18.10. EXERCISES 495

2. Let π : Rn→ Rm be defined as π (x) = xi where i= (i1, i2, ..., im) . What does The-orem 17.6.1 say if π = f in this theorem?

3. In calculus, you found the area of the parallelogram determined by two vectors u,vin R3 by taking the magnitude |u×v| , meaning the Euclidean norm of the crossproduct. Show that you get the same answer by forming(

det(u v

)∗ (u v

))1/2

where here you have(u v

)is the matrix which has u as first column and v as

second column.

4. In calculus, you also found the volume of a parallelepiped determined by the vectorsu,v,w by |u·(v×w)| , the absolute value of the box product. Show this turns outto be the same thing as(

det(u v w

)∗ (u v w

))1/2.

5. Imagine a fluid which does not move. Let B(x,ε) be a small ball in this fluid. Usethe Euclidean norm. Then the force exerted on the ball of fluid is −

∫∂B(x,ε) pndA

where p is the pressure. Here n is the unit exterior normal. Now the force actingon the ball from gravity is −gk

∫B(x,ε) ρdV where ρ signifies the density of the fluid

and k signifies the direction which is up. The vectors i,j are in the direction of thepositive x and y axes respectively. These two forces add to 0 because it is given thatthe fluid does not move. Use the divergence theorem to show that ∇p = ρgk. Thisis a really neat result.

6. Archimedes principle states that when a solid body is immersed in a static fluid, theforce acting on the body by the fluid is directly up and equals the total weight ofthe fluid displaced. Surely this is an amazing result. It doesn’t matter about theshape of the body. Remember that the weight is the acceleration of gravity timesthe mass. Hint: You need to start with the force acting on the body B by the fluidwhich is−

∫∂B pndA. Assume the divergence theorem holds for B. As shown, this is

typically the case.

7. You have a closed region R which is fixed in space. A fluid is flowing through R.The density of this fluid is ρ (t,x) where x gives the coordinates in space and ρ

depends on t because it might change as time progresses. The velocity of this fluidis v (t,x). Then the rate at which the fluid crosses a surface S from one side to theother is

∫S ρv ·ndS where n is the unit normal to the surface which points in the

direction of interest. You can think about this a little and see that it is a reasonableclaim. If, for example, v ·n = 0, then the velocity of the fluid would be parallel tothe surface so it would not cross it at all. Also, the total mass of the fluid which isin R is

∫R ρdV. Assuming anything you like about regularity, which is what we do in

situations like this, explain using the divergence theorem why∫

V∂ρ

∂ t +∇ · (ρv)dV =0. Recall that ∇·F denotes the divergence of F . Now explain why it should be thecase that ∂ρ

∂ t +∇ · (ρv) = 0. This is called the balance of mass equation.

8. The permutation symbol is ε i jk = 1 if the permutation(

1 2 3i j k

)is even and

−1 if this permutation is odd. Such a permutation is odd or even depending on