So i have the object defined as:
$$(x^2+y^2+z^2)^2=x^2-y^2, x\geq0$$
Where density is defined as: $$\rho=\sqrt{(x^2+y^2+z^2)}$$
So i have to calculate mass:
$$M=\iiint_{G} \rho dV$$
So using: $$x=r\cos\phi \cos\theta$$ $$y=r\sin\phi \cos\theta$$ $$z=r\sin\theta$$
I bounded $r$ as $r\in[0,\cos\theta\sqrt(\cos2\phi)]$
Since: $x\geq0$ Therefore : $r\cos\theta\cos\phi \geq0$
Is it okay to bound $\phi\in[-\frac{\pi}{2},\frac{\pi}{2}]$ And then to look at $\theta$ with the the nonequality above so that it is $\geq0$
So i will just have to calculate $$\iiint_{G} r^3cos\theta dV$$ for when i have the correct bounds right?
I just want some help getting the bounds. Thank you in advance.
Note that the equation defining the region requires you to have $x^2-y^2\ge 0.$ This, together with the given condition $x\ge 0$ gives $x\ge |y|,$ or $$-x\le y\le x.$$ It remains to find bounds on $z.$ Solving for $z^2$ gives $$z^2=-(x^2+y^2)+\sqrt {x^2-y^2}.$$ Taking square roots gives you the bounds on $z.$ But this requires that $-(x^2+y^2)+\sqrt {x^2-y^2}\ge 0.$ Studying this inequality should either reinforce what you have before or tell you something new. Also, setting $x=0$ in the original equation shows that the region does not intersect the $yz$ plane, so it lies entirely in the half-space facing you in normal orientation of $xyz$-axes.