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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 [Broken]

#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...

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 [Broken]

#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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