The base of a solid is the circle x^2 + y^2 = 9. Cross sections of the solid perpendicular to the x-axis are equilateral triangles. What is the volume, in cubic units, of the solid?

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The base of a solid is the circle x^2 + y^2 = 9. Cross sections of the solid perpendicular to the x-axis are equilateral triangles. What is the volume, in cubic units, of the solid?

Mathematics
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Would the radius not be 3?
https://i.gyazo.com/f3d88835c49eb9270050edba54fcec2d.png

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Other answers:

this isn't a volume of revolution
Hmmmm
it's kinda like this https://www.youtube.com/watch?v=tmotNcRHr9Q
instead of gluing a bunch of disks together, you're gluing a bunch of equilateral triangles together
|dw:1444448060321:dw|
So it would look like a cone?
Since the base is a circle
kinda, more like the sydney opera house https://upload.wikimedia.org/wikipedia/commons/7/75/Sydney_Opera_House,_botanic_gardens_1.jpg
|dw:1444448113459:dw|
|dw:1444448131880:dw|
hint: solve x^2 + y^2 = 9 for y
\[\sqrt{9-x^2}\]
yep
by symmetry, the distance between point A and B is \(\Large 2\sqrt{9-x^2}\) |dw:1444448401555:dw|
ok, what am I supposed to do with that?
ok imagine that circle is the floor of this weird 3D object
|dw:1444448545776:dw|
we're going to build a triangular roof like so |dw:1444448565044:dw|
each triangle is an equilateral triangle
so what we do really is integrate a bunch of these triangles together to form the solid so we do \[\Large \int_{-3}^{3}Adx\] where A is the area of any given equilateral triangle
area of equilateral triangle with side length s \[\Large A = \frac{\sqrt{3}}{4}*s^2\]
\[\Large A = \frac{\sqrt{3}}{4}*s^2\] \[\Large A = \frac{\sqrt{3}}{4}*\left(2\sqrt{9-x^2}\right)^2\] \[\Large A = \frac{\sqrt{3}}{4}*\left(4(9-x^2)\right)\] \[\Large A = \sqrt{3}(9-x^2)\]
Ok, now I'm following you, but what would I enter for X?
you'll be integrating with respect to x
\[\Large \int_{-3}^{3}Adx = \int_{-3}^{3}\left[\sqrt{3}(9-x^2)\right]dx = ??\]
36sqrt(3)
very good
I have some other questions that I solved, do you mind checking them for me?
alright
Or should I post it separately?
separately to avoid lag and clutter

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