Finding the Volume of a Wedge Using Calculus - Calculus Worksheet Solution

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SUMMARY

The discussion focuses on calculating the volume of a wedge removed from a right circular cylinder log with radius 'a' using calculus. The initial approach incorrectly assumes the wedge is 1/8 of a sphere, leading to a volume of (1/6)πa³. However, the correct method involves using integrals, specifically integrating the function 2ysqrt(a² - y²) from 0 to a, resulting in a volume of (2/3)a³. This highlights the importance of understanding the geometric properties of the wedge in relation to the cylinder.

PREREQUISITES
  • Understanding of integral calculus
  • Familiarity with the volume of a sphere formula, V = (4/3)πr³
  • Knowledge of geometric properties of right circular cylinders
  • Ability to set up and evaluate multiple integrals
NEXT STEPS
  • Study the derivation of the volume of a wedge using integral calculus
  • Learn about cylindrical coordinates and their applications in volume calculations
  • Explore the concept of triple integrals in three-dimensional geometry
  • Investigate the relationship between geometric shapes and their integrals for volume determination
USEFUL FOR

Students studying calculus, educators teaching volume calculations, and anyone interested in applying integral calculus to geometric problems.

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A log having the shape of a right circular cylinder of radius a is lying on its side. A wedge is removed from the log by making a vertical cut and another cut at an angle of 45 degrees, both cuts intersecting at the center of the log. Find the volume of the wedge.

And as soon as I finish typing the problem, I find the scan already online. Oh well.

http://www.mvla.net/teachers/JimM/Calculus%20BC-AP/Documents/Worksheet%205%204.pdf

#17 on the second page here.

Obviously, I'm supposed to use the integral somehow, but my first approach was different. Because the wedge has an angle of 45, I know it is 1/8 of the sphere around it. The volume of a sphere is V= (4/3)pi*r^3, and in this case, you just replace r with a. Now that you have the volume of the sphere, I just multiply it by 1/8 to get (1/6)pi*r^3.

However, this answer is totally different from the one in the book. The answer key has you take the integral from 0 to a of 2sqrt(a^2y^2) which equals (2/3)a^3

I don't understand how that works and how my thinking was wrong...
 
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You mean 2ysqrt(a^2-y^2) which equals (2/3)a^3?

That is not a part of the sphere if you look at it carefully

How they got 2ysqrt(a^2-y^2) ?
Using three integrals,
z goes from y to a
x goes from - sqrt(a^2-y^2) to sqrt(a^2-y^2)
y goes from 0 to a

solving this will give you 2ysqrt(a^2-y^2) = (2/3)a^3
 

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