Integration of a cone to find centre of mass

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SUMMARY

The discussion focuses on calculating the volume of a uniform solid cone with height b and base radius a, specifically deriving the volume of discs at varying heights h above the base. The volume of the cone is expressed as V = π/3 * b * a². The method involves using similar triangles to relate the radius of the disc to its height, followed by integrating the volume of the discs from h = 0 to h = b. The final derived expression confirms the total volume of the cone.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques
  • Familiarity with geometric concepts such as similar triangles
  • Knowledge of the volume formulas for cylinders and cones
  • Basic proficiency in algebra for manipulating equations
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  • Study the principles of integration in calculus, focusing on definite integrals
  • Learn about the properties of similar triangles and their applications in geometry
  • Explore the derivation of volume formulas for various geometric shapes, including cones and cylinders
  • Practice solving problems involving cross-sections of solids of revolution
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Students in mathematics or engineering fields, educators teaching calculus and geometry, and anyone interested in understanding the geometric properties and volume calculations of cones.

rainbowGirl
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Can anyone help me with this question?

A uniform solid cone of height b and base radius a stands on a horizontal table. Find an expression for the volume of the disc at height h above the base. Integrate over all the discs to show that the total volume, V, is given by V =pi/3 * b * a^2
 
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Start by drawing a picture. Seen from the side, the cone looks like a triangle with base 2r and height b. Now draw a line across the triangle at height h (and so b-h from the vertex). Use "similar triangles" to find the base of that triangle, as a function of h. The disk formed by the cone has that base as diameter and thickness dx. I assume you know that the voume of a disk of diameter d and thickness s is pi(d2/4)s.
 


Sure, I can help you with this question! Let's start by visualizing the problem. We have a cone with a height of b and a base radius of a, and we want to find the volume of the disc at a given height h above the base. This disc can be thought of as a cross-section of the cone at that particular height.

To find the volume of this disc, we can use the formula for the volume of a cylinder, which is given by V = πr^2h, where r is the radius and h is the height. In this case, the radius (r) of the disc is proportional to its height (h), and we can use similar triangles to find the relationship between r and h.

If we draw a line from the center of the base of the cone to the edge of the disc, we can see that it forms a right triangle with the height of the cone (b) and the radius of the base (a). The height of this triangle (h) is equal to the height of the disc that we are trying to find the volume of.

Using the Pythagorean theorem, we can find the relationship between h and r:

r^2 + h^2 = a^2

Solving for r, we get:

r = √(a^2 - h^2)

Now, we can substitute this value for r into the formula for the volume of a cylinder:

V = πr^2h

V = π(a^2 - h^2)h

V = π(a^2h - h^3)

We can now integrate this expression over all the discs, from the base of the cone (h = 0) to the top of the cone (h = b):

V = ∫π(a^2h - h^3)dh from h = 0 to h = b

V = π(a^2h^2/2 - h^4/4) from h = 0 to h = b

V = π(a^2b^2/2 - b^4/4) - π(0 - 0)

V = π(a^2b^2/2 - b^4/4)

Simplifying this expression, we get:

V = π/2 * (a^2b^2 - b^4/2)

And finally, using the formula for the volume of a cone (
 

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