+y+z,x-2y-z,2x-y+z)=(x,y,z)$ .

If we let $A = \left( \begin{array}{ccc}1&1&1\\1&-2&-1\\2&-1&1 \end{array} \right)$ then $A(x,y,z)^T = (x+y+z,x-2y-z,2x-y+z)^T$ .
Thus, $A^{-1} (x+y+z,x-2y-z,2x-y+z)^T = (x,y,z)^T$ .

This means, the function $\bold{g}(\bold{x}) = A^{-1}\bold{x}$ (thinking of vectors in $\mathbb{R}^3$ as coloums).

Now, this is convenient because $\bold{g}^{-1}(V) = \{ (x,y,z)| \bold{g}(x,y,z) \in V \} = \{ (x,y,z) | x^2+y^2+z^2 \leq h^2 \}$

The volume of $V$ is given by,
$\iiint_V 1 = \iiint_{\bold{g}^{-1}(V)} |\det D(\bold{g})|$

But, $\det D(\bold{g}) = \det A^{-1} = (\det A)^{-1} = 3^{-1}$

Thus the volume is $\tfrac{4}{9}\pi h^3$ because the volume of a sphere is $\tfrac{4}{3}\pi h^3$
August 14th 2008, 09:24 PM
asi123

Quote:

Originally Posted by ThePerfectHacker

I assume the volume you are trying to find is the set $V=\{ (x,y,z)| (x+y+z)^2+(x-2y-z)^2+(2x-y+z)^2 \leq h^2 \}$ (where $h>0$ ).

The Jacobian theorem says,
$\iiint_R f = \iiint_{\bold{g}^{-1}(R)} (f \circ \bold{g}) \cdot | \det D(\bold{g})|$
Here $\bold{g}:\mathbb{R}^3 \to \mathbb{R}^3$ is a differenciable vector function.
And $\det D (\bold g)$ is the determinant of the Frechet' derivative i.e. the Jacobian.
Of course, $\bold{g}^{-1}(R)$ is the set $\{ (x,y,z)\in \mathbb{R}^3 | \bold{g}(x,y,z) \in R \}$

It is conventient to have $\bold{g}$ be such a function so that $\bold{g}(x+y+z,x-2y-z,2x-y+z)=(x,y,z)$ .

If we let $A = \left( \begin{array}{ccc}1&1&1\\1&-2&-1\\2&-1&1 \end{array} \right)$ then $A(x,y,z)^T = (x+y+z,x-2y-z,2x-y+z)^T$ .
Thus, $A^{-1} (x+y+z,x-2y-z,2x-y+z)^T = (x,y,z)^T$ .

This means, the function $\bold{g}(\bold{x}) = A^{-1}\bold{x}$ (thinking of vectors in $\mathbb{R}^3$ as coloums).

Now, this is convenient because $\bold{g}^{-1}(V) = \{ (x,y,z)| \bold{g}(x,y,z) \in V \} = \{ (x,y,z) | x^2+y^2+z^2 \leq h^2 \}$

The volume of $V$ is given by,
$\iiint_V 1 = \iiint_{\bold{g}^{-1}(V)} |\det D(\bold{g})|$

But, $\det D(\bold{g}) = \det A^{-1} = (\det A)^{-1} = 3^{-1}$

Thus the volume is $\tfrac{4}{9}\pi h^3$ because the volume of a sphere is $\tfrac{4}{3}\pi h^3$

So I was suppose to take $(Jacobian)^{-1}$ , right?
August 14th 2008, 10:32 PM
ThePerfectHacker

Quote:

Originally Posted by asi123

So I was suppose to take $(Jacobian)^{-1}$ , right?

I am not sure what you mean by this.
You probably mean the reciprocal of the determinant you computed.
Maybe you did not understand what I did - I know I do Jacobians my way so maybe it was confusing.
August 14th 2008, 10:39 PM
asi123

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I mean that I calculated that Jacobian with the first equation (in the pic) instead using the second one.
I switched to u=u(x,y,z), v=(x,y,z) and R=(x,y,z) and as I was saying used the first one instead of the second.
And my final answer would have also been $\tfrac{4}{9}\pi h^3$ if I would have taken Jacobian $3^{-1}$ so I think that was my mistake.