Find the volume of the solid in the first octant bounded by the sphere rho =20, the coordinate...
Question:
Find the volume of the solid in the first octant bounded by the sphere {eq}\displaystyle \rho =20, {/eq} the coordinate planes, and the cones {eq}\displaystyle \phi=\frac{\pi}{6} {/eq} and {eq}\displaystyle \phi=\frac{\pi}{3}. {/eq}
Integrals in Spherical Coordinates:
One way to find the volume of a three-dimensional object is with a triple integral over the space occupied by the object: {eq}V = \iiint_E dV {/eq}
Some triple integrals are easier to integrate in spherical coordinates, especially those with circular or spherical symmetry. To convert an integral into spherical coordinates, we need new bounds of integration for {eq}\rho {/eq}, {eq}\theta {/eq}, and {eq}\phi {/eq}. We also need to rewrite the volume differential: {eq}dV = dx \ dy \ dz = \rho^2 \sin \phi \ d\rho \ d\theta \ d\phi {/eq}
Answer and Explanation: 1
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View this answerWe need three sets of integration bounds. We are bounded by the sphere {eq}\rho = 20 {/eq}. So {eq}0 \le \rho \le 20 {/eq}.
We are also bounded by...
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