Volume of ice cream cone triple integral

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SUMMARY

The discussion focuses on calculating the triple integral for the volume between a hemisphere centered at z=1 and a cone with angle α. The radius of the hemisphere is determined as tan(α), and the integral is approached using cylindrical and Cartesian coordinates. The primary challenge arises in converting to spherical coordinates, specifically in deriving the equation for ρ in terms of the angle φ. The correct equation for the hemisphere is identified as x² + y² + (z-1)² = 1, leading to the conclusion that ρ = 2cos(φ) for the spherical coordinate system.

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  • Understanding of triple integrals and volume calculations
  • Familiarity with spherical coordinates and their transformations
  • Knowledge of cylindrical and Cartesian coordinate systems
  • Proficiency in solving equations involving trigonometric functions
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  • Learn how to convert equations between Cartesian and spherical coordinates
  • Study the application of the triple integral in calculating volumes of solids
  • Explore the use of the quadratic formula in solving for variables in integrals
  • Investigate the geometric interpretation of spherical coordinates in calculus
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Students studying calculus, particularly those focusing on multivariable integration, as well as educators teaching volume calculations involving complex shapes like hemispheres and cones.

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Homework Statement


Find the triple integral for the volume between a hemisphere centred at ##z=1## and cone with angle ##\alpha##.

The Attempt at a Solution



What I tried to do first was to get the radius of the hemisphere in terms of the angle ##\alpha##. In this case the radius is ##\tan \alpha##. I already figured out this integral in cylindrical polar coordinates and Cartesian coordinates. I am having a lot of trouble with spherical coordinates. I am trying to get ##\rho## for the hemisphere by drawing the projection of the shape on the xz plane and trying to get a formula for a radial ray that hits the hemisphere. The formula for a circle that is shifted up by one is ##x^2+(z-1)^2=\tan^2 \alpha##, this is the part where I am stuck trying to find the equation for ##\rho## in terms of the angle ##\phi## of the radial ray. The limits for ##\theta## and ##\phi## are really easy, but again ##\rho## I just can't seem to get.
 
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Do you need the integral in spherical coordinates? It is messy, and the other coordinate systems are much easier.

You are missing y2 in your hemisphere formula.
You can find z as function of the two angles, plug it into the equation for the hemisphere (in the right place) and solve for ρ.
 
mfb said:
Do you need the integral in spherical coordinates? It is messy, and the other coordinate systems are much easier.

You are missing y2 in your hemisphere formula.
You can find z as function of the two angles, plug it into the equation for the hemisphere (in the right place) and solve for ρ.
So I will have to use the quadratic formula to solve for ##\rho##?
 
I disagree with mfb on this problem. In fact spherical coordinates is the natural and easiest way to work this problem. You didn't state it in your problem but I assume you are talking about the upper half of the sphere of radius ##1## centered at ##(0,0,1)##. Its equation is ##x^2+y^2+(z-1)^2 = 1## or, expanding it, ##x^2 + y^2 + z^2 = 2z##. If you change that to spherical coordinates you should get ##\rho = 2\cos\phi##. So the ##\rho,~\theta,~\phi## limits are all very easy.
 
How could you fix the radius of the half-sphere to 1?
 
mfb said:
How could you fix the radius of the half-sphere to 1?
I couldn't. I scanned it too quickly and assumed it was the "standard" problem you normally see in calculus books.
 

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