Famous integral's of exponiental

1. Aug 19, 2008

dj023102

Hi guys, i have attached my question. really am stuck with this one. apparently it is a famous integral from poisson. Any ideas on where to start would be good. Cheers

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2. Aug 20, 2008

HallsofIvy

Staff Emeritus
I, and many other people, will not open "Word" files. They are notorious for having viruses.

3. Aug 20, 2008

tiny-tim

… famous French fish …

Hi dj023102!

If you really are stuck, then you have plenty of time to type it out for us …

4. Aug 20, 2008

dj023102

Gee you guys are difficult :)
it's all good just thought it was quicker to use word to type the query up, lets see how i go using latex
evaluate[Infinity ]\int[/-Infinity] [e]^{-x^2}.
does that make sense?
basically integrate e^(-x squared) with upper boundary of infinity and lower boundary of minus infinity.
The tutor wants to image the graph in 3D, the graph of the function
z = e^(-x^2) on the xz-plane. Then rotate this graph around the z-axis to get a bell shaped infinite surface hovering above the xy-plane. The solid chunk is in a space between the xy-plane and the shape looks like a bell.

When you intersect the bell with the infinite hollow cylinder of radius r and the z-axis as its central axis you get a finite hollow cylinder. How do you find the area of this finite hollow cylinder (this cylinder is lacking the caps on both ends, so the area of these caps don’t occur towards the area of the cylinder)

I have no idea where to start, I always thought a cylinder only has volume. Anyone have any ideas where to begin with this one?

5. Aug 20, 2008

Dick

You want the integral of e^(-x^2)*dx over all x, call it I. That's the same as the integral e^(-y^2)*dy. If you multiply them together you get I^2=integral e^(-(x^2+y^2))dxdy in the x-y plane. You can do this integral by changing to polar coordinates. I think that's what you want to do from your description.

6. Aug 21, 2008

dj023102

The question ask what is the area of a finite cylinder of radius r when it intersect the bell when the function z=e^(-x^2) is rotated around the z axis (to make the bell).
The area is A = pi times (f(x))^2
Is that what the question is asking i think, but is f(x) = e^(-x^2) or because it is rotated around the z axis i need to make x the subject so f(z) = square root of -In(z)
Does it really matter?

7. Aug 21, 2008

Dick

You find the area of that cylinder just like you find the area of any other cylinder. It's the height times the circumference.

8. Aug 21, 2008

dj023102

Yeah but which area should i use?

f(x) = e^(-x^2) or because it is rotated around the z axis i need to make x the subject so f(z) = square root of -In(z)

cos after that i need to calculate the volume of the cylinder by integrating the area with lower bound 0 and upper bound infinity. How would i go about using that? but firsty i need to know which function to use, any help there?

9. Aug 21, 2008

Dick

Once you rotate it x^2 becomes r^2=x^2+y^2. So the height of the cylinder is exp(-r^2). What's the radius? You are overthinking this.

10. Aug 21, 2008

dj023102

Ok the height of the exp(-r^2) is, but what about the area? thats is what i having the problem with. i am assuming that i should take the area of the circular disk and times it by the height (exp(-r^2)) and that will give me the volume.

But what is the area i should be multiplying by?

11. Aug 21, 2008

Dick

You want the AREA of the CYLINDER. You will later integrate that to get the volume of the solid. You don't want the volume of the cylinder. Again, area=height times circumference.

12. Aug 21, 2008

dj023102

ah ok! i think i get it. Sorry been looking at the wrong part of the text.
So area of cylinder = (exp(-r^2))*(2*r*pi).
Is that right?

13. Aug 21, 2008

splatter

hey i've been watching this thread for a while, and im pretty sure i have exactly the same assignment.. anyway the one i have is broken up into 6 sections, the 4th of which is really bugging me. It says "express the area of intersection of the bell (the rotated integral) with the plane parallel to, and a distance y, from the xz-plane as a function S(y).
just stuck on how you would put this intersection into writing, or where to start from. any ideas???

14. Aug 22, 2008

dj023102

hang on there i need more help, stuck on the first part of the question. but i am heading in the right direction now am i?

15. Aug 22, 2008

splatter

yeah im pretty sure, thats how i did that part anyway. if you've found C(r) it should be fairly easy to integrate.

16. Aug 22, 2008

splatter

i think i have that part now, but theres some other stuff that follows on from that which is really confusing me, even though it should be fairly straight forward. i THINK i have S(y), and it then says that the integral S(y) over all y is equal to the volume of the bell, given by the integral of the cylinder mentioned earlier in thread. S(y) ends up as "integral e^(-(x^2+y^2))dx over all x" i think, but it then says to find the integral of S(y) over all y, which will contain two integral signs, and separate integral S(y) into the product of two simple integrals, one containing only variable x and the other containing only y. the reason im confused is that i end up with [integral e^(-x^2)dx over all x]*[integral e^(-y^2)dy over all y], which presents the same problem as the initial integral of I=int e^(-x^2) dx over all x....

17. Aug 22, 2008

tiny-tim

Hi dj023102!

(have an integral: ∫ and a squared: ² and a theta: θ )
hmm … you might have said that at the beginning, so that we knew what the context was.

ok … the aim is to find the volume under the bell.

To find a volume, you're probably used to dividing the volume into horizontal slices of thickness dz, and then integrating over z.

But in cases of rotational symmetry (like this), it's easier to use cylindrical slices (or "shells") of thickness dr, and then integrate over r.

The volume of a cylindrical shell is A dr, where A is the area …
that's the only reason you need to know the area!
Hi splatter!

Hint: ∫∫e-x² e-y² dx dy = ∫∫e-r² dx dy …

now change the variables of integration from x and y to r and θ.

18. Aug 22, 2008

Dick

That's exactly right. If you integrate that r=0 to infinity you get the volume of the region under the bell. And, as splatter is finding, that is I^2, where I=integral exp(-x^2)dx.

19. Aug 22, 2008

splatter

Sorry about that :P i see the stupidity now.. hehe
thanks heaps for that though - big help!!! :) i hope i can do it properly, its been making me crazy!

Thanks again

20. Aug 22, 2008

splatter

sorry i just realised i dont understand what you meant by changing the variable y to θ - i don't have any variables except x, y, r, and possibly z.. sorry to be so annoying =/