Calculating Volume of Described Solid with Equilateral Cross-Sections

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The problem involves calculating the volume of a solid with a triangular base defined by vertices (0, 0), (5, 0), and (0, 5), with equilateral triangular cross-sections perpendicular to the y-axis. The initial approach used the line equation x = 4 - y, which was incorrect; the correct equation for the line between (5, 0) and (0, 5) is x = 5 - y. The area of the equilateral triangle cross-section was calculated as A(y) = ((5 - y)^2)/2. After realizing the mistake in the line equation, the correct integral for volume calculation was confirmed. The discussion highlights the importance of accurately determining the base line equation for proper volume computation.
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Homework Statement



The base of S is the triangular region with vertices (0, 0), (5, 0), and (0, 5). Cross-sections perpendicular to the y-axis are equilateral triangles.
Find V of described triangle.

Homework Equations





The Attempt at a Solution


I first wrote the equation of the line in terms of x (x = 4-y), which is the base. Since we are dealing with an equilateral triangle the area of the cross section would be A(y)= ((4-y)^2)/2 and so the integral to calculate the volume is A(y)dy from 0 to 4.
Since I am arriving at the supposedly wrong answer, what am I missing?
 
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Nicolaus said:

Homework Statement



The base of S is the triangular region with vertices (0, 0), (5, 0), and (0, 5). Cross-sections perpendicular to the y-axis are equilateral triangles.
Find V of described triangle.

Homework Equations





The Attempt at a Solution


I first wrote the equation of the line in terms of x (x = 4-y), which is the base.
I don't see where your equation comes from. What's the equation of the line between (5, 0) and (0, 5)?
Nicolaus said:
Since we are dealing with an equilateral triangle the area of the cross section would be A(y)= ((4-y)^2)/2 and so the integral to calculate the volume is A(y)dy from 0 to 4.
Since I am arriving at the supposedly wrong answer, what am I missing?
 
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It's a typo; meant 5-y. I figured out the problem. Thanks for your interest in helping.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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