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Tricky Multiple integral word problem 
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#1
Nov2407, 02:52 PM

P: 12

So Ive been talking with some people about this problem, but I cant seem to find the answer, or even set it up!
"A region in space, when viewed from 3 different views, looks like a circle, looks like a square and looks like a triangle. Describe this object. And then use multiple integration to determine its volume" So if you think of any two shapes its pretty easy. Such as square and triangle = pyramid, circle and square = cylinder and so on. But I couldnt figure out a shape for when you add the third shape. And then Figuring out the integral would be even more difficult. What do you guys think about it? =D 


#2
Nov2407, 03:34 PM

P: 492

you almost had it with the circle+square=cylinder, what if you were to have only a square from the "top/bottom" and the sides a triangle?



#3
Nov2407, 03:43 PM

P: 12

Im still not getting it.. Pictures would help. >.<
So Square base and top. Sides triangle? Where would the circle go? 


#4
Nov2407, 03:45 PM

P: 492

Tricky Multiple integral word problem
it's originally a cylinder, 2r=height, so from 2 sides you'd have a square. and now cut off some part in order to get a triangle from the other 2 sides. So it looks like a wedge with a circular base and square top/bottom.



#5
Nov2407, 04:40 PM

P: 12

OH I think I get it now. So bascially, its 1/2 a cylinder right?
http://img77.imageshack.us/img77/8178/shapehi9.gif So side view is square, bottom is circle and other sides are triangles. Thanks!! But how would I find the volume of that shape? 


#6
Nov2407, 04:47 PM

P: 492

hm..depends how you place the axis's on it, but I think you'll need to add 2 integrals due to the circle you'll have the top and bottom part of it.
and it might be better to cut the cylinder by 2 lines so looking at the triangle side you can cut it at the center of the circle and get 2 symmetric sides so you can just do 2xsome integral to find the volume. 


#7
Nov2407, 04:51 PM

P: 12

Well if I cut the cylinder in half diagonally, couldn't I just take the volume of the cylinder and then divide it by two? When you say cut the cylinder with two lines what do you mean?
And I have no idea how to set up the integral. >.< could you get me started please? 


#8
Nov2407, 04:54 PM

P: 492

o yea you could just do that.
Use cylindrical coordinates to setup your integral and then just multiply it by .5 to find the volume. 


#9
Nov2407, 05:15 PM

P: 12

could you help me set it up? thats the part im really bad at. Many thanks for discussing this problem with me. =)



#10
Nov2407, 05:19 PM

P: 492

well you have [tex]r,\theta,z[/tex] in cylindrical coordinates.
r? theta? z? z being the height and from what we've written you know what it is. so just find the r and the z and you can setup the integral. 


#11
Nov2407, 05:30 PM

P: 12

theta would be 0  2pi?
z would be the height soo lets say 0  a what would r be? 


#12
Nov2407, 05:33 PM

P: 492

r would be 0 to the circle's radius, say R, and the height of the cylinder can be described in terms of R. So you can have the square when looking at it from 2 sides.



#13
Nov2407, 05:41 PM

P: 12

what would the actual integrand be? [tex]rdrdz\theta[/tex]
r  0R theta  02pi z  0R ? 


#14
Nov2407, 05:47 PM

P: 492

no the z=2R because you need the diameter of the circle to = the height in order to get the square when looking at it from that side.
and yes the integrand will just be rdrdzd(theta). 


#15
Nov2407, 05:55 PM

P: 12

[tex] 2\pi (\int \int rdrdz) [/tex]
02R for first integral and 0R for second one. how would I go from there? the rdr is confusing me a little. 


#16
Nov2407, 05:59 PM

P: 492

[tex]\int_0^{2\pi}d\theta\int_0^R rdr\int_0^{2R}dz[/tex]
the rdr just means integrate r w.r.t. r just like: [tex]\int x dx[/tex] 


#17
Nov2407, 06:06 PM

P: 12

I got 2pi r^3 for final answer
after dividing by two. How does that sound? 


#18
Nov2407, 06:09 PM

P: 492

yep that's what I got.
the volume for a cylinder is pi r^2 *h. h=2r >2pir^3 and then *.5 = pi r^3 


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