Stumped on $\int_{-1}^1\int_{-\sqrt{1-x^2}}^\sqrt{1-x^2} \int_{x^2+y^2}^{2-x^2-y^2} dz\;dy\;dx$

Question

I cannot express the following integral in spherical coordinates. It as though I am finding the volume between the solids. Any help will be appreciated.

$$\int_{-1}^1\int_{-\sqrt{1-x^2}}^\sqrt{1-x^2} \int_{x^2+y^2}^{2-x^2-y^2} dz\;dy\;dx$$

I tried to sum different parts, but to no avail.

Answer 1

The domain of integration is equivalent to $$ \left\{ \begin{aligned} 0 &\leq r \leq 1 \\ 0 &\leq \theta \leq 2\pi \\ r^2 &\leq z \leq 2 - r^2 \end{aligned} \right. $$ in cylindrical coordinates $(r, \theta, z)$, where the volume form transforms as $$ \mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z = r\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}z. $$

Answer 2

HINT

Here is the plot of the intersection of the two surfaces (made using Mathematica):

Based on it, we can observe that a cylindrical change of coordinates would be interesting.

Indeed, one has that: \begin{align*} \int_{-1}^{1}\int_{-\sqrt{1 - x^{2}}}^{\sqrt{1 - x^{2}}}\int_{x^{2} + y^{2}}^{2 - x^{2} - y^{2}}\mathrm{d}z\mathrm{d}y\mathrm{d}x & = \int_{0}^{1}\int_{0}^{2\pi}\int_{r^{2}}^{2 - r^{2}}r\mathrm{d}z\mathrm{d}\theta\mathrm{d}r \end{align*}

Can you take it from here?

Answer 3

Ok, first say that if you want to calculate the volume, it must be done using cylindrical coordinates, because in spherical coordinates it is very ugly.

The easiest thing here is to find $\theta$, this is $$ 0 \leq \theta \leq 2 \pi $$

now to find $\phi$ and $\rho$ you have to separate the integrals into 2 parts. To understand why, let's see what the surfaces look like in the $xz$ plane.

then here it separates at an angle into two $$ 0 \leq \phi \leq \pi/4 , \quad \pi/ 4 \leq \phi\leq \pi/2 $$

for $0 \leq \phi\leq \pi/4$ we must find $\rho$, in this case

\begin{align*} z = 2 - x^2 - y^2 \Rightarrow \rho^2 \sin^2(\phi) + \rho \cos(\phi) - 2 = 0 \end{align*}

solving the quadratic equation we are left with $$ A(\phi) = \rho = \frac{-\cos \phi\pm \sqrt{\cos^2 \phi + 8\sin^2 \phi}}{2\sin^2 \phi} \implies A(\phi) = \rho = \frac{-\cos \phi + \sqrt{\cos^2 \phi + 8 \sin^2 \phi}}{2\sin^2 \phi} > 0 $$

then the first integral is

$$ \int_{0}^{2\pi} \int_{0}^{\pi /4} \int_{0}^{A(\phi)} \rho^2 \sin(\phi)\,d\rho\,d\phi\,d\theta $$

now, for $\pi/4 \leq \phi\leq \pi/2$ we must find $\rho$, in this case \begin{align*} z = x^2+ y^2 \Rightarrow \rho \cos(\phi) = \rho^2 \sin^2(\phi) \Rightarrow \rho^2 \sin^2(\phi) - \rho \cos(\phi) = 0 \Rightarrow \rho = \frac{\cos(\phi)}{\sin^2(\phi)} \end{align*}

then the second integral is $$ \int_{0}^{2\pi} \int_{\pi/4}^{\pi/2} \int_{0}^{\frac{\cos(\phi)}{\sin^2(\phi)}} \rho^2 sin(\phi)\,d\rho\,d\phi\,d\theta $$ Finally, the volume is $$ V = \int_{0}^{2\pi} \int_{0}^{\pi /4} \int_{0}^{A(\phi)} \rho^2 \sin(\phi)\,d\rho\,d\phi\,d\theta + \int_{0}^{2\pi} \int_{\pi/4}^{\pi/2} \int_{0}^{\frac{\cos(\phi)}{\sin^2(\phi)}} \rho^2 \sin(\phi)\,d\rho\,d\phi\,d\theta $$

Finally we can enter everything into Wolfram and verify that everything is fine

where indeed the sum of the last two integrals is $\pi$.