Triple integral (spherical coordinate)

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SUMMARY

The discussion focuses on calculating the volume of the solid D that lies above the cone defined by the equation z = (x² + y²)^(1/2) and below the sphere z = (x² + y² + z²). The integration process involves substituting cos(φ) = u, with the upper limit for u being 1/√2 when φ ranges from 0 to π/4. Additionally, the volume calculation requires considering the intersection of the sphere, centered at (0,0,1/2) with a radius of 1/2, and the cone at z = 1/2. The integration must account for both the hemisphere and the cone to accurately determine the volume.

PREREQUISITES
  • Spherical coordinates and their applications in volume integration
  • Understanding of the equations of cones and spheres in three-dimensional space
  • Knowledge of integration techniques, particularly in polar and spherical coordinates
  • Familiarity with the concept of limits in integration
NEXT STEPS
  • Study the method of triple integrals in spherical coordinates
  • Learn how to set up and evaluate integrals involving cones and spheres
  • Explore the concept of volume calculation using polar coordinates as a comparison
  • Review the process of changing variables in integrals, specifically trigonometric substitutions
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on calculus and geometric applications, as well as educators teaching integration techniques in three-dimensional contexts.

naspek
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find the volume of the solid D that lies above the cone z = (x^2 + y^2)^1/2
and below the sphere z = (x^2 + y^2 + z^2)

i've done the integration until i need to substitute cos phi = u..
however.. i don't know to change the range..

http://imageshack.us/photo/my-images/839/spherical.jpg/"
 
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[Edit] Never mind the below paragraph. Your work is correct; I misread the equation at first. All you have to do is put in the correct u limit in your last equation. Since u = cos(φ) and φ goes from 0 to pi/4, your upper u limit is 1/sqrt(2). Be careful with your lower limit for u, it isn't 0. Just finish your integration.I don't think your integral is set up correctly. I see a sphere centered at (0,0,1/2) with radius 1/2 intersecting the cone at z = 1/2. So you apparently want the volume under the top half of the sphere but above the cone. Are you working this in two parts, the hemisphere and the cone with a flat base? If so, for the cone, ρ goes from 0 to z=1/2 which is ρcos(φ)=1/2 or ρ = (1/2)sec(φ). And you have to add the hemisphere.
 
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