Two spheres, triple integration, not their intersection

Question

Two spheres, one of radius $1$ and one of radius $\sqrt{2}$, have centres that are $1$ unit apart. Find the volume of the smaller region that is outside one sphere and inside the other.

Can use either spherical or cylindrical coordinates. The $2$ spheres can be anywhere in three dimensional space. Apparently using the correct coordinates will lead into an easy integration. Been tossing around the two and haven't seen any of them working out to be easy.

Answer 1

You can avoid integrals.

1) choose the coordinates such as the two sphere have equations $x^2+y^2+z^2=2$ and $(x-1)^2+y^2+z^2=1$

2) By simmetry, intersecting the circumpherences $x^2+y^2=2 $ and $(x-1)^2+y^2=1$ , you can find the radius $a$ of the two spherical cups that delimit your volume.

3) find the volumes of the cups by the formula $$ V=\frac{\pi h}{6}\left(3a^2+h^2 \right) $$ where the height $h$ of the cups can be easily find by the geometry

For an anologous problem see: Calculate volume of intersection of 2 spheres

Answer 2

The region whose volume $V$ is to be found is obtained by revolving the shaded area around the $y$-axis. For $x$ between $-1$ and $1$, the upper bounding curve has equation $y=\sqrt{1-x^2}$, while the lower bounding curve has equation $y=\sqrt{2-x^2}-1$. Using shell integration we may calculate $V$: $$V=2\pi\int_0^1x(\sqrt{1-x^2}-\sqrt{2-x^2}+1)\ dx$$ $$=2\pi\int_0^1x\sqrt{1-x^2}-x\sqrt{2-x^2}+x\ dx$$ $$=2\pi\left[-\frac13(1-x^2)^{3/2}+\frac13(2-x^2)^{3/2}+\frac12x^2\right]_0^1$$ $$=2\pi\left(\frac56-\frac{2\sqrt2-1}3\right)$$ $$=\frac{(7-4\sqrt2)\pi}3$$

Answer 3

$x^2 + y^2 + z^2 = 2\\ x^2 + y^2 + (z-1)^2 = 1$

Would be an equation for two spheres with centers at (0,0,0) and (0,0,1) (1 unit apart), one of radius $\sqrt2$, and one of radius 1

Expand and subtract one from the other to find the the points of intersection.

$2z - 1 = 1\\ z = 1$

$x^2 +y^2 = 1$

Would you rather do this in spherical or cylindrical coordinates?

In cylindrical:

$x = r \cos \theta\\ y = r \sin \theta\\ z = z\\ dx\;dy\;dz = r \;dz\;dr\;d\theta$

Make these substitutions into the equations of the two spheres, and the circle formed by their intersection to find the limits of integration.

$z = \sqrt {2 - r^2}\\ z = \sqrt {1-r^2}+1\\ r = 1$

$\int_0^{2\pi}\int_0^{1}\int_{\sqrt {2-r^2}}^{\sqrt {1 - r^2}+1} r \;dz\;dr\;d\theta$

In spherical:

$x = \rho \cos \theta\sin\phi\\ y = \rho \sin \theta\sin\phi\\ z = \rho \cos\phi\\ dx\;dy\;dz = \rho^2 \sin\phi \;d\rho\;d\phi\;d\theta$

The limits tend to be harder to find in spherical, but the integrations tend to be easier.

Update________

The equations of the spheres, and the circle where the two intersect.

$x^2 + y^2 + z^2 = 2\\ x^2 + y^2 + (z-1)^2 = 1\\ z= 1, x^2 + y^2 = 1$

Substitute our new coordinate system.

$\rho^2 \sin^2\phi\cos^2\theta + \rho^2 \sin^2\phi\sin^2\theta + \rho^2 \cos^2\phi = 2\\ \rho^2 \sin^2\phi\cos^2\theta + \rho^2 \sin^2\phi\sin^2\theta + \rho^2 \cos^2\phi - 2\rho\cos\phi + 1 = 1\\ \rho\cos\phi= 1, \rho^2 \sin^2\phi\cos^2\theta + \rho^2 \sin^2\phi\sin^2\theta = 1$

Simplify:

$\rho^2 = 2\\ \rho^2 - 2\rho\cos\phi = 0\\ \rho\cos\phi= 1, \rho^2 \sin^2\phi = 1$

$\rho = \sqrt 2\\ \rho = 2\cos\phi\\ \frac {\rho\sin \phi}{\rho\cos \phi} = \tan \phi= 1\\ \phi = \frac {\pi}{4}$

Now, it may help to sketch these curves. Do it in 2D. Just look at the picture in the xz plane, and that should give you some idea of what is going on for $\rho$ and $\phi.$

\update______

$\int_0^{2\pi}\int_0^{\frac\pi4}\int_{\sqrt 2}^{2\cos\phi} \rho^2 \sin\phi \;d\rho\;d\phi\;d\theta$