$B = \left\lbrace \, (x, \; y, \; z) \; \left| \; \; \; \rule{0pt}{12pt} \right. (3 x + 3 y + z)^2 \;+\; (3 x + y + 3 z)^2 \;+ \; ( x + 3 y + 3 z)^2 \; \leq \; 4 \; \right\rbrace$
Find the volume of $B$.
My attempt :
$u = 3 x + 3 y + z, v = 3 x + y + 3 z, w = x + 3 y + 3 z \implies u^2+v^2+w^2\leq 4$.
The jacobian is $|J|=28$ hence:
$u=r\cos\theta \sin \varphi, v=r\sin\theta \sin\varphi, w=r\cos\phi$
The jacobian is $r^2\sin \varphi$.
$\displaystyle \iiint_{B} \mathrm dx\,\mathrm dy\,\mathrm dz =\iiint 28 \,\mathrm du\,\mathrm dv\,\mathrm dw = \int _{-2}^2\:\int _0^{2\pi }\:\int _0^{\frac{\pi }{2}}\:28r^2\sin\phi \:\mathrm d\phi \:\mathrm d\theta \:\mathrm dr $
My solution isn't correct and I can't figure out.
Any help is welcome , thanks !
If you use the jacobian transformation correctly, what you would have is :-
$$\int_{u^{2}+v^{2}+w^{2}\leq 4}\frac{1}{28}\,dudvdw$$
Which is $$\frac{1}{28}\cdot\text{Volume of sphere of radius 2}=\frac{4\cdot 2^{3}\pi}{3\cdot28}=\frac{8\pi}{21}$$
If you want to use spherical polar coordinates to calculate the volume , then the limits for integral would be $0\leq r\leq 2$ , $0\leq \theta<2\pi$ and $0\leq \varphi \leq \pi$
I suggest you look at the below picture to see why it is the case:-