Multivariable Calculus Midterm 2 Solutions John Ross

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1 Multivariable Calculus Midterm Solutions John Ross Problem.: False. The double integral is not the same as the iterated integral. In particular, we have shown in a HW problem (section 5., number 9) that the iterated integral can exist while the double integral does not. Problem.: False. One example is the Mobius strip, which is nonorientable but parametrizable. (The parametrization is given in Example 8, page 49-43) Problem.3: False. Any function which has a single point of discontinuity will provide us with a counterexample. { : (x, y) = (, ) More concretely: let f(x, y) = : otherwise Then for a compact subset D R, we have either (, ) / D or (, ) D. In the first case, we have f(x, y)da = since the function is everywhere in D D. In the second case, we set up a limit of Riemann Sums: D f(x, y)da = lim a f(x, y ) a (here, a is the area of each little square in the Riemann sum). But f(x, y ) will equal in every square except possibly in the square containing (, ). Thus, every term of our sum will equal except, perhaps, the term contains the origin. This term could have f(x, y ) equal to or. Thus, the Riemann sum would equal or a. Either way, as a, we get our limit approaches.

2 Problem : We have learned that the mass of some solid V is the integral of the density function (the integral is taken over the region V). In this problem, V is a 3-dimensional solid, so we need the following triple integral: V ρ(x, y, z) dv = x+y z dzdydx x y = = 3 4xy dydx = x dx Problem 3: The curve C is found by taking the intersection of the surfaces x = cos(z) and y = sin(z). Thus, given any z, our curve will have x-coordinate cos(z) and y-coordinate sin(z). So we can parametrize our curve as C(t) = (cos(t), sin(t), t), t The length of C (in fact, the length of any curve) is found by computing the scalar line integral ds. We get: C C ds = C dt = ( sin(t)) + cos (t) + dt = dt = Problem 4: To compute surface area, we first want to parametrize the surface. Since this is the graph of a function, our usual parametrization gives us: X(s, t) = (s, t, f(s, t)) = (s, t, s + t ). T s = (,, s) T t = (,, t) T s T t = ( s, t, ) Since the function is only defined for x + y, the region we wish to integrate over (in the s-t plane) is the disc of radius. This gives us the following surface integral:

3 X ds = X T s T t ds = s (s) + (t) s + dtds We convert this into polar coordinates to simplify (don t forget the Jacobian factor of r!), and then use u-substitution: = = 3π 3 π 4r + r dθdr = π r 4r + dr Problem 5: We don t immediately know how to integrate sin(y 3/ ), so we want to change our order of integration. We rewrite our bounds: 4 becomes 4 y x= y=x y= x= (If this is not clear, draw the picture of the region we are integrating over.) Since we re only changing the order of integration, there is no Jacobian term. Our new integral yields: 4 y sin(y 3/ ) dxdy = 4 y/ sin(y 3/ ) dy = 3 ( cos(8)) Problem 6: This problem requires a change of variables to solve. By drawing a picture of the region, we see that the region is naturally traced out by lines from the origin (ay = x, as a ranges from to 4) and by the curves xy = b (b ranges from to 3). Thus, our change of coordinates should be: u = x y, u 4 v = xy, v 3 3 u3/ Thus, we wish to solve the integral 4 J dvdu (notice that our change of variables also made our integrand easier - a clue that we re doing the right thing!) A quick computation lets us write x and y in terms of u and v: 3

4 x = uv, y = u v So the Jacobian of the transformation is ] J =det [ v uv v u v/u u uv u v/u Thus, our integral becomes: = u 4 3 u3/ u dvdu = 4 3 u/ dvdu 4 3 (8 8) Problem 7: We are asked to evaluate a vector line integral. The vector field is given by F = ( xe sin(y x), e sin(y x) ). The curve can be any simple, oriented curve with initial point (, ) and terminal point (, ). Since the curve is not specified, we hope that F has path independent line integrals (otherwise, we have no hope of answering this problem). So first we must check that F has PILI. Once we ve settled that, we know any path from (, ) to (, ) will give us the same answer: thus, we can choose a specific path that makes our integration easy. Notice F is defined and C on all of R. Since R is a simply connected domain, we recall that F having PILI is the same as F being conservative, which is the same as N x M y =. So we check: N x M y = x esin(y x) y ( xesin(y x) ) = xe sin(y x) ( xe sin(y x) ) = Thus, we have PILI. Now, we want our curve C to make the integral as simple as possible. Let s take the curve y = x (doing so will kill the exponent). This curve can be parametrized as C(t) = (t, t ), t. So we compute: C F d s = ( tesin(t t ), e sin(t t ) ) (, t) dt = ( t, ) (, t) dt = 4

5 Problem 8: The curve C will be some sort of closed curve (similar to a circle or ellipse), looping around the z-axis with a counter-clockwise orientation. Notice that F is not defined on the z-axis, but is defined and is C everywhere else. Furthermore, a quick computation yields that F = (,, (x +y ) x +(x +y ) y (x +y ) ) = (,, ) So let s choose a new, really simple curve oriented counterclockwise around the z-axis (call it C ). We can create a surface S, that doesn t cross the z-axis, and that has boundary equal to C +( C ) (ie, the boundary of S is the C curve (with it s current orientation) and the C curve (with it s orientation reversed). We can then orient S such that it is consistent with C and C (the surface will look like a funny, deformed cylinder). Then, by Stoke s theorem, we see C+( C ) F d s = S ( F ) d S = So C F d s = C F d s. Thus, we can compute the vector line integral using our new, easier curve C. Let s let C be the circle of radius ɛ, living parallel to the x-y plane at height z = ζ. (Note that you could have chosen a different C on your midterm). Then C has parametrization (ɛ cos(t), ɛ sin(t), ζ) from t π. We then compute: C F d s = π ( ɛ sin(t) ɛ, + ɛ cos t ɛ, e ζ ) ( ɛ sin(t), ɛ cos(t), ) dt = π ɛ sin(t) + ɛ sin (t) ɛ + ɛ cos(t) + ɛ cos (t) ɛ + dt = π ɛ sin(t) + ɛ cos(t) dt = π 5