Finding Volume Inside Sphere and Cone: Limits of Integration Explained

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SUMMARY

The discussion focuses on calculating the volume inside both the sphere defined by the equation x² + y² + z² = 1 and the cone described by z = √(x² + y²). The solution involves triple integration in spherical coordinates, yielding a volume of (2/3)π(1 - 1/√2). The limits of integration are established as θ = (0, 2π), φ = (0, π/4), and ρ = (0, 1), with φ's upper limit determined by the intersection of the sphere and cone at z = 1/√2, leading to a zenith angle of π/4.

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Homework Statement


The textbook I have is horrible in explaining this stuff in the chapter so please bare with me.

So we have the information: Find the volume V inside both the sphere x^2 + y^2 + z^2 = 1 and the cone z =sqrt(x^2 + y^2)


Homework Equations


Where the heck do I began? I know this will require triple integration in spherical coordinates and the solution is (2/3)*pi*1^3(1-1/sqrt(2))



The Attempt at a Solution


I know the theta integrand is (0,2pi) because there are no bounds, but I have no idea where to find the limits of integration for the phi and zenith integrands. :(

EDIT (1): Just solved for the intersections of the sphere and cone and got z = 1/sqrt(2))
EDIT (2): Since phi = 1 and z = (1/sqrt(2)), it's safe to conclude that the zenith angle is pi/4...but what would be it's upper limit?
EDIT (3): I believe the limits for phi is 0 to pi/4. But I'm not quite sure why though? I think I got the integrands all figured out. I believe it's: for p = (0,1), phi = (0, pi/4) and theta = (0,2pi).

So overall, could someone just briefly explain to me why phi's limits of integration is from 0 to pi/4. thanks a bunch :D
 
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Looking at the figure from the side (along the negative y-axis), you should see a circle and a "V" (y= 0 so the equations are [itex]x^2+ z^2= 1[/itex], a circle, and [itex]z= \sqrt{x^2}= |x|[/itex]). The "co-latitude", [itex]\phi[/itex], goes from directly up the z-axis, [itex]\phi= 0[/itex], to the line z= x, which bisects the right angle between the x and z axes and so makes angle [itex]\pi/4[/itex] with the z-axis.
 
stratusfactio said:

Homework Statement


The textbook I have is horrible in explaining this stuff in the chapter so please bare with me.

I'm sorry. Undressing is forbidden in this forum. :eek:
 

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