Setting up Triple Integrals over a bounded region

[forestmine]

Homework Statement

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.

Homework Equations

∫∫∫dzrdrdθ

∫∫∫\rho^2*sin\phid\rhod\phid\theta

∫∫∫dxdyzdz

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(\rhosin\phisin\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!

 

[LCKurtz]

Homework Statement

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), (you mean $\sqrt{x^2+y^2}$)
the cylinder x^2+y^2=1 and the coordinate planes in rectangular, cylindrical, and spherical coordinates.

Homework Equations

∫∫∫dzrdrdθ

∫∫∫\rho^2*sin\phid\rhod\phid\theta

∫∫∫dxdyzdz

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?

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

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)?

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:

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

For my cartesian coordinates, I said that my z limits of integration should be the same, I think.

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

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.

Yes. Try writing the integrals

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(\rhosin\phisin\vartheta) and then multiplied by \rho^2*sin\phi.

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 $\phi$.

 

[forestmine]

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

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 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 $\phi$.

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.

 

[LCKurtz]

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:

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 $\pi/2$.

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:


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

Again, your z limits are wrong. You want the outside of the cone (below the cone).

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 $\rho$.

$\rho$ 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.

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.

$\phi$ 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.

 

[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.

 

[LCKurtz]

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.

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?

 

[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?

 

[LCKurtz]

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


Here are my new integrals.

Cartesian Coordinates:

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

Good.

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.

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.

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


Spherical Coordinates:

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

And the last one is good.

 

[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.

 

[LCKurtz]

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.

Yes. You're welcome.