2

Calculate the volume of the area enclosed by the surface \begin{equation} x^{(2/3)}+y^{(2/3)}+z^{(2/3)} = a^{(2/3)} \end{equation} where $a > 0$ is a constant.

However I'm not sure where to begin.

Nosrati
  • 29,995
  • I would try looking at the image of the space under the diffeomorphism $f(x,y,z) = (x^3, y^3, z^3)$, as the image of the set under that map is just the usual unit ball, and see if the Jacobian of it isn't too bad to deal with. (It's not a diffeomorphism at 0, but single points don't matter when integrating.) – Gunnar Þór Magnússon Mar 04 '17 at 09:59

2 Answers2

4

$$ V: (\frac{x_1}{a})^{2/3}+(\frac{x_2}{a})^{2/3}+(\frac{x_3}{a})^{2/3}\le1$$ $$y_i=(\frac{x_i}{a})^{1/3}$$ $$\frac{D(x_1,x_2,x_3)}{D(y_1,y_2,y_3)}=27a^3y_1^2y_2^2y_3^2$$ $$V=\int_V\mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3=\int27a^3y_1^2y_2^2y_3^2\mathrm{d}y_1\mathrm{d}y_2\mathrm{d}y_3$$ $$\left\{ \begin{array}{ccc} y_1&=&r\cos(\theta_1)\\ y_2&=&r\sin(\theta_1)\cos(\theta_2)\\ y_3&=&r\sin(\theta_1)\sin(\theta_2) \end{array} \right .$$ $$\frac{D(y_1,y_2,y_3)}{D(r,\theta_1,\theta_2)}=r^2\sin(\theta_1)$$ So $$V=\int^{\pi/2}_0\mathrm{d}\theta_1\int^{\pi/2}_0\mathrm{d}\theta_2\int_0^1 27a^3(r\cos(\theta_1))^2(r\sin(\theta_1)\cos(\theta_2))^2(r\sin(\theta_1)\sin(\theta_2))^2r^2\sin(\theta_1)\mathrm{d}r\\ =\int^{\pi/2}_0\mathrm{d}\theta_1\int^{\pi/2}_0\mathrm{d}\theta_2\int_0^1 27a^3r^8\cos^2(\theta_1)\sin^5(\theta_1)\cos^2(\theta_2)\sin^2(\theta_2)\mathrm{d}r$$ $$V=3a^3\frac{\Gamma(3/2)\Gamma(6/2)}{2\Gamma(9/2)}\frac{\Gamma(3/2)\Gamma(3/2)}{2\Gamma(6/2)}=3/4a^3\frac{(\Gamma(3/2))^3}{\Gamma(9/2)}=\frac{a^3}{70}\pi$$

xyz
  • 701
1

I want to remind you the answer above is incorrect, altough almost correct. The volume calcuted there is 1/8'th of the actual volume, because 'xyz' lets both angles only range from $0$ to $\pi/2$, which means he only accounts for the 1 first octant.

Because the volume is symmetrical in all three axis, you can just multiply the outcome by 8, or calculate the integral \begin{align*} &3a^3\int_0^{\pi}\int_0^{2\pi}\cos^2\theta\sin^2\theta\cos^2\varphi\sin^5\varphi \text{ d}\theta\text{ d}\varphi\\ ={}& \dfrac{4a^3}{35}\pi \end{align*}

(PS See you in college tonight ;))

Rich_Rich
  • 430