Troubleshooting Surface Integral Issues

[Lancelot59]

I'm not sure why, but I'm having issues with these in general. Specifically surface integrals over vector fields.

The function is zk. The surface is the paraboloid z=x²+y² between the planes z=1 and z=2.

I parametrized it like so:
$$\vec{r}(u,v)=(u,v,u^{2}+v^{2})$$
$$\vec{T_{u}}=(1,0,2u)$$
$$\vec{T_{v}}=(0,1,2v)$$
$$\vec{N}=(-2u,-2v,1)$$

putting r into F:
$$\vec{F}(\vec{r}(u,v))=(u,v,u^{2}+v^{2})$$

The integral then wound up being

$$\int_{D}\int \vec{F(\vec{r})} \cdot \vec{N} = \int_{D}\int u^{2}+v^{2} dv du = \int^{2}_{0} \int^{2\pi}_{0} r^{2} * r dr d\theta = 16\pi$$

Which is apparently wrong. What's going on here?

 

[LCKurtz]

I'm not sure why, but I'm having issues with these in general. Specifically surface integrals over vector fields.

Surface integrals aren't "over vector fields". They are over surfaces.

The function is zk. The surface is the paraboloid z=x²+y² between the planes z=1 and z=2.

I parametrized it like so:
$$\vec{r}(u,v)=(u,v,u^{2}+v^{2})$$
$$\vec{T_{u}}=(1,0,2u)$$
$$\vec{T_{v}}=(0,1,2v)$$
$$\vec{N}=(-2u,-2v,1)$$

putting r into F:
$$\vec{F}(\vec{r}(u,v))=(u,v,u^{2}+v^{2})$$

The integral then wound up being

$$\int_{D}\int \vec{F(\vec{r})} \cdot \vec{N} = \int_{D}\int u^{2}+v^{2} dv du = \int^{2}_{0} \int^{2\pi}_{0} r^{2} * r dr d\theta = 16\pi$$

Which is apparently wrong. What's going on here?

You haven't used the fact that you only want the part of the surface between the planes z = 1 and z = 2. And were you given an orientation for N and checked it?

 

[Lancelot59]

Surface integrals aren't "over vector fields". They are over surfaces.

I mixed up my words there. Sorry.

You haven't used the fact that you only want the part of the surface between the planes z = 1 and z = 2. And were you given an orientation for N and checked it?

I forgot to post that. N was to be the outward normal, and by logic what I have there makes sense.

I figured out my problem. I used a bad set of limits, and made a dumb integration error. The issue I really have is picking limits for these problems. I apparently have a tendency for turning things into boxes that shouldn't be.

 

[LCKurtz]

I mixed up my words there. Sorry.


I forgot to post that. N was to be the outward normal, and by logic what I have there makes sense.

Are you sure about that? Draw a picture of the paraboloid with an outward normal and think about whether the z component should be positive or negative.

 

[Lancelot59]

Are you sure about that? Draw a picture of the paraboloid with an outward normal and think about whether the z component should be positive or negative.

Ah. So my normal should be (2u,2v,-1). So I just get the negative of what I had before. Thanks!

 

[LCKurtz]

Ah. So my normal should be (2u,2v,-1). So I just get the negative of what I had before. Thanks!

You're welcome. You always should check whether your normal direction agrees with the orientation. Also, in this problem your vector field is directed upward, which is opposite to the orientation, so you can check that your answer comes out negative.

 

[Lancelot59]

You're welcome. You always should check whether your normal direction agrees with the orientation. Also, in this problem your vector field is directed upward, which is opposite to the orientation, so you can check that your answer comes out negative.

Opposite because the k component of the normal is negative?

 

[HallsofIvy]

You said the vector function to be integrated was $F(x,y,z)= z\vec{k}$. In terms of your paremeters, u and v, that would be $F(u, v)= (u^2+ v^2)\vec{k}$, NOT $F(u, v)= u\vec{i}+ v\vec{j} + (u^2+ v^2)\vec{k}$ as you have.

Your figure is a paraboloid opening upward. The outward normal will have negative z component and so should be $2u\vec{i}+ 2v\vec{j}- \vec{k}$.

 

[Lancelot59]

Wouldn't I just have a line up the z axis then?

 

[LCKurtz]

Opposite because the k component of the normal is negative?

Opposite because at each point on the surface your vector field F = zk is pointing upwards which is across the surface in the opposite direction from the orientation, which is downwards. Also note Hall's correction which error I didn't notice since somehow you got the integrand right anyway.

 

[HallsofIvy]

Wouldn't I just have a line up the z axis then?

No, the fact that the vector function to be integrated is zk means the vector field points straight up. That has nothing to do with the surface of integration (which is, after all, a surface, not a line.)

 

[Lancelot59]

No, the fact that the vector function to be integrated is zk means the vector field points straight up. That has nothing to do with the surface of integration (which is, after all, a surface, not a line.)

OH! I thought you had changed the parameterization of the function. That's what I was talking about. I see what you have there now. Makes sense.

Opposite because at each point on the surface your vector field F = zk is pointing upwards which is across the surface in the opposite direction from the orientation, which is downwards. Also note Hall's correction which error I didn't notice since somehow you got the integrand right anyway.

That is strange...I must be some kind of genius!