The base of a solid in the xy-plane is the circle x2 + y2 = 16. Cross sections of the solid perpendicular to the y-axis are equilateral triangles. What is the volume, in cubic units, of the solid?

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The base of a solid in the xy-plane is the circle x2 + y2 = 16. Cross sections of the solid perpendicular to the y-axis are equilateral triangles. What is the volume, in cubic units, of the solid?

Mathematics
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Do you know what the solid is?
no I don't know how to start this problem
|dw:1438722075391:dw|

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They gave you x^2 + y^2 = 16 = r^2 That is a circle radius 4 centered at the origin, in the XY-plane...
you recognize that?
yes
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bad drawing, but that is a slice of the shape perpendicular to the Y axis
saying it is equilateral, means that is a Right Circular Cone
okay so you need the volume formula for a right circular cone correct?
yes, just remember, it is the area of the base times 1/3 the height
ok
need to figure the height
Loooking at the cross section at the XZ axis head on... the base is 2 times the radius
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THe y axis is pointing out from the page, that is the origin
Equilateral triangle, so the diagonals are the same as the base, 2 radiuses
radii
r=4, use pythagorean theorem to figure the height
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You get it all?
yes so far
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you can only put in the 2r because they told you the cross sections are equilateral
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pelletty drawing, but hope you see where those measurments come from
yes it is understood
k, so you know the radius and height, you can calculate the volume \[V = \frac{ 1 }{ 3 }\pi*r^2*h\]
r = 4 h = sqrt(8^2 - 4^2)
The way you set it up is multiply 1/3 by pi(4)^2(sqrt(8^2-4^2))?
yes, if you simplify the height first, it is... \[h = \sqrt{64-16} = 4\sqrt{3}\]
okay thank you very much for the explanation
\[V = \frac{ 1 }{ 3 }*\pi*4^2*4\sqrt{3}\]
welcome, that was a fun prob.. hope ya get it

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