# Confused by this integral

1. Dec 5, 2005

### Moneer81

Hello,

It has a been a while since college let alone since I've done any serious calculus, so I hope you can help explain this integral in a simple way that I can understand:

refer to the attached figure. I am trying to find the integral of z dV
where dV is increment of volume.
which could be written as integral of z dx dy dz

now I am trying to understand how to integrate z over dx and dy. the integral will be a circle of a certain radius but how do we exactly integrate z over dx and over dy ?

please no spherical/cylindrical coordinates. thanks a lot.

File size:
6.1 KB
Views:
74
2. Dec 5, 2005

### Tide

When you integrate over x and y, the variable z is fixed in the integrand and only comes into play in the limits of integration. You may find it more convenient to use cylindrical coordinats.

3. Dec 5, 2005

### BerkMath

You must evaluate int(int(int(z)))dV where V is the region enclosed by the positive cone in the xyz plane. In general, if V is defined by: a<x<b, h_1(x)<y<h_2(x), g_1(x,y)<z<g_2(x,y) then int(int(int(f(x,y,z)))dV=int(int(int(f(x,y,z),z,g_1(x,y),g_2(x,y)),y,h_1(x),h_2(x)),x,a,b). In your case, f(x,y,z)=z. g_1(x,y)=0 g_2(x,y)=+sqrt((h/r)(x^2+y^2)) h_1(x)= -sqrt(r^2-x^2) h_2(x)= +sqrt(r^2-x^2) and a=-r b=r. Plug and chug.

4. Dec 6, 2005

### Moneer81

bear with me here,
so are you saying that

$$\int z dx = z \int_{z=0}^{z=h} dx$$

which is equal to $$z . \left[ \ x \right]_{z=0}^{z=h}$$

??

thanks again

Last edited: Dec 6, 2005
5. Dec 6, 2005

### Edwin

would have to agree with Tide,
Because you will need to integrate over a region D that is a circular disc centered at the origin, your best bet is to convert to cylindrical coordinates. This will make your life a lot easier.

To find volume you will need to do a tripple integral.
Your domain will be theta is between 0 and 2(pi)
r is between 0 and 1, and z is between 0 and z.
your tripple integral will simply be,
the integral (z)dz evaluated from z=0 to z=c.

The tripple integral comes out to (pi)*(c^2)/2, where c is the height. This is for the volume of a cylinder with radius r = 1 and height z = c.

Hope this helps a little.

Best Regards,

Edwin

6. Dec 6, 2005

### Moneer81

I do understand that using cylindrical coordinates will be much easier but the book I got this example from used rectangular coordinates and I am just trying to understand this integral for the practice and in case I come across something similar in the future. This is why I am sticking to rectangular coordinates.

thanks again

7. Dec 7, 2005

### Tide

If you insist on using cartesian coordinates then start by limiting yourself to one octant (based on symmetry) so your integral should look like this:

$$I = 4\int_{0}^{h} z dz \int_{0}^{\frac {R}{h} z} dy \int_{0}^{\sqrt {(Rz/h)^2-y^2}} dx$$

As an exercise you may want to try doing it using cylindrical coordinates anyway.

8. Dec 17, 2005

### Moneer81

Thanks a lot Tide.

Can you clarify how did you get the limits of integration of the dy and the dx integral? And also where did the 4 come from?

9. Dec 19, 2005

### Tide

Gosh, you want me to remember what I did two weeks ago?? ;)

Seriously, what you want to note is that at a fixed z-level, the surface is bounded by $x^2 + y^2 = r^2$ where r is the radius at that height. The radius, of course, depends on height and is given by $r = R z/h$ being 0 at the vertex and R at the top (z = h).

To set up the integral, we'll add up the contributions from infinitesimal disks whose volume is dx dy dz (you did insist on using cartesian coordinates!) so the integrand is z dx dy dz. I.e. the volume of each disk is their height (dz) time their cross sectional area (dx dy). We'll focus on the area part first.

Using symmetry, I chose to integrate over a single quadrant in x-y (hence, the factor of 4). Since $x^2 + y^2 = r^2$ we can integrate first over x from x = 0 to $x = \sqrt {r^2 - y^2}$. Once you have completed that integration, proceed to the y integration which extends from 0 to r. But $r = R z/h$ which is what I wrote for the upper limit on the y integration.

Finally, z goes from 0 to h which I use for the limits of the z integration.

Last edited: Dec 19, 2005