Calculating volume using multiple integrals

Question

So I have this restriction $$B=\{(x,y,z)\,\,|\,\, (4x+4y+2z)^2+(4x+2y+4z)^2+(2x+4y+4z)^2\leq1\}$$ so I am trying to do integration by substitution: $$u=4x+4y+2z$$ $$v=4x+2y+4z$$ $$w=2x+4y+4z$$ $$\left|\det\left(\frac{\partial(u,v,w)}{\partial(x,y,z)}\right)\right|=40$$ so $$u^2+v^2+w^2\leq1$$ but now I am not sure how to proceed with $$\int\int\int_Bdxdydz = \int\int\int_{newB}(something)dudvdw$$ because I have to use this 40 somehow, but I am not entirely sure how. Please help! *edit: the answer is : $$\int\int\int_Bdxdydz=\int\int\int_{newB}\frac 1 {40} dudvdw = \frac \pi {30}$$

Answer 1

$$B=\{(x,y,z)\,\,|\,\, (4x+4y+2z)^2+(4x+2y+4z)^2+(2x+4y+4z)^2\leq1\}$$ If you consider the change of variable:

$$u=4x+4y+2z$$

$$v=4x+2y+4z$$

$$w=2x+4y+4z$$ You find that: \begin{cases} x=\dfrac{6}{5}u-\dfrac{4}{5}v-\dfrac{3}{10}w\\ y=-\dfrac{4}{5}u+\dfrac{7}{10}v+\dfrac{1}{5}w \\ z=-\dfrac{3}{10}u+\dfrac{1}{5}v+\dfrac{1}{5}w \end{cases} When you compute the Jacobian associated to this diffeomorphism you obtain: $$\det (J_{\varphi})=\dfrac{1}{40} $$ Now, the new domain is: $$B'=\{(u,v,w)\in \mathbb{R}^3 \,|\, u^2+v^2+w^2\leq1\}$$ It follows that $$\iiint_B\mathrm{d}x\mathrm{d}y\mathrm{d}z=\iiint_{B'}\frac 1 {40} \mathrm{d}u\mathrm{d}v\mathrm{d}w = \frac{1}{40} \text{Vol}(B')=\frac{\pi}{30}$$

Answer 2

Let the linear map $T$ given by $$ T(x,y,z):=4\left(x+y+\frac1{2}z,x+\frac1{2}y+z,\frac1{2}x,y,z\right)=(u,v,w) $$ Therefore $T$ can be represented by the matrix $$ [T]:=4\begin{bmatrix} 1&1&\frac1{2}\\1&\frac1{2}&1\\\frac1{2}&1&1 \end{bmatrix}\implies \det [T]=4^3\cdot \left(\frac{3}{2}-2-\frac1{8}\right) =-4^3\cdot \frac{5}{8}=-40 $$ so $T$ is invertible, and we have that

$$ \begin{align*} \int_{A}d(u,v,w)&=\int_{A}d(T(x,y,z))\\&=\int_{T^{-1}(A)}|\det \partial T(x,y,z)|d(x,y,z)\\&=\int_{T^{-1}(A)}|\det [T]|d(x,y,z)\\ &=40 \int_{B}d(x,y,z) \end{align*} $$ as when $T$ is linear we have that $\partial T\mathbf{v}=T$ for every vector $\mathbf{v}$, and in this case we have that $T^{-1}(A)=B$ for $A:=\mathbb{B}((0,0,0),1)$.∎