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natali 33 [55]
3 years ago
7

Find the volume of a smaller wedge cut from a sphere of radius 66 by two planes that intersect along a diameter at an angle of π

/6
Mathematics
1 answer:
djverab [1.8K]3 years ago
4 0

Answer:

The answer is "\frac{\pi a^3}{9}".

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.

\theta = \frac{\pi}{6}

All appropriate region could then be indicated in spherical coordinates

E= {(\rho, \theta, \phi) : 0 \geq \rho \geq a, 0 \geq \theta \geq \frac{\pi}{6}, 0 \geq \phi \geq \pi }

Calculating the volume:

\to v(E)=\int \int_{E} \int dV\\

            =\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}

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