Let the base of a solid be the first quadrant plane region bounded by y=1-x^2/25 , the x-axis, and the y-axis. Suppose that cross sections perpendicular to the x-axis are equilateral triangles sitting on the base. Find the volume of the solid.
I'm completely lost on this, i'm not even sure where to start!
Thanks
I'm completely lost on this, i'm not even sure where to start!
Thanks
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So first, draw the domain. (see link)
Now, we were given that the cross sections perpendicular to the x-axis will be equilateral triangles, so we need the length of the base. This will be the difference between the function and the x-axis, or just y = f(x).
If we do a little algebra on the triangle area formula,
A = ½ b h
Equilateral triangles can be broken into 2 triangles of (30-60-90)° angles, which have proportions (s/2, s, s√(3)/2):
A = [2 * 1/2] * [(s/2) * (s√(3)/2)]
A = s²√(3)/4
In this case, our side length was already determined to be y = f(x), so we can just plug that in:
A = y²√(3)/4
Then we need to integrate over the x-axis from 0 to 5.
V = ∫ A(x) dx
V = √(3)/4 ∫ (1 - x²/25)² dx[0,5]
And I hope you can take it from there. Just expand the square and integrate term-by-term with the power rule.
Now, we were given that the cross sections perpendicular to the x-axis will be equilateral triangles, so we need the length of the base. This will be the difference between the function and the x-axis, or just y = f(x).
If we do a little algebra on the triangle area formula,
A = ½ b h
Equilateral triangles can be broken into 2 triangles of (30-60-90)° angles, which have proportions (s/2, s, s√(3)/2):
A = [2 * 1/2] * [(s/2) * (s√(3)/2)]
A = s²√(3)/4
In this case, our side length was already determined to be y = f(x), so we can just plug that in:
A = y²√(3)/4
Then we need to integrate over the x-axis from 0 to 5.
V = ∫ A(x) dx
V = √(3)/4 ∫ (1 - x²/25)² dx[0,5]
And I hope you can take it from there. Just expand the square and integrate term-by-term with the power rule.