# Calculate the volume bounded by surface

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

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

• 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!
– 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.

https://gyazo.com/fb0003801540888e101442e4cc56a320