anonymous
  • anonymous
The base of a solid in the xy-plane is the circle x2 + y2 = 16. Cross sections of the solid perpendicular to the y-axis are equilateral triangles. What is the volume, in cubic units, of the solid?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
https://gyazo.com/916a3efc162f658ce2b5d82443c845a4
anonymous
  • anonymous
You're already here, that was quick.
zepdrix
  • zepdrix
Ok no spinning business on this one. That's nice. Do you understand how to visualize this one? Maybe I can draw it out.

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zepdrix
  • zepdrix
|dw:1449604336848:dw|
zepdrix
  • zepdrix
|dw:1449604378694:dw|And if we look at a slice, perpendicular to the y-axis
zepdrix
  • zepdrix
|dw:1449604421538:dw|we get equilateral triangles coming up
anonymous
  • anonymous
Basically a cone.
zepdrix
  • zepdrix
Sort of. It's more like... a shark fin? Hmm I can't think of what would describe it well. All of those triangles give us like a sharp ledge across the middle going down. And it's still rounded along the outside. Hah, maybe I'm thinking of it wrong XD
zepdrix
  • zepdrix
Area of an equilateral triangle, \(\large\rm A=\dfrac{\sqrt{3}}{4}s^2\) where s is the side length.
zepdrix
  • zepdrix
Volume of a triangular prism would simply involve multiplying the area of that triangle by the thickness. With integration we're imagining that this has "thickness" dy. So the volume of our sliced out triangular prism would be given by\[\large\rm dv=\frac{\sqrt3}{4}s^2~dy\]
zepdrix
  • zepdrix
And then I guess we're just "adding up" all of those triangular prisms.\[\large\rm V=\int\limits dv\] \[\large\rm V=\int\limits_{y=-4}^{4}\frac{\sqrt3}{4}s^2~dy\]
zepdrix
  • zepdrix
The side length is a little tricky.
zepdrix
  • zepdrix
|dw:1449604805072:dw|If we write our function in terms of x
zepdrix
  • zepdrix
\[\large\rm x^2+y^2=4^2\]\[\large\rm x=\pm\sqrt{4^2-y^2}\]
zepdrix
  • zepdrix
So that side length is just going to be `two of these square roots`, ya? Something like that...
zepdrix
  • zepdrix
\[\large\rm V=\int\limits\limits_{y=-4}^{4}\frac{\sqrt3}{4}\left(2\sqrt{4^2-y^2}\right)^2~dy\]Ahhh hopefully I did that properly.
zepdrix
  • zepdrix
Aw you ran off XD Well hopefully you can look at this later and maybe make sense of it.
anonymous
  • anonymous
@zepdrix Openstudy is glitching
anonymous
  • anonymous
I am still viewing it.
zepdrix
  • zepdrix
Oh some of that crazy stuff going on? Lame
anonymous
  • anonymous
Let me solve the integral you set up.
anonymous
  • anonymous
https://gyazo.com/5c003dcc88d03f95d8e6f8376a9b8999 I had to find a workaround as the main page won't load.
zepdrix
  • zepdrix
When the main page fails to load, going back to the question through your profile sometimes works.
anonymous
  • anonymous
Got 147.8
zepdrix
  • zepdrix
Same. Do you have answer choices for this one? :O
anonymous
  • anonymous
Which is C.
zepdrix
  • zepdrix
Ooo neato! So we likely did it correctly! :)
anonymous
  • anonymous
Yessir, Thanks for the help. I had another question.....bu it won't let me post it.

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