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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 @zepdrix

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would the radius not be 3?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2this isn't a volume of revolution

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2it's kinda like this https://www.youtube.com/watch?v=tmotNcRHr9Q

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2instead of gluing a bunch of disks together, you're gluing a bunch of equilateral triangles together

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444448060321:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So it would look like a cone?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since the base is a circle

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2kinda, more like the sydney opera house https://upload.wikimedia.org/wikipedia/commons/7/75/Sydney_Opera_House,_botanic_gardens_1.jpg

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444448113459:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444448131880:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2hint: solve x^2 + y^2 = 9 for y

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2by symmetry, the distance between point A and B is \(\Large 2\sqrt{9x^2}\) dw:1444448401555:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, what am I supposed to do with that?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2ok imagine that circle is the floor of this weird 3D object

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444448545776:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2we're going to build a triangular roof like so dw:1444448565044:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2each triangle is an equilateral triangle

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2so 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
 one year ago
Best ResponseYou've already chosen the best response.2area of equilateral triangle with side length s \[\Large A = \frac{\sqrt{3}}{4}*s^2\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large A = \frac{\sqrt{3}}{4}*s^2\] \[\Large A = \frac{\sqrt{3}}{4}*\left(2\sqrt{9x^2}\right)^2\] \[\Large A = \frac{\sqrt{3}}{4}*\left(4(9x^2)\right)\] \[\Large A = \sqrt{3}(9x^2)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, now I'm following you, but what would I enter for X?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you'll be integrating with respect to x

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large \int_{3}^{3}Adx = \int_{3}^{3}\left[\sqrt{3}(9x^2)\right]dx = ??\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have some other questions that I solved, do you mind checking them for me?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Or should I post it separately?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2separately to avoid lag and clutter
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