Centre of mass of the shape formed when a hemisphere is attached to a cone(an ice-cream cone shape)

Asked Oct 27 '15 at 11:55

Active Oct 27 '15 at 22:14

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What is the centre of mass of a hemisphere attached to a cone? I know that the bounds for the integration with respect to z must be the two equations( lower bound being the equation of the cone and the upper bound being the equation of the hemisphere, both with z as the subject) itself but I can't seem to be able to integrate that so I am unable to find out the centre of mass of the figure

Equation of the sphere(in this case hemisphere) $x^2+y^2+z^2=16$

Equation of cone is:$(x^2+y^2)=z^2$

The bounds for both x and y and from -3 to 3

Assuming that the density is a constant=1

The formulas are:

edited Oct 27 '15 at 12:24

asked Oct 27 '15 at 11:55

Maths2468

1

The figure doesn't look like it's a hemisphere attached to the cone, more like a spherical cap. Possibly the whole thing is intended to be a spherical sector. – hmakholm left over Monica Oct 27 '15 at 12:02
Where is the centre of the sphere? What is the radius of the sphere? – zoli Oct 27 '15 at 12:06
The equations are given, and I am sorry for removing the image( I could not find a good one and nor does my grapher allow me to remove parts of the graph that I don't need, so that I can make the shape look like an ice-cream cone. – Maths2468 Oct 27 '15 at 12:09
What have you tried so far? Have you computed the circle where the two objects intersect? – John Hughes Oct 27 '15 at 12:12
I just typed that in:) – Maths2468 Oct 27 '15 at 12:13
I don't see an equation for the curve of intersection; perhaps I'm misreading. Anyhow, I'd recommend using polar coordinates for this problem, with the integral over "radius" being the innermost integral. – John Hughes Oct 27 '15 at 12:29
I understand that, but what do i take as the bounds for integration with respect to z when using spherical coordinates. – Maths2468 Oct 27 '15 at 12:31
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Determine the mass and the centre of mass both for a cone and a hemisphere, then apply a weighted mean. – CiaPan Oct 27 '15 at 12:43
Would that work here? – Maths2468 Oct 27 '15 at 12:46
Related http://math.stackexchange.com/questions/495520/center-of-mass-of-semi-sphere and http://math.stackexchange.com/questions/820594/center-of-mass-of-a-right-circular-cone – cgiovanardi May 20 '16 at 00:44

Centre of mass of the shape formed when a hemisphere is attached to a cone(an ice-cream cone shape)

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