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.

- anonymous

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- jim_thompson5910

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

- anonymous

It's \[1-x/2\] and \[-1+x/2\]

- jim_thompson5910

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

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## More answers

- jim_thompson5910

|dw:1444623238793:dw|

- jim_thompson5910

|dw:1444623252668:dw|

- jim_thompson5910

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

- jim_thompson5910

|dw:1444623358917:dw|

- anonymous

x=2 , the upper limit

- jim_thompson5910

yes x = 0 is the lower limit

- jim_thompson5910

|dw:1444623523502:dw|

- jim_thompson5910

|dw:1444623536918:dw|

- anonymous

The length of the black line?

- jim_thompson5910

yeah

- jim_thompson5910

sorry I should have put it on the other drawing

- jim_thompson5910

|dw:1444623622402:dw|

- 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

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

- jim_thompson5910

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

- jim_thompson5910

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

- 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

not quite

- jim_thompson5910

|dw:1444624174284:dw|

- 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

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

- jim_thompson5910

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

- jim_thompson5910

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

- jim_thompson5910

imagine it from a different perspective

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

Okay..

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

\[\sqrt3/4(any side)^2\]

- jim_thompson5910

yep

- 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

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

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

- 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

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

- jim_thompson5910

correct, you're welcome

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