Answer:
The answer is "[tex]\frac{\pi a^3}{9}[/tex]".
Step-by-step explanation:
please find the complete question in the attached file.
Let the sphere center be (0,0,0) and then let the intersection diameter lie all along the z-axis.
So one of the collision plans is the xz-plane the other is the path via an xz-plane angle.
[tex]\theta = \frac{\pi}{6}[/tex]
All appropriate region could then be indicated in spherical coordinates
[tex]E= {(\rho, \theta, \phi) : 0 \geq \rho \geq a, 0 \geq \theta \geq \frac{\pi}{6}, 0 \geq \phi \geq \pi }[/tex]
Calculating the volume:
[tex]\to v(E)=\int \int_{E} \int dV\\[/tex]
[tex]=\int_{0}^{\frac{\pi}{6}} \int_{0}^{\pi} \int_{0}^{a} \rho^2 \sin \phi d \rho d \phi d \theta\\\\ =\int_{0}^{\frac{\pi}{6}} d \theta \int_{0}^{\pi} \sin \phi d \int_{0}^{a} \rho^2 d \rho\\\\= [\theta]^{\frac{\pi}{6}}_{0} [-\cos \phi]^{\pi}_{0} [\frac{\rho^3}{3}]^{a}_{0}\\\\= \frac{\pi}{6} [1+1] \frac{a^3}{3}\\\\=\frac{\pi a^3}{9}[/tex]
