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