anonymous
  • anonymous
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.
Mathematics
chestercat
  • chestercat
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jim_thompson5910
  • jim_thompson5910
f(x) is this? \[\Large f(x) = \frac{1-x}{2}\] or this? \[\Large f(x) = 1-\frac{x}{2}\]
anonymous
  • anonymous
It's \[1-x/2\] and \[-1+x/2\]
jim_thompson5910
  • jim_thompson5910
so those 1s aren't part of the fraction?

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jim_thompson5910
  • jim_thompson5910
|dw:1444623238793:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1444623252668:dw|
jim_thompson5910
  • jim_thompson5910
the base of the solid is this triangle shaded |dw:1444623339028:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1444623358917:dw|
anonymous
  • anonymous
x=2 , the upper limit
jim_thompson5910
  • jim_thompson5910
yes x = 0 is the lower limit
jim_thompson5910
  • jim_thompson5910
|dw:1444623523502:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1444623536918:dw|
anonymous
  • anonymous
The length of the black line?
jim_thompson5910
  • jim_thompson5910
yeah
jim_thompson5910
  • jim_thompson5910
sorry I should have put it on the other drawing
jim_thompson5910
  • jim_thompson5910
|dw:1444623622402:dw|
anonymous
  • anonymous
Sorry, I don't really know where the line meets the x axis. Is the line at x=1 or x=2?
jim_thompson5910
  • jim_thompson5910
pick any point on the line sloping downward what is the y coordinate of this point?
jim_thompson5910
  • jim_thompson5910
what is the y coordinate of this point? in terms of x
jim_thompson5910
  • jim_thompson5910
that y coordinate would be 1 - (x/2), right?
anonymous
  • 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 \[\sqrt3/4\int\limits_{0}^{2}(1-x/2)^2-(-1+x/2)^2\] ?
jim_thompson5910
  • jim_thompson5910
not quite
jim_thompson5910
  • jim_thompson5910
|dw:1444624174284:dw|
jim_thompson5910
  • 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|
anonymous
  • anonymous
Oh, I see. Then if 2-x is the distance, what is my next step?
jim_thompson5910
  • jim_thompson5910
imagine the shaded region in the drawing |dw:1444624423818:dw| forms the floor of this 3D figure
jim_thompson5910
  • jim_thompson5910
if we look at it from a different angle, we will see |dw:1444624456882:dw|
jim_thompson5910
  • jim_thompson5910
imagine it from a different perspective
jim_thompson5910
  • jim_thompson5910
lay out that line that is 2-x units long |dw:1444624481399:dw|
jim_thompson5910
  • jim_thompson5910
then form an equilateral triangle |dw:1444624510896:dw| each side of the equilateral triangle is 2-x units long
anonymous
  • anonymous
Okay..
jim_thompson5910
  • jim_thompson5910
hopefully you're able to picture what is going on in 3D
jim_thompson5910
  • jim_thompson5910
what is the area of an equilateral triangle (in general)?
anonymous
  • anonymous
\[\sqrt3/4(any side)^2\]
jim_thompson5910
  • jim_thompson5910
yep
jim_thompson5910
  • jim_thompson5910
So in this case, \[\Large A = \frac{\sqrt{3}}{4}s^2\] \[\Large A = \frac{\sqrt{3}}{4}(2-x)^2\]
jim_thompson5910
  • 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
anonymous
  • anonymous
Oh, I see. That actually does make sense. But wouldn't the upper limit be 2 instead of 3?
jim_thompson5910
  • jim_thompson5910
oh right, not sure how 3 snuck in there like that. so yeah it should be \[\Large \frac{\sqrt{3}}{4}\int_{0}^{2}(4-4x+x^2)dx\]
anonymous
  • anonymous
\[(2\sqrt3)/3\] is my answer. Thanks!
jim_thompson5910
  • jim_thompson5910
correct, you're welcome

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