EXAM 2 SOLUTIONS 1. If z = e r cos θ, r = st, θ = s 2 + t 2, find the partial derivatives dz ds chain rule. Write your answers entirely in terms of s and t. dz and dt using the Solution. This is a routine application of the chain rule. dz = dz dr + dz dθ ds dr ds dθ ds = te r cos(θ) ser sin(θ) = te st cos( s 2 + t 2 ) sest sin( dz = dz dr + dz dθ dt dr dt dθ dt = se r cos(θ) ter sin(θ) = se st cos( s 2 + t 2 ) test sin( 2. Use Lagrange Multipliers to find the minimum and maximum values of f(x) = x 2 y subject to the constraint g(x, y) = x 2 + 2y 2 = 6. Solution. We first need to find the gradients of the two functions and set them equal to each other (with a factor of λ): f =< 2xy, x 2 > g =< 2x, y > 2xy = 2xλ x 2 = yλ The first λ equation requires either y = λ or x =. First, x = in the constraint yields y = ± 3. The other option, y = λ, leads to x 2 = y 2 in the second equation. This plugged into the constraint yields 6y 2 = 6, or y = ±1. In turn, x = ±2. In summary, the points to be tested in f(x, y) are the following: f(±2, 1) = f(±2, 1) = f(, ± 3) = Thus the maximum is and the minimum is. 1
Note. The most common mistake on this problem was forgetting to consider x = when looking at the first λ equation. This is necessary to do, as it represents a legitimate solution to the system of equations! Those points did not turn out to be min/maxes this time, but in general they might, so I took off points for not doing this step. Note. It is not necessary to perform the calculations from 12.8 in this problem to find local min/maxes. The problem only asked to find the global max and min, not other extreme values. The constraint represents a curve within 2 dimensions. The 12.8 extreme value methods are only needed for a 2-D region in 2 dimensions or a 3-D space in 3 dimensions. 3. Evaluate the triple integral below. Solution. π 1 y 2 y sin xdzdydx π π [yz sin x] sqrt1 y2 dydx sqrt1 y 2 y sin xdydx Set u = 1 y 2, du = 2ydy to integrate with respect to y. 1 2 2 3 π [(1 y 2 ) 3/2 sin x] 1 dx 1 π sin xdx = 1 3 3 [ cos x]π = 1 3 (1 + 1) = 2 3 Note. This is a very typical kind of integral you would be expected to evaluate in this course. It requires basic knowledge of what an integral with respect to 1 variable is, plus a basic u-substitution which should be easy by now from Calc 1 and 2. If you got this problem wrong (besides a silly mistake or missing a constant or something) you MUST come see me if you are serious about this class. You cannot pass this class without knowing basic Calc 1 concepts!. Evaluate the following integral by first sketching the region of integration implied by the limits, then reversing the order of integration. 2
y ye x2 x 3 dxdy The outside integral is with respect to y, which goes from to 1. The inner integral for x has two bounds, the first from left to right which is y(i.e.y = x 2 ) and then the line x = 1. So the region is this: Changing order of integration, the outside integral is with respect to x, which ranges from to 1. Then y ranges from a lower bound of y = to y = x 2. So the new integral is: x 2 ye x2 x dydx 3 Use a u-substitution u = x 2, du = 2xdx. 1 2 [ y2 e x2 2x 3 ]x2 dx xe x2 dx 1 [ex2 ] 1 = 1 (e 1) Note. The change of variables is necessary to calculate the integral. Without a change of variables, the dx integral would require lots of integration by parts. Note. This is another very important concept going forward, as I have stressed in class over and over. If you do not understand how I reversed the order of integration, 3
you MUST see me if you are serious about this class. You cannot simply switch the integrals and switch the differentials. Doing this gives you a variable on the outer integral limit! This is never a good thing since it means your final answer will have variables in it, not just a number. Also, you cannot reverse the order of integration by just solving all the limits for the opposite variable. I saw a lot of people calculate the inner integral from x 2 to 1. An accurate diagram is necessary to actually see which functions should be in which limits. 5. Find the center of mass of the portion of the sphere x 2 + y 2 + z 2 = which is in the first octant (x >, y >, z > ) if it has density function δ(x, y, z) = 1 + x 2 + y 2 + z 2 Solution. First it is necessary to find the mass of the sphere piece. The mass can be found by integrating the density function over the sphere piece, S (1 + x 2 + y 2 + z 2 )dv Both the region of integration and the density function suggested switching to spherical coordinates. Since the integral is only over the first octant of the sphere, the limits are θ π/2 and φ π/2. The sphere s radius is 2. Remember to include the standard spherical coordinates differential. 2 π/2 (1 + ρ 2 )ρ 2 sin(φ)dρdθdφ [(ρ 3 /3 + ρ 5 /5) sin(φ)] 2 dθdφ (8/3 + 32/5) sin(φ)dθdφ (π/2)(8/3 + 32/5) sin(φ)dφ (π/2)(8/3 + 32/5)(1) 1.2 Now to calculate one of the center of mass coordinates, say x, calculate the same integral except with an extra variable multiplied at the beginning, then divide by mass. So, for example, the x integral will have x multiplied by the density function. Since we re doing the calculation in spherical coordinates, we must use x = ρ sin(φ) cos(θ). 2 (1 + ρ 2 )ρ 3 sin 2 (φ) cos(θ)dρdθdφ
[(ρ / + ρ 6 /6 sin 2 (φ) cos(θ)] 2 dθdφ π/2 ( + 32/3) sin 2 (φ) cos(θ)dθdφ ( + 32/3) sin 2 (φ)dφ ( + 32/3)[φ/2 1 sin(2φ)]π/2 = π ( + 32/3) 11.52 x = 11.52/1.2 =.89 The other two coordinates actually turn out to be the same. The reason for this is the original situation was entirely symmetric with respect to x, y, and z. Actually calculating the components using integrals will support this. Note. Some people tried to convert to cylindrical coordinates instead. Ths is a valid thing to do, but it doesn t really make the integral any easier to calculate so is not recommended. The fact that x 2 + y 2 + z 2 shows up twice in the problem is a solid tell that spherical coordinates are the best choice. Note. A very common mistake was setting up the integral in rectangular coordinates and integrating from to 2 in all three variables. This demonstrates to me a lack of understanding in what the limits of integration mean. Integrating from to 2 would mean calculating the mass of a cube with 2 units per side, not a sphere of radius 2! Note. Some people switched to spherical coordinates and attempted to find ρ and θ instead of x. Although this was a clever idea, it doesn t quite work and gives the wrong answer for reasons that are hard to describe. To put it simply, the mass distribution in x, y, z is a different question than the mass distribution along a radial vector. You can check for yourself by calculating ρ to be about 1.6, while the ρ value of the real center of mass is 3(.89) 1.1. Bonus Problem (1 Pts.) Use the transformation x = u 2, y = v 2, z = w 2 to calculate the volume inclosed by the surface x + y + z = 1 and the planes x =, y =, and z =. Solution. First it is necessary to calculate the Jacobian for this transformation. 5
dx dx dx du dv dw dy dy dy du dv dw dz dz dz = 2u 2v 2w = 8uvw du dv dw The region of integration changes to the region bounded by the planes u =, v =, w =, and u + v + w = 1. The original desired integral was a volume integral, i.e. integrating the constant function 1. This function is still 1 in uvw coordinates. Thus, u 8 u u v u uvwdwdvdu uv(1 u v) 2 dvdu uv + u 3 v + uv 3 2u 2 v 2uv 2 + 2u 2 v 2 dvdu u(1 u) 2 /2 + u 3 (1 u) 2 /2 + u(1 u) / u 2 (1 u) 2 2u(1 u) 3 /3 + 2u 2 (1 u) 3 /3du u/12 u 2 /3 + u 3 /2 u /3 + u 5 /12du (1/2 1/9 + 1/8 1/15 + 1/72) = 1/9 Note. The simplification of the mess with all the (1 u) 2 terms can be made a little easier by first factoring out the biggest thing you can from all of the terms, which is u(1 u) 2. It actually turns out that the entire expression can be written as 1/12(u)(1 u). This can be integrated by parts without having to multiply it all out. 6