Calculate the volume bounded by the surface $$(x^2+y^2+z^2)^2 = a^2(x^2+y^2-z^2)$$

Using the spherical coordinates $$\begin{cases} x = rcos\varphi cos\theta & \\ y = rsin\varphi cos\theta \\z = rsin\theta \end{cases}$$ and substituting those into the original equation we get $$r^2 = a^2(cos^2\theta - sin^2\theta) = a^2cos2\theta$$ and from that $$0 \leq r \leq a\sqrt{cos2\theta}$$

Calculating Jacobian gives us $$ J = r^2cos\theta$$

Given all that the target volume could be calculated as

$$V = \int_{0}^{2\pi}\,d\varphi \int_{0}^{\pi}\,d\theta \int_{0}^{a\sqrt{cos2\theta}}\,r^2cos\theta dr$$

But this yields incorrect result, moreover the supposed answer should be calculated given the following integral $$V = 8\int_{0}^{\frac{\pi}{2}}\,d\varphi \int_{0}^{\frac{\pi}{4}}\,d\theta \int_{0}^{a\sqrt{cos2\theta}}\,r^2cos\theta dr $$

But I have trouble understanding where do the integrating boundaries for $\varphi$ and $\theta$ come from.

I undesrtand that given the fact that the surface and therefore target solid are symmetrical, we can integrate over a certain part of the solid and then multiply the result by a proper constant, but if we use the following bounds for $\varphi$ and $\theta$

$$0 \leq \varphi \leq \pi/2 \\ 0 \leq \theta \leq \pi/4$$

how come we multiply by 8 and not by 16?

Any tips on what I'm doing wrong ?


Note the plot of $r=a\sqrt{\cos2\theta}$ below

enter image description here

with range $\theta \in [-\frac\pi4, \frac\pi4]$. Thus, the volume integral is set up as $$V = \int_{0}^{2\pi}\,d\varphi \int_{-\frac\pi4}^{\frac\pi4}\,d\theta \int_{0}^{a\sqrt{cos2\theta}}\,r^2\cos\theta dr\\ =8 \int_{0}^{\frac\pi2}\,d\varphi \int_{0}^{\frac\pi4}\,d\theta \int_{0}^{a\sqrt{cos2\theta}}\,r^2\cos\theta dr $$

  • $\begingroup$ I see, for some reason I thought theta ranged from 0 to pi and not from -pi/4 to pi/4 Such a simple blunder, thank you very much! $\endgroup$
    – toss
    Dec 5 '20 at 23:32

If you multiply by $16$, you would get all possible values of $\theta$, from $0$ to $\pi$. What is the limit of integration for $r$ when $\theta=\pi/2$? You get $\cos(2\theta)=-1$, so the square root is not defined (in real numbers). In fact for $\theta\in(\pi/4,\pi/2)$ your upper limit is not defined.


The reason is that $$\int_{0}^{\frac{\pi}{2}}d\varphi\int_{0}^{\frac{\pi}{4}} d\theta\int_{0}^{a\sqrt{\cos (2\theta)}} r^2\cos\theta dr$$

Is only in the first octant. We multiply by 8 to cover all 8 octants. The screenshot of the graph will make it obvious that your $\theta$ bounds run from $0$ to $\frac{\pi}{4}$ if you run through the first octant, and the bounds you specify are all 8 octants.



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