# Tricky Multiple integral word problem

by numberonenacho
Tags: integral, multiple, tricky, word
 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
 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?
 P: 12 Im still not getting it.. Pictures would help. >.< So Square base and top. Sides triangle? Where would the circle go?
 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.
 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?
 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.
 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?
 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.
 P: 12 could you help me set it up? thats the part im really bad at. Many thanks for discussing this problem with me. =)
 P: 492 well you have $$r,\theta,z$$ 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.
 P: 12 theta would be 0 - 2pi? z would be the height soo lets say 0 - a what would r be?
 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.
 P: 12 what would the actual integrand be? $$rdrdz\theta$$ r - 0-R theta - 0-2pi z - 0-R ?
 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).
 P: 12 $$2\pi (\int \int rdrdz)$$ 0-2R for first integral and 0-R for second one. how would I go from there? the rdr is confusing me a little.
 P: 492 $$\int_0^{2\pi}d\theta\int_0^R rdr\int_0^{2R}dz$$ the rdr just means integrate r w.r.t. r just like: $$\int x dx$$
 P: 12 I got 2pi r^3 for final answer after dividing by two. How does that sound?
 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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