anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@jim_thompson5910 @zepdrix
anonymous
  • anonymous
Would the radius not be 3?
anonymous
  • anonymous
https://i.gyazo.com/f3d88835c49eb9270050edba54fcec2d.png

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More answers

jim_thompson5910
  • jim_thompson5910
this isn't a volume of revolution
anonymous
  • anonymous
Hmmmm
jim_thompson5910
  • jim_thompson5910
it's kinda like this https://www.youtube.com/watch?v=tmotNcRHr9Q
jim_thompson5910
  • jim_thompson5910
instead of gluing a bunch of disks together, you're gluing a bunch of equilateral triangles together
jim_thompson5910
  • jim_thompson5910
|dw:1444448060321:dw|
anonymous
  • anonymous
So it would look like a cone?
anonymous
  • anonymous
Since the base is a circle
jim_thompson5910
  • jim_thompson5910
kinda, more like the sydney opera house https://upload.wikimedia.org/wikipedia/commons/7/75/Sydney_Opera_House,_botanic_gardens_1.jpg
jim_thompson5910
  • jim_thompson5910
|dw:1444448113459:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1444448131880:dw|
jim_thompson5910
  • jim_thompson5910
hint: solve x^2 + y^2 = 9 for y
anonymous
  • anonymous
\[\sqrt{9-x^2}\]
jim_thompson5910
  • jim_thompson5910
yep
jim_thompson5910
  • jim_thompson5910
by symmetry, the distance between point A and B is \(\Large 2\sqrt{9-x^2}\) |dw:1444448401555:dw|
anonymous
  • anonymous
ok, what am I supposed to do with that?
jim_thompson5910
  • jim_thompson5910
ok imagine that circle is the floor of this weird 3D object
jim_thompson5910
  • jim_thompson5910
|dw:1444448545776:dw|
jim_thompson5910
  • jim_thompson5910
we're going to build a triangular roof like so |dw:1444448565044:dw|
jim_thompson5910
  • jim_thompson5910
each triangle is an equilateral triangle
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
area of equilateral triangle with side length s \[\Large A = \frac{\sqrt{3}}{4}*s^2\]
jim_thompson5910
  • jim_thompson5910
\[\Large A = \frac{\sqrt{3}}{4}*s^2\] \[\Large A = \frac{\sqrt{3}}{4}*\left(2\sqrt{9-x^2}\right)^2\] \[\Large A = \frac{\sqrt{3}}{4}*\left(4(9-x^2)\right)\] \[\Large A = \sqrt{3}(9-x^2)\]
anonymous
  • anonymous
Ok, now I'm following you, but what would I enter for X?
jim_thompson5910
  • jim_thompson5910
you'll be integrating with respect to x
jim_thompson5910
  • jim_thompson5910
\[\Large \int_{-3}^{3}Adx = \int_{-3}^{3}\left[\sqrt{3}(9-x^2)\right]dx = ??\]
anonymous
  • anonymous
36sqrt(3)
jim_thompson5910
  • jim_thompson5910
very good
anonymous
  • anonymous
I have some other questions that I solved, do you mind checking them for me?
jim_thompson5910
  • jim_thompson5910
alright
anonymous
  • anonymous
Or should I post it separately?
jim_thompson5910
  • jim_thompson5910
separately to avoid lag and clutter

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