Wikipedia:Reference desk/Archives/Mathematics/2011 July 26

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July 26

Triple Trouble

Greetings, and great to be back after all this time. I am having trouble with a mathematics question where we are asked to use a triple integral to find the volume of the ellipsoid, where x²/a²+ y²/b²+ z²/c² = 1 .

I have looked at all the ways to do it, and am still confused, having been working on this for days. If I integrate, where does the pi term come from to form the volume which I do know to be pi times abc  ? Chris the Russian Christopher Lilly 07:45, 26 July 2011 (UTC)

Generally it would come as a result of the arcsin function which comes up when you integrate factors of the form ${\sqrt {1-x^{2}}}$. -- Meni Rosenfeld (talk) 09:02, 26 July 2011 (UTC)

It may be easier to start with determining the area of the ellipse x²/a²+ y²/b² = 1 using a double integral, similar in principle but with fewer steps.→86.155.185.195 (talk) 10:24, 26 July 2011 (UTC)

The difficult bit is working out the limits on the integrals. Once you've done that, it's all fairly standard integrals. Say you integrate with respect to x first. You need to find the limits of x within which x²/a²+ y²/b²+ z²/c² < 1. --Tango (talk) 19:04, 26 July 2011 (UTC)

I would recommend changing from Cartesian coordinates to spherical coordinates. — Fly by Night (talk) 19:17, 26 July 2011 (UTC)

Use the change of variables u=x/a, v=y/b, w=z/c. The Jacobian is abc, so the volume is abc times the volume of the unit sphere: $4\pi abc/3$.Sławomir Biały (talk) 21:41, 26 July 2011 (UTC)

I thought about something similar, i.e. using a special linear transformation to change the ellipsoid to a sphere and then applying the formula for the volume of a sphere. But the question wants derivation of the formula of an ellipsoid using integration. There is a general formula for the volume of an ellipsoid. If the OP can't use that, then we can assume that s/he can't use the formula for the volume of a sphere. — Fly by Night (talk) 22:13, 26 July 2011 (UTC)

Ok, in principle one could evaluate the triple integral

$\int _{-a}^{a}\int _{-{\sqrt {b^{2}-x^{2}/a^{2}}}}^{\sqrt {b^{2}-x^{2}/a^{2}}}\int _{-{\sqrt {c^{2}-x^{2}/a^{2}-y^{2}/b^{2}}}}^{\sqrt {c^{2}-x^{2}/a^{2}-y^{2}/b^{2}}}\,dz\,dy\,dx.$

--Sławomir Biały (talk) 00:16, 27 July 2011 (UTC)

Ouch! There's no need for language like that ;-) Wouldn't spherical coordinates be the natural choice? Trying to integrate as you suggest presents all kinds of problems. The signs of the arguments of the radicals is one. Besides that, the integrals themselves are less than elegant, e.g.:

$\int {\sqrt {1-t^{2}}}\,\operatorname {d} \!t={\frac {1}{2}}t{\sqrt {1-t^{2}}}+{\frac {1}{2}}\arcsin t\,.$

I won't even try to write the answer to

$\int \int {\sqrt {1-u^{2}-v^{2}}}\,\operatorname {d} \!u\,\operatorname {d} \!v\,,$

but it involves logs, inverse tangents and absolute values. P.S. Did you miss the upper limit from the middle integral? — Fly by Night (talk) 00:18, 27 July 2011 (UTC)

Well, I think the natural coordinate system to carry out the integration is the one I suggested. Compute the integral of the Jacobian over the unit ball in the uvw space. (That integral can be done using spherical coordinates, or just with the volume formula.). But I thought you were saying we weren't allowed to change coordinates, though. Hence I suggested the nasty brute-force approach. Sławomir Biały (talk) 01:50, 27 July 2011 (UTC)

I guess it's just a matter of taste. I prefer to see the volume of a unit sphere, centre the origin, is expressed in spherical coordinates as

$V=\int _{0}^{2\pi }\int _{0}^{\pi }\int _{0}^{1}r^{2}\sin \theta \,\operatorname {d} \!r\,\operatorname {d} \!\theta \,\operatorname {d} \!\varphi ={\frac {4\pi }{3}}\pi \,.$

instead of the Cartesian coordinate system that you suggested

$V=\int _{-1}^{1}\int _{-{\sqrt {1-x^{2}}}}^{\sqrt {1-x^{2}}}\int _{-{\sqrt {1-x^{2}-y^{2}}}}^{\sqrt {1-x^{2}-y^{2}}}\!\!\!\operatorname {d} \!z\,\operatorname {d} \!y\,\operatorname {d} \!x\,.$

I suppose that one might argue that your expression is easier to calculate than mine. — Fly by Night (talk) 03:41, 27 July 2011 (UTC)

You misunderstand me. I'm not opposed to using spherical coordinates to calculate $\iiint abc\,dV$ after one applies the Jacobian change of variables (although it isn't necessary to do so for hopefully obvious reasons). The reason I wrote the above is that you seemed to suggest that changes of coordinates were off limits. That just leaves us with the original Cartesian integral, which can (in principle) be calculated using elementary methods by brute force. Sławomir Biały (talk) 11:41, 27 July 2011 (UTC)

Added: Although I see I probably misunderstood your reply as well. Reading it again, I see you meant that we can't use the standard formula for the volume of the sphere. (Instead we should calculate this volume by some standard method: spherical coordinates being the easiest, and cylindrical the next easiest.) Ok. I guess we agree, then. Sławomir Biały (talk) 11:46, 27 July 2011 (UTC)

Thanks all so much, and he, that is I, am very impressed by all of this, which should be of much use indeed. Chris the Russian Christopher Lilly 01:20, 27 July 2011 (UTC)

(edit conflict) My solution would be to make a volume presrving, linear change or coordinates. The equation (x/a)² + (y/b)² + (z/c)² = 1 gives an ellipsoid that cuts the x–axis at x = ±a, it cuts the y–axis at y = ±b and it cuts the z–axis at z = ±c. We can make a linear transformation of xyz–space that preserves the x–, y–, and z–axes; which also preserves volume. Such a transformation is given by (x,y,z) → (αx,βy,γz) where α, β and γ are real numbers such that the product αβγ = 1. We want the linear transforation to carry the ellipsoid to a sphere. A sphere has an equation of the form x² + y² + z² = r² for some real number r > 0. We must choose α, β and γ so that the equation (x/a)² + (y/b)² + (z/c) = 1 is transformed into the equation x² + y² + z² = r². If we make the linear transformation (x,y,z) → (rx/a,ry/b,rz/c) then the image of the set of points given by the equation (x/a)² + (y/b)² + (z/c) = 1 will be transformed into the set of points given by the equation x² + y² + z² = r² or, if you prefer (x/r)² + (y/r)² + (z/r)² = 1. We still don't know what r is. We can determine r from the fact that we want (x,y,z) → (rx/a,ry/b,rz/c) to be volume preserving. The determinant of (x,y,z) → (rx/a,ry/b,rz/c) is given by r³/abc and r³/abc = 1 if and only if r = (abc)^1/3. Thus, your problem reduces to evaluating the volume of the sphere x² + y² + z² = (abc)^2/3. To solve this you can use the formula for the volume of a solid of revolution. The formula (which can easily be proven if necessary) is as follows:

$\pi \int _{\ell _{1}}^{\ell _{2}}\!\!f(x)^{2}\,\operatorname {d} \!x$

gives the volume of the solid given by rotating the graph y = ƒ(x), for $\ell _{1}\leq x\leq \ell _{2}$, given in the xy–plane by a full 2π radians about out the x–axis in xyz–space. As a result, the volume of the sphere (and so the volume of the ellipsoid) is given by calculating

$V=\pi \int _{-k}^{k}\left({\sqrt {k^{2}-x^{2}}}\right)^{2}\,\operatorname {d} \!x={\frac {4}{3}}\pi k^{3}\,,$

where k = (abc)^1/3. It follows that V = 4⁄3πabc. I know that it might seem long winded; but I wanted to explain everything properly. The main details are regarding the linear transformation, which is the easiest bit once you understand it. But it's not always so clear for the the first time. Anyway, I hope this helps. — Fly by Night (talk) 01:46, 27 July 2011 (UTC)

Yes, thank You. I see the key is in trying to simplify the ellipsoid equation into that of a sphere, but being in mind of the differences between the two. To be sure, as seems obvious, an ellipsoid is just a skewed kind of sphere, where in the end it makes sense that their volumes should be of a similar nature. I have spent most of this week on this problem, which it appears is one given to unsuspecting maths students all over the world, designed to get us to think. From this I can begin to understand how the triple integral works, and this shall assist me a lot in giving a proper answer, thank You.Chris the Russian Christopher Lilly 23:36, 27 July 2011 (UTC)

Eigenvalues of real skew-symmetric matrices are imaginary. Does this work in the other direction?

All eigenvalues of real skew-symmetric matrices are imaginary or zero. Does this work in the other direction, i.e. are all real matrices whose eigenvalues are imaginary or 0 similar to a real skew-symmetric matrix? I.e. if all eigenvalues of some real matrix A are imaginary, does this imply the existence of a transformation P such that P^(-1)AP is skew-symmetric? 83.134.166.168 (talk) 16:56, 26 July 2011 (UTC)

No. Any nilpotent matrix is a counterexample. Sławomir Biały (talk) 21:39, 26 July 2011 (UTC)

Thanks! Apparently nilpotent matrix have only zero complex eigenvalues. What if all eigenvalues are imaginary and there's at least one nonzero complex conjugate eigenvalue pair? Is that a sufficient condition for the matrix to be similar to a real skew-symmetric matrix? (clearly if the matrix is real and all its eigenvalues are imaginary, then the polynomial whose zeros are eigenvalues factors like lambda^(number of zero eigenvalues)*(lambda^2+c1)*(lambda^2+c2)*... with all c real and positive, so the eigenvalues do come in complex conjugate pairs ) 83.134.166.168 (talk) 06:18, 27 July 2011 (UTC)

No. You need to assume diagonalizability as well, since any skew symmetric matrix is diagonalizable. If you assume the eigenvalues of a diagonalizable matrix are imaginary and come in conjugate pairs, then the matrix is similar to a skew symmetric matrix. (Hopefully the reason for this is clear.) Sławomir Biały (talk) 10:46, 27 July 2011 (UTC)