Find volume of the body given by the following inequalities: $$D= \begin{cases}\left(\frac{x}{5}\right)^6+\left(\frac{|y|}{4}\right)^7 < 1, \\[1ex] \left(\frac{x}{5}\right)^6+\left(\frac{|z|}{3}\right)^\frac{42}{29} < 1. \end{cases} $$
These seem to be two intersecting cylinders of radius $1$.
Substitute $$ x=5u^\frac{1}{3}, \\ y=4v^\frac{2}{7}, \\ z=3w^\frac{29}{21}. $$
Then, multiplying by the absolute value of the Jacobian while moving to the new coordinates, $$ \int \int \int _{D}dx \ dy \ dz = \int z'_w \ dw \int y'_v \ dv \int x'_u \ du = \int dz(w) \int dy(v) \int dx(u) . $$
$$D'= \begin{cases}u^2+v^2 < 1, \\[1ex] u^2+w^2 < 1. \end{cases} $$
To calculate, we need to bother with the bounds:
$$ \int_{-1}^1 dx(u) \int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}} dy(v) \ \int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}} dz(w) = \int_{-1}^1 2 * 3(1-u^2)^\frac{29}{21*2} dx(u) \int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}} \ dy(v) = \int_{-1}^1 2 * 3(1-u^2)^\frac{29}{21*2} \ * \ 2*4(1-u^2)^\frac{2}{7*2} dx(u), $$
and I am not sure if what I have done above was legal. Either way,
$$ 2^4*3*5*\frac{1}{3}\int_{-1}^{1} (1-u^2)^\frac{5}{6}u^{-\frac{2}{3}} du, $$
$$u:=\sqrt{t},$$
$$ \int_{0}^{1} (1-t)^\frac{5}{6} \ t^{-\frac{1}{3}} \cdot \frac{1}{2\sqrt{t}} \ dt = \frac{1}{2} \int_{0}^{1} (1-t)^\frac{5}{6} \ t^{-\frac{5}{6}} \ dt = \frac{1}{2} \ В \left( \frac{1}{6}, \frac{11}{6} \right). $$
And here I am stuck. In classroom, we only did beta-gamma functions with integers as a way to help in calculation of similar integrals. I have no in-depth knowledge of beta-gamma functions. We'll need to solve a similar problem in class: it's time-restricting, and my approach seems swampy.
Is what was written above technically correct?
How do I calculate the last integral?
Is there a fast way to solve this?
Wanting to calculate the following triple integral:
$$ ||\Omega|| := \iiint\limits_{\Omega} 1\,\text{d}x\,\text{d}y\,\text{d}z $$
where is it:
$$ \Omega := \left\{ (x,\,y,\,z) \in \mathbb{R}^3 : \left(\frac{x}{5}\right)^6 + \left(\frac{|y|}{4}\right)^7 \le 1, \; \left(\frac{x}{5}\right)^6 + \left(\frac{|z|}{3}\right)^{\frac{42}{29}} \le 1 \right\}, $$
for symmetrical reasons we can reduce to the following calculation:
$$ ||\Omega|| = 2^3\iiint\limits_{\Omega_{\text{sym}}} 1\,\text{d}x\,\text{d}y\,\text{d}z $$
where is it:
$$ \Omega_{\text{sym}} := \left\{ (x,\,y,\,z) \in \mathbb{R}^3 : \left(\frac{x}{5}\right)^6 + \left(\frac{y}{4}\right)^7 \le 1, \; \left(\frac{x}{5}\right)^6 + \left(\frac{z}{3}\right)^{\frac{42}{29}} \le 1, \; x \ge 0, \; y \ge 0, \; z \ge 0 \right\}. $$
So, opting for the following coordinate transformation:
$$ \Phi : \begin{cases} x = 5\,u^{1/3} \\ y = 4\,v^{2/7} \\ z = 3\,w^{29/21} \end{cases} \; \; \; \; \; \; \text{with} \; (u,\,v,\,w) \in \left[0,\,1\right] \times \left[0,\,\sqrt{1-u^2}\right] \times \left[0,\,\sqrt{1-u^2}\right] $$
and Jacobian $J_{\Phi} = \frac{1160}{147}\frac{w^{8/21}}{u^{2/3}\,v^{5/7}}$, all this translates into the following calculation:
$$ \begin{aligned} ||\Omega|| & = 8 \int_0^1 \int_0^{\sqrt{1-u^2}} \int_0^{\sqrt{1-u^2}} J_{\Phi}\,\text{d}u\,\text{d}v\,\text{d}w \\ & = \frac{9280}{147} \int_0^1 u^{-2/3}\,\text{d}u \int_0^{\sqrt{1-u^2}} v^{-5/7}\,\text{d}v \int_0^{\sqrt{1-u^2}} w^{8/21}\,\text{d}w \\ & = \frac{320}{7} \int_0^1 u^{-2/3}\,\left(1-u^2\right)^{29/42}\,\text{d}u \int_0^{\sqrt{1-u^2}} v^{-5/7}\,\text{d}v \\ & = 160 \int_0^1 u^{\color{blue}{-2/3}}\,\left(1-u^{\color{red}{2}}\right)^{\color{green}{5/6}}\,\text{d}u \\ & = 160 \int_0^1 \frac{1\color{blue}{-\frac{2}{3}}+\color{red}{2}\cdot\color{green}{\frac{5}{6}}}{1\color{blue}{-\frac{2}{3}}+\color{red}{2}\cdot\color{green}{\frac{5}{6}}}\,u^{-2/3}\,\left(1-u^2\right)^{5/6}\,\text{d}u \\ & = 80 \int_0^1 \left(\frac{1}{3}\,u^{-2/3}\,\left(1-u^2\right)^{5/6} + \frac{5}{3}\,u^{-2/3}\,\left(1-u^2\right)^{-1/6}\,\left(1 - u^2\right)\right)\text{d}u \\ & = 80 \int_0^1 \left(\frac{1}{3}\,u^{-2/3}\,\left(1-u^2\right)^{5/6} - \frac{5}{3}\,u^{4/3}\,\left(1-u^2\right)^{-1/6} + \frac{5}{3}\,u^{-2/3}\,\left(1-u^2\right)^{-1/6}\right)\text{d}u \\ & = 80 \int_0^1 \frac{1}{3}\,\left(1-6\,u^2\right)\,u^{-2/3}\,\left(1-u^2\right)^{-1/6}\,\text{d}u + 80 \int_0^1 \frac{5}{3}\,u^{-2/3}\,\left(1-u^2\right)^{-1/6}\,\text{d}u \\ & = 80\left[u^{1/3}\,\left(1-u^2\right)^{5/6}\right]_{u=0}^{u=1} + 400 \int_0^1 \frac{1}{3}\,u^{-2/3}\,\left(1-u^2\right)^{-1/6}\,\text{d}u \\ & = 400 \int_0^1 \left(1-x^6\right)^{-1/6}\,\text{d}x \\ & = 400 \int_0^{\infty} \left(1+y^6\right)^{1/6}\,\frac{1}{\left(1+y^6\right)^{7/6}}\,\text{d}y \\ & = 400 \int_0^{\infty} \frac{1}{1+y^6}\,\text{d}y \\ & = \color{gold}{\frac{400}{3}\,\pi}\,. \end{aligned} $$