- Let V be such that the divergence theorem holds. Show that
where n is the exterior normal and

is defined in Problem 5. Here ∇^{2}u ≡∇⋅and in rectangular coordinates this is just ∑_{i=1}^{n}u_{xixi}. - In calculus, you found the area of the parallelogram determined by two vectors u,v in
ℝ
^{3}by taking the magnitude, meaning the Euclidean norm of the cross product. Show that you get the same answer by forming^{1∕2}where here you haveis the matrix which has u as first column and v as second column. - In calculus, you also found the volume of a parallelepiped determined by the vectors
u,v,w by , the absolute value of the box product. Show this turns out to be the same thing as
- Show that ∇⋅= 0. Now show ∫
_{∂V }∇× v ⋅ ndA = 0 where V is a region for which the divergence theorem holds and v is a C^{2}vector field. You could also use Stoke’s theorem to verify this. ∇× v is the curl of v. In the new notation,_{i}= ε_{ijk}∂_{j}v_{k}where ∂_{j}is an operator which means to take the partial derivative with respect to x_{j}. It is understood here that the coordinates are rectangular coordinates. The first part of this is real easy if you remember the big theorem about equality of mixed partial derivatives. - Imagine a fluid which does not move. Let Bbe a small ball in this fluid. Use the Euclidean norm. Then the force exerted on the ball of fluid is
where p is the pressure. Here n is the unit exterior normal. Now the force acting on the ball from gravity is −gk∫

_{B}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 the positive x and y axes respectively. These two forces add to 0 because it is given that the fluid does not move. Use the divergence theorem to show that ∇p = ρgk. This is a really neat result. - Archimedes principle states that when a solid body is immersed in a static fluid, the
force acting on the body by the fluid is directly up and equals the total weight of the
fluid displaced. Surely this is an amazing result. It doesn’t matter about the shape of
the body. Remember that the weight is the acceleration of gravity times the mass.
Hint: You need to start with the force acting on the body B by the fluid which is
−∫
_{∂B}pndA. Assume the divergence theorem holds for B. As shown, this is typically the case. - You have a closed region R which is fixed in space. A fluid is flowing through R. The
density of this fluid is ρwhere x gives the coordinates in space and ρ depends on t because it might change as time progresses. The velocity of this fluid is v. Then the rate at which the fluid crosses a surface S from one side to the other 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 reasonable claim. If, for example, v ⋅ n = 0, then the velocity of the fluid would be parallel to the surface so it would not cross it at all. Also, the total mass of the fluid which is in R is ∫_{R}ρdV. Assuming anything you like about regularity, which is what we do in situations like this, explain using the divergence theorem whyRecall that ∇⋅F denotes the divergence of F. Now explain why it should be the case that

This is called the balance of mass equation.

- Let α
_{n}be the volume of the unit ball in ℝ^{n}. (Euclidean norm) Then by the divergence theorem,Explain why the unit exerior normal to this ball is x∕

. Then explain whyFind areas of spheres of radius r in ℝ

^{n}for various values of n. For example, the one dimensional area of the circle of radius 1 is 2π by definition. Note that it is the derivative of the area of the unit disk which is πr^{2}. The two dimensional area of a sphere of radius r is 4πr^{2}and so forth. Also explain why if B_{n}is the ball in ℝ^{n}and B_{n+1}is the ball in ℝ^{n+1}, you haveLet t = r cosθ and change the variables to get

- Prove the following theorem involving the gamma function. Recall that
In the situation of Problem 8, α

_{n}=. To show this, verify that it is true for n = 1,2. Recall that from Problem 3, Γ= αΓand that Γ=. Now prove by induction using the result of Problem 8. Here is a start: For n > 1,and so

Now from Problem 8 and induction,

Now use the result of that problem again to solve for the integral in terms of some α

_{k}and use Γ= αΓ. - Let U be a bounded open set in ℝ
^{n}and suppose u ∈ C^{2}∩Csuch that ∇^{2}u ≥ 0 in U. Then letting ∂U = U∖U, it follows that max= max. The symbol ∇^{2}is the Laplacian. Thus ∇^{2}u = ∑_{i}u_{xixi}. In terms of repeated index summation convention, ∇^{2}u = u_{,ii}. Hint: Suppose this does not happen. Then there exists x_{0}∈ U such that u> max. Since U is bounded, there exists ε > 0 such thatTherefore, u

+ ε^{2}also has its maximum in U because for ε small enough,for all x ∈ ∂U. Now let x

_{1}be the point in U where u+ ε^{2}achieves its maximum. Now recall the second derivative test from single variable calculus. Explain why at a local maximum of f you must have ∇^{2}f ≤ 0. Apply this to the function x → u+ ε^{2}at the point x_{1}and get a contradiction. This is called the weak maximum principle. - Let U be a bounded open set for which Green’s theorem holds. Let C be the oriented
boundary consistent with Green’s theorem. Show
- From the little generalization for the divergence theorem given at the end which allows
one to consider piecewise C
^{1}surfaces, Green’s theorem holds for piecewise smooth curves bounding an open set U which lies locally on one side of its boundary. (Note this condition eliminates the curve crossing itself.) Now suppose you have a closed polygonal curve going from→→=oriented such that the direction of motion is in the direction k × n where n is the unit outer normal from the divergence theorem. Show that if U is this enclosed polygon, thenThis is a pretty remarkable result. Just draw a few such polygons and think how you would find their area without it.

- Let f : ℂ → ℂ. Thus if x + iy ∈ ℂ, f= u+ ivwhere u is the real part and v is the imaginary part. As in one variable calculus, we define
and we say that this derivative exists exactly when this limit exists. Consider h = it and then let h = t. Take limits in these two ways and conclude that if f

^{′}exists, then it is given byThus you have the Cauchy Riemann equations u

_{x}= v_{y},v_{x}= −u_{y}. Show that if u,v are both C^{1}, and these Cauchy Riemann equation hold, then the function will be differentiable. When this happens, we say the function is analytic. (In fact it can be shown that if the limit of the difference quotient exists, then these real and imaginary parts will automatically be continuous and the function will be analytic.) - For a function f : ℂ → ℂ which is continuous and γ : → Γ ⊆ ℂ where Γ is a piecewise smooth curve, we define the contour integral
Show that this equals F

−Fif f is analytic, this for some function F. In particular, if Γ is a closed curve, then ∫_{Γ}fdz = 0. This is Cauchy’s theorem from complex analysis.

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