## anonymous one year ago I keep getting 0 as the volume? Giving a medal to whoever can explain it to me correctly! Find the volume of the solid whose base is bounded by the lines F(x) = 1-x/2 , g(x) = -1+x/2 , and x=0. The cross sections perpendicular to the x-axis are equilateral triangles.

1. jim_thompson5910

f(x) is this? $\Large f(x) = \frac{1-x}{2}$ or this? $\Large f(x) = 1-\frac{x}{2}$

2. anonymous

It's $1-x/2$ and $-1+x/2$

3. jim_thompson5910

so those 1s aren't part of the fraction?

4. jim_thompson5910

|dw:1444623238793:dw|

5. jim_thompson5910

|dw:1444623252668:dw|

6. jim_thompson5910

the base of the solid is this triangle shaded |dw:1444623339028:dw|

7. jim_thompson5910

|dw:1444623358917:dw|

8. anonymous

x=2 , the upper limit

9. jim_thompson5910

yes x = 0 is the lower limit

10. jim_thompson5910

|dw:1444623523502:dw|

11. jim_thompson5910

|dw:1444623536918:dw|

12. anonymous

The length of the black line?

13. jim_thompson5910

yeah

14. jim_thompson5910

sorry I should have put it on the other drawing

15. jim_thompson5910

|dw:1444623622402:dw|

16. anonymous

Sorry, I don't really know where the line meets the x axis. Is the line at x=1 or x=2?

17. jim_thompson5910

pick any point on the line sloping downward what is the y coordinate of this point?

18. jim_thompson5910

what is the y coordinate of this point? in terms of x

19. jim_thompson5910

that y coordinate would be 1 - (x/2), right?

20. anonymous

Yeah, the equation for y in terms of x is 1/(x/2). If the 1-x/2 and -1+x/2 weren't the same when squared, wouldn't the equation just be $\sqrt3/4\int\limits_{0}^{2}(1-x/2)^2-(-1+x/2)^2$ ?

21. jim_thompson5910

not quite

22. jim_thompson5910

|dw:1444624174284:dw|

23. jim_thompson5910

subtract the two (top - bottom) to find the distance [1-(x/2)] - [-1+(x/2)] 1-(x/2) + 1-(x/2) 2-2*(x/2) 2-x so no matter what x value you pick (from x = 0 to x = 3), the distance is 2-x |dw:1444624259423:dw|

24. anonymous

Oh, I see. Then if 2-x is the distance, what is my next step?

25. jim_thompson5910

imagine the shaded region in the drawing |dw:1444624423818:dw| forms the floor of this 3D figure

26. jim_thompson5910

if we look at it from a different angle, we will see |dw:1444624456882:dw|

27. jim_thompson5910

imagine it from a different perspective

28. jim_thompson5910

lay out that line that is 2-x units long |dw:1444624481399:dw|

29. jim_thompson5910

then form an equilateral triangle |dw:1444624510896:dw| each side of the equilateral triangle is 2-x units long

30. anonymous

Okay..

31. jim_thompson5910

hopefully you're able to picture what is going on in 3D

32. jim_thompson5910

what is the area of an equilateral triangle (in general)?

33. anonymous

$\sqrt3/4(any side)^2$

34. jim_thompson5910

yep

35. jim_thompson5910

So in this case, $\Large A = \frac{\sqrt{3}}{4}s^2$ $\Large A = \frac{\sqrt{3}}{4}(2-x)^2$

36. jim_thompson5910

You need to compute $\Large \int_{0}^{3}Adx$ $\Large \int_{0}^{3}\frac{\sqrt{3}}{4}(2-x)^2dx$ $\Large \frac{\sqrt{3}}{4}\int_{0}^{3}(2-x)^2dx$ $\Large \frac{\sqrt{3}}{4}\int_{0}^{3}(4-4x+x^2)dx$ I'll let you finish up

37. anonymous

Oh, I see. That actually does make sense. But wouldn't the upper limit be 2 instead of 3?

38. jim_thompson5910

oh right, not sure how 3 snuck in there like that. so yeah it should be $\Large \frac{\sqrt{3}}{4}\int_{0}^{2}(4-4x+x^2)dx$

39. anonymous

$(2\sqrt3)/3$ is my answer. Thanks!

40. jim_thompson5910

correct, you're welcome