Volume of a cone using integrals

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SUMMARY

The discussion focuses on calculating the work required to pump water from an inverted cone-shaped tank with a height of 6 m and a base radius of 1.5 m. The integral used for this calculation incorporates the density of water, gravitational acceleration (g = 9.8 m/s²), and the area of cross-sections of the cone. The cross-sectional area is derived as π * (y²)/16, which accounts for the changing radius as water is pumped from different heights within the cone. Understanding the relationship between the radius and height using trigonometry is essential for solving this problem.

PREREQUISITES
  • Understanding of integral calculus, specifically in the context of work and volume calculations.
  • Familiarity with the concepts of density and gravitational force.
  • Knowledge of cross-sectional area calculations for geometric shapes, particularly circles.
  • Basic trigonometry to relate the radius and height of the cone.
NEXT STEPS
  • Study the derivation of the cross-sectional area of a cone using similar triangles.
  • Learn about the application of integrals in physics, particularly in calculating work done against gravity.
  • Explore the concept of density in fluid mechanics and its role in work calculations.
  • Practice solving similar problems involving different shapes and their respective integrals.
USEFUL FOR

Students studying calculus, physics enthusiasts, and anyone involved in engineering or architecture who needs to understand fluid dynamics and work calculations in conical structures.

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Homework Statement



A water tank is shaped like an inverted cone with height 6 m and base radius 1.5 m

If the tank is full, how much work is required to pump the water to the level of the top of the tank and out of the tank?


Homework Equations



Integral of ( density * g (acceleration due to gravity) * A(y) (area of a cross section ) * change in y )

The Attempt at a Solution



The integral is from zero to 6 since this strange cone-shaped tank is 6 m high. g = 9.8 m/s^2, times the change in y which is (y-6) since the cone is 6m high

My problem is that I have no idea how to compute the cross-sectional area of a cone. The cuts are circles which have an area of pi*r^2. The tricky thing with this problem though is the radius does not remain constant from top to bottom.

My solutions manual gives the area as pi * (y^2)/16. I have no idea how they got to this. Any help would be very much appreciated.
 
Physics news on Phys.org
Draw the cone on your paper as a triangle facing down. Pick some point along the center axis. The radius is the distance from the center to the edge, and the height is the distance from the bottom to the current location.

Now you can use trigonometry to solve for the radius in terms of the height.
 

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