Find the volume of a rotationally symmetric 3D body

Question

let $S$ be the set of points $(x,y,z)$ satisfying $ x^2 + y^2 + z^2 = 1 $ and $ 0\leq z \leq \frac{1}{\sqrt{2}} $ let $D$ be the $ 3 D $ body obtained by taking the union of all segments connecting $(0,0,0)$ to the points of $S$ Find the volume of the body.

Here is my trial :

Let $g(t) = (0,0,0) + t(x,y,z)$ such that t $ \in [0,1]$ by the giving $ S $ in the question we get that : $ D =$ {$0 \leq x^2 + y^2 + z^2 \leq 1$}

so we want to do triple integral on $ D$ to find the volume : $$ \int_0^{2\pi} \int_{\pi/4}^{\pi/2} \int_0^1 r^2 \sin(\phi) {\ dr} \ d{\phi }\ d{\theta} $$

the thing is when i move to spherical coordinates i get different answer than the basic coordinates .

Spherical Answer : $ \frac{\sqrt{2}\pi}{3} $

x,y,z coordinates Answer $ \frac{5\pi}{6\sqrt{2}} $

Answer 1

The integral in cartesian co-ordinates looks like this: $I=I_1-I_2$

$\displaystyle I_1=\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\sqrt{1-x^2-y^2}\ dx\ dy=2\pi/3\\I_2=\displaystyle\int_{-\frac1{\sqrt2}}^{\frac1{\sqrt2}}\int_{-\sqrt{\frac12-y^2}}^{\sqrt{\frac12-y^2}}\sqrt{1-x^2-y^2}-\sqrt{x^2+y^2}\ dx\ dy$

$I_1$ is just the volume of the hemsiphere. $I_2$ gives the volume of the inverted cone with spherical base that needs to be subtracted from $I_1$.

The answer obtained is the same as the one produced by the spherical integral: $\sqrt2\pi/3$.

Answer 2

$$\int\limits_{z=0}^{1/\sqrt{2}} \pi \left( (1 - z^2) - z^2 \right)\ dz = \int\limits_{z=0}^{1/\sqrt{2}} \pi (1 - 2 z^2)\ dz = \frac{\sqrt{2} \pi }{3}$$

Annular disk of inner radius $z$ and outer radius $\sqrt{1 - z^2}$ and thickness $dz$ (in blue).