Multivariable calculus problems

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Multivariable calculus problems


1 Introduction These exercises serve as homework for our lesson on the two dimensional diver- gence theorem, which says∫∫ Ω div(F⃗ )dA = ∫ ∂Ω F⃗ · n⃗ds, where F⃗ is some differentiable vector field in Ω and n⃗ is the outward pointing unit normal vector at each point on the boundary ∂Ω of Ω. If the components of F are F = ⟨P,Q⟩ and the boundary is positively oriented, meaning you follow the boundary curve in such a way that the interior of the region is on your left, then n⃗ds = ⟨dy,−dx⟩ and so this can be written∫∫ Ω ∂P ∂x + ∂Q ∂y dA = ∫ ∂Ω P (x, y)dy −Q(x, y)dx. Alternatively, we can replace P and Q respectively with Q and −P in that formula to get∫ ∂Ω F⃗ · dx⃗ = ∫ ∂Ω P (x, y)dx+Q(x, y)dy = ∫∫ Ω ∂Q ∂x − ∂P ∂y dA, which when written this way is called Green’s Theorem. Ultimately the 2D- Divergence and Green’s Theorems are the same theorem but phrased in slightly different notation. A couple comments: ˆ I can’t stress enough just how powerful this theorem is. Clever choices of F and Ω lead to so many useful results in math. ˆ If the boundary has multiple components, you add up the integral over each component. ˆ Careful with orientation! Especially if the region has an inner and outer boundary component. Always remember that Ω should be on your left when you walk along the boundary curve. ˆ Since we have to work with tangent and normal vectors to curves, I’ll remind you that if you rotate a vector ⟨a, b⟩ clockwise by 90 degrees you get the vector ⟨b,−a⟩. Keep this in mind when setting up your formulas. In particular, if a positively oriented curve is given, we would have Tds = ⟨dx, dy⟩ and so the outward pointing unit normal (times ds) is the clockwise 90 degree rotation of that, giving us nds = ⟨dy,−dx⟩. 1 2 Problems Exercise 1 Consider the vector field F⃗ = −y⃗i+ x⃗j x2 + y2 1. Show that ∂ ∂x ( x x2 + y2 ) = ∂ ∂y ( −y x2 + y2 ) . 2. Show that, as long as x > 0, ∇(arctan(y/x)) = F . Because of this, we can (formally) think of F as the gradient of the angle function: F = ∇θ, where x = r cos θ and y = r sin θ are polar coordinates. Note that θ is not actually a well defined function in the entire xy-plane, but is well defined if we restrict its domain to things like x > 0 or to a simply connected region away from the origin. 3. Suppose a curve γ in the xy-plane starts at the point with polar coordi- nates (r1, θ1) and ends at a nearby point with polar coordinates (r2, θ2). Keeping the previous comment in mind, use the fundamental theorem of line integrals to show ∫ γ F⃗ · dx⃗ = θ2 − θ1 4. If Ω is all of R2 except for the origin, show that F is NOT the gradient of a function on Ω by directly computing∫ γ F · Tds where γ = ⟨cos(t), sin(t)⟩, 0 ≤ t ≤ 2π. Remember if F was a gradient in the region this would have to be zero, but it isn’t. 5. What is ∫ γ F · Tds for the curves γ = ⟨R cos(t), R sin(t)⟩, 0 ≤ t ≤ 2π? Does your answer depend on R? 6. What is ∫ γ F ·Tds for the curves γ = ⟨cos(nt), sin(nt)⟩, 0 ≤ t ≤ 2π? Does your answer depend on n? 7. Now apply Green’s theorem with P = −y(x2+y2)−1 and Q = x(x2+y2)−1, where Ω has two nested boundary curves γ1 and γ2. What does this say about ∫ γ1 F · Tds and ∫ γ2 F · Tds? 2 Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Jess Boling Exercise 2 If we use Green’s theorem with P = 0 and Q = x we get∫ ∂Ω xdy = Area(Ω) Use this to find the area of the region enclosed by the curve x = t(1− t) y = t3(1− t) where 0 ≤ t ≤ 1. Again, remember in Green’s theorem that the boundary has to be positively oriented. Exercise 3 Use the divergence theorem to compute the value of ∫ γ x2ds, where γ is the circle x2 + y2 = 1. As a hint, try to find a vector field F where x2 = F · n Notice that the outward unit normal at the point (x, y) on this circle is simply n = ⟨x, y⟩. Exercise 4 Use Green’s theorem to show that the x-coordinate of the center of mass of a uniformly dense region Ω is x̄ = 1 Area(Ω) ∫ ∂Ω x2 2 dy while the y-coordinate is ȳ = 1 Area(Ω) ∫ ∂Ω −y2 2 dx Yet again, remember in Green’s theorem that the boundary has to be positively oriented. Exercise 5 If f(x, y) is a twice continuously differentiable function its Lapla- cian is defined to be ∆f = div(∇f) = ∂ 2f ∂x2 + ∂2f ∂y2 Laplacians are very important because they serve as diffusion terms in a variety of phenomena, from electromagnetism to stock prices. For example, heat flow through an object is governed by the equation ∂tf = ∆f . 1. If f(x, y) = x2 + y2 what is ∆f? 2. If f(x, y) = x3 − 3xy2 what is ∆f? 3. If f(x, y) = ex sin(y) what is ∆f? 3 4. If f(x, y) = ex sin(2y) what is ∆f? 5. Use the Divergence theorem to prove the formula∫∫ Ω ∆fdA = ∫ ∂Ω ∇f · n⃗ds 6. Let’s apply this to Ω = {x⃗|f(x⃗) ≤ c} and assume that ∂Ω = {x⃗|f(x⃗) = c}. Note this assumption isn’t true for all functions. In this case, show ∫∫ {f≤c} ∆fdA = ∫ {f=c} |∇f |ds As a hint, what do you recall about level sets and gradients? Exercise 6 Suppose that f and g are smooth functions. Show∫∫ Ω (f∆g +∇f · ∇g) dA = ∫ ∂Ω f∇g · n⃗ds This is called Green’s first identity. Hint: Consider F = f∇g and use the Divergence theorem and product rule. Exercise 7 Use Green’s first identity to show that, if f and g are both zero on ∂Ω, then ∫∫ Ω f∆gdA = ∫∫ Ω g∆fdA You just proved that ∆ is what is called a self adjoint operator on these sort of functions, a concept which is very important in linear algebra. 4 Introduction Problems
Answered 2 days AfterMay 07, 2022

Answer To: Multivariable calculus problems

Anil answered on May 10 2022
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