Understanding the 4-Ball: ||x|| ≤ r

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Homework Statement



What is the x-simple description of a 4-ball = { ||x|| \leq r }

Homework Equations



It's a 4-ball, so isn't the equation x^{2} + y^2 +z^2 +w^2 ?

The Attempt at a Solution


For my limits, I got x is between \pm \sqrt{1-z^2} and y is between -1 and 1, and z is between \pm \sqrt{1-x^2-y^2}, and w is between \pm \sqrt{1-x^2-y^2-z^2} Is this right? Thanks!
 
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buuumppp...please help! Any advice is good!
 
stanford1463 said:

Homework Statement



What is the x-simple description of a 4-ball = { ||x|| \leq r }

Homework Equations



It's a 4-ball, so isn't the equation x^{2} + y^2 +z^2 +w^2 ?
That couldn't possibly be the equation. An equation always states that two expressions have the same value. I see only one expression, and even more to the point, I don't see an equals sign.

What does x-simple description mean?
stanford1463 said:

The Attempt at a Solution


For my limits, I got x is between \pm \sqrt{1-z^2} and y is between -1 and 1, and z is between \pm \sqrt{1-x^2-y^2}, and w is between \pm \sqrt{1-x^2-y^2-z^2} Is this right? Thanks!

Where did the 1 come from?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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