# Setting up Triple Integrals over a bounded region

1. Jul 15, 2012

### forestmine

1. The problem statement, all variables and given/known data

Set up triple integrals for the integral of f(x,y,z)=6+4y over the region in the first octant that is bounded by the cone z=(x^2+y^2), the cylinder x^2+y^2=1 and the coordinate planes in rectangular, cylindrical, and spherical coordinates.

2. Relevant equations

∫∫∫dzrdrdθ

∫∫∫$\rho$^2*sin$\phi$d$\rho$d$\phi$d$\theta$

∫∫∫dxdyzdz

3. The attempt at a solution

I think this problem is mainly giving me a hard time due to that it is in the first octant. I'm still having a hard time visualizing things in three dimensions.

First thing I did was sketch the region. I have the cylinder with a radius of 1 centered on the z-axis. My cone, starts at the origin, and extends to a maximum radius of 1 where it touches the cylinder. (Now this is where I get a little confused...) The axes all bound the region as well, so the z axis cuts the cone and the cylinder in half, and then the half remaining is once again cut in half by the x and y axes?

Ok. So given that's at least remotely correct, I began by setting up the integral in cylindrical coordinates.

For my z limits of integration, I said from z=(x^2+y^2)^1/2, which is z=r to z=1. For my r limits, from 0 to 1. And I'm confused about $\theta$, since we're dealing with the "first octant," so I'm thinking it's from 0 to pi/2.

My function itself, 6+4y, is not in the appropriate coordinate system. Does the function in the integral become 6+4(rsin$\theta$)?

For my cartesian coordinates, I said that my z limits of integration should be the same, I think. Now when I look at the circle on the xy-plane, since we're looking in only one octant, I'm essentially looking at a quarter of the circle, right? So my x-limits are from x=0 to x=(1-y^2)^1/2. And my y limits are simply 0 to 1.

Lastly, spherical coordinates.

For my limits for $\rho$, I have from 0 to 1. $\phi$ is from 0 to pi/4, and $\theta$ is from 0 to pi/2.

Now as far as the function inside the integral, I already have 6+4y, which converts to 6+4($\rho$sin$\phi$sin$\vartheta$) and then multiplied by $\rho$^2*sin$\phi$.

Whew. Alright.

Hopefully I'm somewhat on the right track, at least. Hope my formatting is easy enough to follow. Thanks so much!

2. Jul 15, 2012

### LCKurtz

It isn't the axes that cut the region in half, it is the zy and zx planes that do.

Yes, you have the right idea. It isn't difficult to write that in latex. Your integral would look like this
$$\int_{0}^{2\pi}\int_{0}^{1}\int_{r}^{1} 6+4r\sin\theta dzdrd\theta$$
Here's how that is written:
Code (Text):

$$\int_{0}^{2\pi}\int_{0}^{1}\int_{r}^{1} 6+4r\sin\theta dzdrd\theta$$

But expressed with only x,y,z's in the limits. No r.

Yes. Try writing the integrals
I assume you are describing limits for the order of integration $d\rho d\theta d\phi$. If you think about $\rho$, it would go from the origin to the $\rho$ on the cylinder, which clearly depends on $\phi$. And your region is outside the cone, so you need to re-think your values for $\phi$.

3. Jul 15, 2012

### forestmine

Should your theta limits be from 0 to $$\theta/2$$ since it's only the first octant?

So this is what my integral in cartesian coordinates looks like:

$$\int_{0}^{1}\int_{0}^{\sqrt{1-y^{2}}}\int_{\sqrt{x^{2}+y^{2}}}^{1} 6+4y dzdxdy$$

I guess I don't quite understand $\rho$, in that case. I know that it refers to the distance from the origin, but since we're in three dimensions here, I'm confused about how to think about $\rho$.

As for $\phi$, from what I understand, it's 0 at the positive z-axis and pi at the negative z axis. That's where I got pi/4 from for the phi limits of integration.

Last edited: Jul 15, 2012
4. Jul 15, 2012

### LCKurtz

WOOPS! I made a mistake telling you that was right. Your z limits describe the inside of the cone but you want the outside. And your remark below is correct except for the typo. Yes, for the first octant the upper limit for $\theta$ should be $\pi/2$.

Again, your z limits are wrong. You want the outside of the cone (below the cone).
$\rho$ goes from the origin to any place on the cylinder. You need to put the cylinder into spherical coordinates to see what $\rho$ on the cylinder is.
$\phi$ goes from $\phi$ on the cone to $\phi$ on the xy plane. You need to figure out the vertex angle of the cone to know its $\phi$ value. Hint: It's constant.

5. Jul 15, 2012

### forestmine

Hm, I am thoroughly confused now, haha. I thought the region was the inside of the cone?
What I'm picturing: If I drew a cylinder and placed a cone directly inside it, so that the top of the cone meets the top of the cylinder, the region I'm looking at is inside the cone.

6. Jul 15, 2012

### LCKurtz

The cone has its point at the origin and opens up. If the region were inside the cone, how could it be bounded by the cylinder which is outside the cone?

7. Jul 16, 2012

### forestmine

Yeah, you're absolutely right. I realized that right after I submitted that last post.

Here are my new integrals.

Cartesian Coordinates:

$$\int_{0}^{1}\int_{0}^{\sqrt{1-y^{2}}}\int_{0}^{\sqrt{x^{2}+y^{2}}} 6+4y dzdxdy$$

Cylindrical Coordinates:

I'm not sure about this one, particularly the r limits. It starts at the origin, and goes out to the edge of the cylinder and the edge of the cone. If I describe it as the cylinder, I'd have r variables in the limits for r, which doesn't seem right.

$$\int_{0}^{\pi/2}\int_{0}^{r^{2}}\int_{0}^{r} r(6+4rsinθ) dzdrdθ$$

Spherical Coordinates:

$$\int_{0}^{\pi/2}\int_{\pi/4}^{\pi/2}\int_{0}^{1/sin(\phi)} \rho^{2}sin{\phi}(6+4(\rho sin(\phi) sin(\theta)) d\rho d\phi d\theta$$

How do they look?

Last edited: Jul 16, 2012
8. Jul 16, 2012

### LCKurtz

Good.

No, it isn't right as you suspect. You can never have the variable of integration in the limits for that variable in this type of problem. Once you have integrated in the z direction you need to look at the shadow of your figure in the $xy$ plane to get the $r,\theta$ limits.

And the last one is good.

9. Jul 16, 2012

### forestmine

Ah ok. How's this?

$$\int_{0}^{\pi/2}\int_{0}^{1}\int_{0}^{r} r(6+4rsinθ) dzdrdθ$$

Thanks again for all the help by the way. Definitely starting to feel a bit better about these conversions.

10. Jul 16, 2012

### LCKurtz

Yes. You're welcome.