Question

Just question 6![

In a bathtub, the velocity of water near2 the drain is given by the vector field (22 +1)2222 + 1 where r, y, and z are measured in centimeters and (0, 0,0) is at the center of the drain. 1. Rewriting F as follows, describe in words how the water is moving: Consider each of the three terms in equation (4). (Look at some plots.) For fixed z, what is the flow like? How does the flow change with z? 2. The drain in the bathtub is a disk in the ry-plane with center at the origin and radius 1 cm. Compute the flux of F through this disk, oriented down. Give units for your answer. What is the meaning of this number (in terms of the tub water)? 3. Find the divergence of F. (Simplify your answer, if possible 4. Find the net flux of water out through the surface of a sphere of radius 1 centered at the origin, oriented outward. You may compute the flux integral directly or use an appropriate theorem. 5. Set up and evaluate a flux integral to find the net flux of water through the bottom 2,oriented downwards --,/1 _ 2.2-2,2. Orient half of the sphere. (Set up and evaluate the double integral in spherical coordinates; do not use a theorem here.) 6. For the vector field G - where C is the edge of the drain oriented clockwise when viewed from above

Just question 6!

Answer 1

Since the plane of the drain is simply in x-y plane with origin at the center, this implies z = 0 for this disk.

Thus the vector field $\vec{G}$ in this disk is

$\vec{G} = \frac{y}{2} \vec{i} -\frac{x}{2} \vec{j}+ \frac{x^2+y^2}{2} \vec{k}$

For the line integral about the edge of the drain $\vec{dr} = dx \vec{i} + dy \vec{j} + 0\vec{ k}$ as the vector $\vec{dr}$ is embedded in the disk.

cG.dr Jc (d- dy)

If is the radius of the curve over which the line integral is taken, the radius vector $\vec{r}$ can be written in terms of polar coordinates.

Hence, $\vec{r} = a\cos{\theta}\vec{i} + a\sin{\theta}\vec{j} = x \vec{i} + y\vec{j}$

thus $\vec{dr} = -a\sin{\theta}d\theta\vec{i} + a\cos{\theta}d\theta\vec{j} = dx \vec{i} + dy\vec{j}$

Thus,

a cos

2na に。

cG.dr Jc (d- dy)
= $-\pi a^2$