How to find the volume of the unit sphere above $z=1/\sqrt2$?

I have managed to find the limit for $\rho$ as $(1/\sqrt2) \sec \phi$ to $1$ and for $\theta$ as $0$ to $2\pi$

but I can't figure out the limits for $\phi$.

Can anyone help?

This is the problem by the way:


In spherical coordinates, the region can be written as $$ E=\{(\rho,\theta,\phi)\;|\; 0 \le \theta \le 2\pi , \frac{1}{\sqrt{2}\cos\phi}\le \rho \le 1,0 \le \phi\le \cos^{-1}\left( \frac{1}{\sqrt{2}}\right)(=\frac{\pi}{4})\} $$ It follows that the volume equals $$ \iiint_E \rho^2\sin \phi\; dV = \frac{2\pi}{3}-\frac{\pi}{\sqrt{2}}+\frac{\pi}{6\sqrt{2}} $$


  • To specifically answer your question, the upper bound for $\phi$ comes from equations $$ z=\frac{1}{\sqrt{2}}=\rho \cos \phi, \quad \rho=1 $$
  • Cylindrical coordinates are also a good option here: $$ E=\{(r,\theta,z)\;|\; 0 \le \theta \le 2\pi , 0\le r\le \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \le z \le \sqrt{1-r^2} \} $$

With a generic sphere of radius R,if you slice perpendicular to the z axis the slices are circles of radius $\sqrt{R^2-z^2}$ and the area of each slice is $\pi(R^2-z^2)$

Volume of a sphere = $\pi\int_{-R}^R (R^2 - z^2) dz$

The cap of the unit sphere above $z = \dfrac{1}{\sqrt2}$

$\pi \int_{\frac{1}{\sqrt2}}^1 (1 - z^2) dz$

So, you still want to do a triple integral...

$\int_0^{2\pi}\int_0^{\pi/4}\int_{\frac{\sec\phi}{\sqrt2}}^{1} \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta$

$\int_0^{2\pi}\int_0^{\pi/4}\frac{1}{3}\rho^3 \sin\phi\, d\phi\, d\theta|_{\frac{\sec\phi}{\sqrt2}}^{1}$

$\int_0^{2\pi}\int_0^{\pi/4}\frac{1}{3}\sin\phi\ (1-\frac{1}{2^{3/2}\cos^3\phi})\, d\phi\, d\theta$

$\int_0^{2\pi}\int_0^{\pi/4}\frac{1}{3}\sin\phi\ d\phi\, d\theta$ -$\int_0^{2\pi}\int_0^{\pi/4}\dfrac{\sin\phi}{2^{3/2}\cos^3\phi}\, d\phi\, d\theta$

and each of those should be pretty straightforward.

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