Did I set up my integral correctly?

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The discussion focuses on using spherical coordinates to calculate the volume between two spheres and above a cone. The original setup included an integral from 0 to 0.322 for φ, but it was advised to adjust this to the arcsine of the intersection radius for a more precise result. Participants emphasized the importance of presenting solutions in a typed format rather than handwritten, suggesting that LaTeX could enhance clarity. The conversation also touched on the challenges of creating clear visual representations of mathematical work. Overall, the thread highlights the need for accurate integral limits and clearer communication in mathematical discussions.
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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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