# Tricky Multiple integral word problem

1. ### numberonenacho

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. ### bob1182006

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. ### numberonenacho

12
Im still not getting it.. Pictures would help. >.<
So Square base and top. Sides triangle? Where would the circle go?

4. ### bob1182006

492
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.

12
6. ### bob1182006

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. ### numberonenacho

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. ### bob1182006

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. ### numberonenacho

12
could you help me set it up? thats the part im really bad at. Many thanks for discussing this problem with me. =)

10. ### bob1182006

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.

11. ### numberonenacho

12
theta would be 0 - 2pi?
z would be the height soo lets say 0 - a
what would r be?

12. ### bob1182006

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. ### numberonenacho

12
what would the actual integrand be? $$rdrdz\theta$$
r - 0-R
theta - 0-2pi
z - 0-R

?

14. ### bob1182006

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. ### numberonenacho

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.

16. ### bob1182006

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: $$\int x dx$$

17. ### numberonenacho

12
I got 2pi r^3 for final answer
after dividing by two. How does that sound?

18. ### bob1182006

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

19. ### numberonenacho

12
cool! Thanks a lot for all your help. ^_^