Did I set up my integral correctly?

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Homework Help Overview

The original poster seeks to determine the volume inside a sphere of radius 4, outside a sphere of radius 2, and above the cone defined by z=3√(x^2+y^2) using spherical coordinates.

Discussion Character

  • Exploratory, Assumption checking, Mixed

Approaches and Questions Raised

  • The original poster attempts to set up the integral for the volume calculation and discusses the intersection of the cone and the sphere. Some participants question the limits of integration, suggesting a different approach for determining φ.

Discussion Status

Participants are actively discussing the setup of the integral, with some providing guidance on the limits of integration. There is no explicit consensus on the best approach yet, as different interpretations are being explored.

Contextual Notes

There are mentions of preferences for typed solutions over handwritten work, indicating a potential constraint in how solutions are presented in the forum.

qq545282501
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Homework Statement


use spherical coordinates to find volume inside a sphere of radius 4 and outside sphere of radius 2, and inside/above the cone z=3√(x^2+y^2).

Homework Equations


[/B]
x=rcosθ =ρsinφcosθ , y=rsinθ =ρsinφsinθ
z=ρcosφ
r= ρsinφ

The Attempt at a Solution


replacing z^2 in the sphere of radius 4 with the cone z^2=9(x^2+y^2) I get the radius of the intersection of √( 8/5), using this radius and known ρ ( 4), φ=inverse sin of (r/ρ) i found φ to be 0.322 or around 18.4 degrees.

V= ∫ 0 to 2 pi ∫0 to 0.322 ∫2 to 4 (ρ^2sinφ) dρdφdθ or shown in the whiteboard :

Snapshot.jpg
 
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I agree with your calculation.
 
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qq545282501 said:

Homework Statement


use spherical coordinates to find volume inside a sphere of radius 4 and outside sphere of radius 2, and inside/above the cone z=3√(x^2+y^2).

Homework Equations


[/B]
x=rcosθ =ρsinφcosθ , y=rsinθ =ρsinφsinθ
z=ρcosφ
r= ρsinφ

The Attempt at a Solution


replacing z^2 in the sphere of radius 4 with the cone z^2=9(x^2+y^2) I get the radius of the intersection of √( 8/5), using this radius and known ρ ( 4), φ=inverse sin of (r/ρ) i found φ to be 0.322 or around 18.4 degrees.

V= ∫ 0 to 2 pi ∫0 to 0.322 ∫2 to 4 (ρ^2sinφ) dρdφdθ or shown in the whiteboard :

View attachment 91741

You should not take the integral from 0 to 0.322, but rather, take it from 0 to ## p = \arcsin \left( \frac{\sqrt{8/5}}{4} \right)##, because that will give you an exact answer in terms of a nifty formula, rather than a decimal approximation.

I future, could you PLEASE avoid posting all your really messy handwritten work (as you have done in several recent posts)? Just type it out, which you could probably do almost as quickly as you can make images of the handwritten work and upload the files, etc. This Forum greatly prefers typed solutions, although it has so far not forbidden pictures of handwriting. It would be useful to you now, and in the future, to learn how to use LaTeX; but even the type of ASCII plain text you wrote above is acceptable also.
 
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Ray Vickson said:
I future, could you PLEASE avoid posting all your really messy handwritten work (as you have done in several recent posts)?

I'm guessing the OP is creating those images with a mouse on the "whiteboard". One of the relatively recent "enhancements" of the PF site that don't make it better.
 
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Ray Vickson said:
You should not take the integral from 0 to 0.322, but rather, take it from 0 to ## p = \arcsin \left( \frac{\sqrt{8/5}}{4} \right)##, because that will give you an exact answer in terms of a nifty formula, rather than a decimal approximation.

I future, could you PLEASE avoid posting all your really messy handwritten work (as you have done in several recent posts)? Just type it out, which you could probably do almost as quickly as you can make images of the handwritten work and upload the files, etc. This Forum greatly prefers typed solutions, although it has so far not forbidden pictures of handwriting. It would be useful to you now, and in the future, to learn how to use LaTeX; but even the type of ASCII plain text you wrote above is acceptable also.

Ok, thank you, will do. but was my handwriting that bad :frown:
 

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