<s>[INST] Find the volume of the areas bounded by following: 
*

*$(x^2+y^2+z^2)^2=a^2(x^2+y^2-z^2)$ with $a = const$

*$z=x^2+y^2, z^2=2(x^2+y^2), xy=a^2 , xy=2a^2,x=2y, 2x=y$ and $x > 0, y> 0$


First of all, I have never heard of this first geometric shape defined in (1)
This was given as a homework assignment but we haven't done assignment like this one, just simpler equations, inequalities. Does anyone know how this is done, or know a set of similar problems online that are done that I can read through.. Help very much needed. 
 [/INST] 1.
Start by making the replacement $x' = x/a, y' = y/a, z' = z/a$, then the equation becomes $(x'^2+y'^2+z'^2)^2=(x'^2+y'^2-z'^2)$, and $dx\,dy\,dz = a^3\,dx'\,dy'\,dz'$, so the volume is related by $V = a^3V'$. We can assume $a=1$, and get the more general case by multiplying the volume by $a^3$.
Next, switch to cylindrical coordinates: $$x = r\cos \theta\\y= r\sin\theta$$ Then $dV = r\,dr\,d\theta\,dz$ and the volume is
$$V = \iiint dV = \iiint r\,dr\,d\theta\,dz$$
The equation defining the volume becomes $$(r^2 + z^2)^2 = r^2 - z^2$$. Note that does not involve $\theta$ at all, so $\theta$ is independent of $r$ and $z$. Hence its limits are $[0, 2\pi]$ and we can integrate it out:
$$V = 2\pi\int_z\int_r rdr\,dz$$
If we do the integration on $r$ first, holding $z$ constant, then integrate $z$, we can make this substitution: $u = r^2 + z^2$, which gives $u^2 = u - 2z^2$. So $du = 2rdr$ and the limits on $u$ are $$u_l=\frac12 - \frac{\sqrt{1 - 8z^2}}2 \le u \le \frac12 + \frac{\sqrt{1 - 8z^2}}2 = u_u$$
And this also provides the limits on $z$, which must have $8z^2 \le 1$. So $$\begin{align}V &= 2\pi\int_z\int_{u_l}^{u_u} \frac12du\,dz\\&=\pi\int_zu_u - u_l\,dz\\&=\pi\int_{-\sqrt{2}/4}^{\sqrt{2}/4}\sqrt{1 - 8z^2}\,dz\end{align}$$
I'll leave the rest to you.


*The first graph is of $r= \sqrt{x^2 + y^2}$ vs $z$. The second is the $x-y$ plane (where $a = 1$). If you take the solid of revolution of the first about the $z$-axis, and the intersect it with the prism based on the second extended orthogonally up and down for all $z$, the resultant volume is what you are calculating.
Now making the cylindrical substitutions into the various restrictions gives:


*

*$z = r^2$

*$z^2 = 2r^2$ (this is a problem - maybe your prof mean $z = 2r^2$, not $z^2 = 2r^2$?)

*$r^2\sin\theta\cos\theta = a^2 \implies r^2\sin 2\theta = 2a^2$

*$r^2\sin\theta\cos\theta = 2a^2 \implies r^2\sin 2\theta = 4a^2$

*$r\cos\theta = 2r\sin\theta \implies \tan\theta = 1/2$

*$2r\cos\theta = r\sin\theta \implies \tan\theta = 2$


Which gives us the following limits:


*

*$r^2 \le z \le \sqrt2r$

*$\frac{\sqrt2a}{\sqrt{\sin 2\theta}} \le r \le \frac{2a}{\sqrt{\sin 2\theta}}$

*${1\over2} \le \tan \theta \le 2$


The volume is given by $$V = \iiint dV =\iiint dz\,rdr\,d\theta$$
Now $$\int_{r^2}^{\sqrt2r}dz = r^2 - \sqrt2r$$
$$\int r^3 - \sqrt2r^2\,dr = \left.\frac{r^4}4 - \frac{\sqrt2}{3}r^3\right|_{\sqrt2a/\sqrt{\sin 2\theta}}^{2a/\sqrt{\sin 2\theta}} = \frac{3a^4}{\sin^22\theta} -\frac{(8\sqrt2-4)a^3}{3\sin^{3/2}2\theta}$$
(note that the power of $r$ in the integral increases because of the extra $r$ already there from the Jacobian). 
The $\sin^22\theta$ term can be integrated easily, but the term in $\sin^{3/2}2\theta$ does not have a nice antiderivative. In fact Wolfram Alpha expresses in terms of the elliptic integral of the 2nd kind. 
I strongly suspect there are errors in the statement of the problem. If you didn't make them, your professor did. There is no nice closed form answer.
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