## 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 x-axis are equilateral triangles. What is the volume, in cubic units, of the solid?

1. anonymous

@jim_thompson5910 @zepdrix

2. anonymous

Would the radius not be 3?

3. anonymous
4. jim_thompson5910

this isn't a volume of revolution

5. anonymous

Hmmmm

6. jim_thompson5910

7. jim_thompson5910

instead of gluing a bunch of disks together, you're gluing a bunch of equilateral triangles together

8. jim_thompson5910

|dw:1444448060321:dw|

9. anonymous

So it would look like a cone?

10. anonymous

Since the base is a circle

11. jim_thompson5910

kinda, more like the sydney opera house https://upload.wikimedia.org/wikipedia/commons/7/75/Sydney_Opera_House,_botanic_gardens_1.jpg

12. jim_thompson5910

|dw:1444448113459:dw|

13. jim_thompson5910

|dw:1444448131880:dw|

14. jim_thompson5910

hint: solve x^2 + y^2 = 9 for y

15. anonymous

$\sqrt{9-x^2}$

16. jim_thompson5910

yep

17. jim_thompson5910

by symmetry, the distance between point A and B is $$\Large 2\sqrt{9-x^2}$$ |dw:1444448401555:dw|

18. anonymous

ok, what am I supposed to do with that?

19. jim_thompson5910

ok imagine that circle is the floor of this weird 3D object

20. jim_thompson5910

|dw:1444448545776:dw|

21. jim_thompson5910

we're going to build a triangular roof like so |dw:1444448565044:dw|

22. jim_thompson5910

each triangle is an equilateral triangle

23. 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

24. jim_thompson5910

area of equilateral triangle with side length s $\Large A = \frac{\sqrt{3}}{4}*s^2$

25. jim_thompson5910

$\Large A = \frac{\sqrt{3}}{4}*s^2$ $\Large A = \frac{\sqrt{3}}{4}*\left(2\sqrt{9-x^2}\right)^2$ $\Large A = \frac{\sqrt{3}}{4}*\left(4(9-x^2)\right)$ $\Large A = \sqrt{3}(9-x^2)$

26. anonymous

Ok, now I'm following you, but what would I enter for X?

27. jim_thompson5910

you'll be integrating with respect to x

28. jim_thompson5910

$\Large \int_{-3}^{3}Adx = \int_{-3}^{3}\left[\sqrt{3}(9-x^2)\right]dx = ??$

29. anonymous

36sqrt(3)

30. jim_thompson5910

very good

31. anonymous

I have some other questions that I solved, do you mind checking them for me?

32. jim_thompson5910

alright

33. anonymous

Or should I post it separately?

34. jim_thompson5910

separately to avoid lag and clutter