# Showing that a surface is a disk? Help please.

1. Nov 6, 2009

### Susanne217

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

Given the vectorfield

$$\textbf{F}(r) = y^2\bold{i} + z^2\bold{j} + x^2\bold{k}$$ and the surface S defined by a mapping

$$(\rho,\phi) \in [0,1] \times [0,2\pi] \mapsto \bold{r}(\rho,\phi) = 4\bold{i} + (1+\rho cos(\phi))\bold{j} + (2+\rho sin(\phi))\bold{k}$$

show that S is a disk parallel with the plane yOz with the center at $$\bold{r}_c = 4\bold{i} + \bold{j} + 2\bold{k}$$

2. Relevant equations

3. The attempt at a solution

I can see that I need to rewrite S is a circle and then I take y and z component of S and write them as a circle. That I get :)

But how do I use this to show that S can be viewed as plane yOz parallel with a disk?

Which theorem do I use?

2. Nov 6, 2009

### Staff: Mentor

No theorem. Use your brain. In parametric form your surface S is x = 4, y = 1 + r cos(t), z = 2 + r sin(t) (using r instead of rho, and t instead of theta). It's pretty easy to show that for fixed r, these equations describe a circle of radius r, centered at (4, 1, 2). If r is allowed to vary, the circles fill in a disk of radius 1.

3. Nov 6, 2009

### Susanne217

If I then run the math I get that

$$\bold{F}(r(\rho,\phi)) = 4^2 + (2+r\cdot sin(t))^2 + (1+r\cdot cos(t))^2 = 18 + r^2 +2r \cdot cos(t)+2\cdot r \cdot sin(t) +18 = x^2 + y^2$$

where rcos(t) = x and rsin(t) = y thus

$$18 + r^2 +2r \cdot cos(t)+2\cdot r \cdot sin(t) +18 = x^2 + y^2 \leftrightarrow x^2 + y^2 -18 -2x - 2y = r^2$$

Last edited: Nov 6, 2009
4. Nov 6, 2009

### Staff: Mentor

I don't know why you would want to come up with the equation of a sphere first. The parametric equations I gave describe exactly the disk that you want, and it is parallel to the y-z plane.

The problem seems to have extra information in it, namely the business about the vector field. This equation
$$(\rho,\phi) \in [0,1] \times [0,2\pi] \mapsto \bold{r}(\rho,\phi) = 4\bold{i} + (1+\rho cos(\phi))\bold{j} + (2+\rho sin(\phi))\bold{k}$$

5. Nov 6, 2009

### Susanne217

do I need to show that the dot-product between the circle in yz plane and some line x=??? is different from zero and thus they are parallel?

6. Nov 6, 2009

### Staff: Mentor

Every point on the disk has an x-coordinate of 4. That means that the disk is parallel to the plane x = 0 (the y-z plane).

7. Nov 7, 2009

### Susanne217

okay to generate the circle in YoZ plane I set x = 0

and get the circle

$$(1+rcos(t))^2 + (2+ rsin(t))^2 = r^2 + 2r \cdot cos(t) + 2r \cdot sint(t) +2$$ and this is then $$y^2 + z^2 = r^2 = -2rcos(t)- 2rsin(t) - 2$$

That if I put rsin(t) = y and rcos(t) = z

that I get $$r^2 = (y-1)^2 + (z-1)^2 = r^2 = 1$$ if I insert (1,2) in that disk it gives 1. But if this is my disk how do i show that this disk i parallel with let say x=4?

Last edited: Nov 7, 2009
8. Nov 7, 2009

### HallsofIvy

Staff Emeritus
No matter what r and t are, x is always 4. The disk is in the plane x= 4. A normal vector to that plane is <1, 0, 0>. That is also normal to the YZ plane.

9. Nov 8, 2009

### Susanne217

I don't have any of the fancy math software like Mathlab, Maple etc. But I would like to know is there any website where I can do a computer drawing of this for free?

10. Nov 8, 2009

### Staff: Mentor

From an earlier post, x = 4, y = 1 + rcos(t), and z = 2 + rsin(t).
==> x = 4, y - 1 = rcos(t), and z - 2 = rsin(t)
==> x = 4 and (y - 1)2 + (z - 2)2 = r2(cos2(t) + sin2(t)) = r2

For each value of r such that 0 <= r <= 1, these parametric equations represent a circle centered at (4, 1, 2) of radius r. Each circle is in the plane x = 4, which is parallel to the y-z plane. Taken together, these circles constitute a circular disk of radius 1, centered at (4, 1, 2).

You don't need "fancy software" like Matlab or Maple to be able to do what I've done above. These software packages can be helpful at times, but if you're not able to carry out analysis like what's above, these tools might be a hindrance to your understanding.