Volume of cylinders using double integrals

  • #1

Homework Statement


1. Find the volume bounded by the cylinder y^2 + z^2 = 4 and the planes x=2y, x=0, z=0 in the first octant.

2. Find the volume bounded by the cylinders x^2 + y^2 = r^2 and y^2 + z^2 = r^2

Homework Equations





The Attempt at a Solution


Both questions are from the section before polar/cylindrical coordinates and triple integrals so we have to use double integrals in cartesian coordinates only.

1. I drew a graph on the xy plane with x=2y. When z=0, y^2 = 4, y=2. So the boundary is the triangle between x=2y and y=2.
the boundary of integration is {(x,y)|0<=x<=4. x/2<=y<=2}, is this right? I think the limits for x are right but y I'm not too sure.

If I continue, then I get V = ∫ ∫f(x,y)dydx = ∫ ∫sqrt(4-y^2)dydx with the limits above. The inner integral gets a bit tricky with trig substitutions. I made y = 2sinu. After a few steps, I got ∫sqrt(4-y^2)dy = y/2*sqrt(4-y^2) + 2arcsin(y/2). Plugging in the limits for y...I got π + x/2 -(x/4)sqrt*4-x^2/4) - arcsin(x/4)
And if I integrate that with respect to x from 0 to 4 I get another messy integral with many terms in the solution. One of the term is 2sqrt(4-x^2). But that is undefined when x = 4! So I did something wrong. I don't think it was the actual integration that I messed up on because I used wolfram to double check.

2. The boundary I got is {(x,y)|-r <= x <= r, -sqrt(r^2 - x^2) <= y <= sqrt(r^2 - x^2)}
The integral is pretty similar to the one above but it's very long and complex with many square roots and arcsines to integrate. Did I make the limits wrong because it seems too complex...

The questions would be a lot simpler in polar form I think.
 
Last edited:

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
i think you need to be a bit careful with setting up the integral

First if you are only integrating over 2 variables, the first step is to decide what a volume element is
dV(x,y) = f(x,y)dxdy

the f(x,y) actually comes form the first step in a triple integral & will be the extent of z for a given x & y.

For the first case, drawing a picture of the "wedge" cut form the cylinder, i think f(x,y) should depend on both x & y. So maybe have another look at your initial volume element.

I also think for that problem, it would be easier to set up the integration in terms x & z. as the volume element should only depend on x in that case (and will given by the difference in y between the planes x = 0 & x = 2y)
[tex] \int_a^b (\int_{g(x)}^{h(x)} f(x) dz) dx [/tex]

note that by setting these up in this way you are pretty much doing a triple integral, just shortcutting the first step anyway
 

Related Threads on Volume of cylinders using double integrals

Replies
3
Views
2K
Replies
2
Views
5K
Replies
3
Views
12K
Replies
14
Views
3K
Replies
3
Views
913
Replies
5
Views
1K
Replies
1
Views
2K
Replies
2
Views
1K
  • Last Post
Replies
8
Views
8K
  • Last Post
Replies
2
Views
2K
Top