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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) = 1x/2 , g(x) = 1+x/2 , and x=0. The cross sections perpendicular to the xaxis are equilateral triangles.
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) = 1x/2 , g(x) = 1+x/2 , and x=0. The cross sections perpendicular to the xaxis are equilateral triangles.

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3f(x) is this? \[\Large f(x) = \frac{1x}{2}\] or this? \[\Large f(x) = 1\frac{x}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's \[1x/2\] and \[1+x/2\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3so those 1s aren't part of the fraction?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3dw:1444623238793:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3dw:1444623252668:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3the base of the solid is this triangle shaded dw:1444623339028:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3dw:1444623358917:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x=2 , the upper limit

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3yes x = 0 is the lower limit

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3dw:1444623523502:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3dw:1444623536918:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The length of the black line?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3sorry I should have put it on the other drawing

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3dw:1444623622402:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, I don't really know where the line meets the x axis. Is the line at x=1 or x=2?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3pick any point on the line sloping downward what is the y coordinate of this point?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3what is the y coordinate of this point? in terms of x

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3that y coordinate would be 1  (x/2), right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, the equation for y in terms of x is 1/(x/2). If the 1x/2 and 1+x/2 weren't the same when squared, wouldn't the equation just be \[\sqrt3/4\int\limits_{0}^{2}(1x/2)^2(1+x/2)^2\] ?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3dw:1444624174284:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3subtract the two (top  bottom) to find the distance [1(x/2)]  [1+(x/2)] 1(x/2) + 1(x/2) 22*(x/2) 2x so no matter what x value you pick (from x = 0 to x = 3), the distance is 2x dw:1444624259423:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I see. Then if 2x is the distance, what is my next step?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3imagine the shaded region in the drawing dw:1444624423818:dw forms the floor of this 3D figure

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3if we look at it from a different angle, we will see dw:1444624456882:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3imagine it from a different perspective

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3lay out that line that is 2x units long dw:1444624481399:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3then form an equilateral triangle dw:1444624510896:dw each side of the equilateral triangle is 2x units long

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3hopefully you're able to picture what is going on in 3D

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3what is the area of an equilateral triangle (in general)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sqrt3/4(any side)^2\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3So in this case, \[\Large A = \frac{\sqrt{3}}{4}s^2\] \[\Large A = \frac{\sqrt{3}}{4}(2x)^2\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3You need to compute \[\Large \int_{0}^{3}Adx\] \[\Large \int_{0}^{3}\frac{\sqrt{3}}{4}(2x)^2dx\] \[\Large \frac{\sqrt{3}}{4}\int_{0}^{3}(2x)^2dx\] \[\Large \frac{\sqrt{3}}{4}\int_{0}^{3}(44x+x^2)dx\] I'll let you finish up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I see. That actually does make sense. But wouldn't the upper limit be 2 instead of 3?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3oh right, not sure how 3 snuck in there like that. so yeah it should be \[\Large \frac{\sqrt{3}}{4}\int_{0}^{2}(44x+x^2)dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(2\sqrt3)/3\] is my answer. Thanks!

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3correct, you're welcome
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