Calculus IV. Exam 2 November 13, 2003

Transcription

1 Name: Section: Calculus IV Math 1 Fall Professor Ben Richert Exam November 1, Please do all your work in this booklet and show all the steps. Calculators and note-cards are not allowed. Problem Possible points Score Total 7

2 Problem 1. (1 pts.) A 15-ft-by-15-ft hole has been excavated in Cal Poly canyon. You take depth measurements every 5 ft (starting from the southwest corner of the hole),and come up with the following table The next day the hole fills with water during a torrential rain. Estimate the amount of water in the hole using m = n = subdivisions (you don t need to add it all up, just write a sum which gives the estimate). We can estimate the volume of the hole by setting up the Riemann sum f(x i, y j ) A i=1 j=1 where A = x y = = 5, x i = 5i, and y j = 5j for i, j = 1,,. The result is 5( ). There are, of course, many other ways to do this. For instance, you could let x i = 5(i 1) for i = 1,,. You obtain a different estimate in that case. 1

3 Problem. (1 pts.) Compute the surface area of the function f(x, y) = x + y above the region between the two circles x + y = 1 and x + y =. Let D be the region in question. Then to compute the surface area of the function f(x, y) above the region D, we compute the integral (f x ) + (f y ) + 1 da. D In polar coordinates D = { {r, θ} 1 r, θ π }, and it is not difficult to show that f x (x, y) = x and f y (x, y) = y, so (f x ) + (f y ) + 1 = x + y + 1 = r + 1. Thus the integral we want to compute is π r + 1 r drdθ. 1 Let u = r + 1 so that du = r, or du = r dr. Making these substitutions, we get π 5 u du dθ, where we have changed the limits of the inner integral to u limits. Then π 5 du π u dθ = π 6 u 5 dθ = 6 (5 ) dθ = π 6 (5 ) square units. 1

4 Problem. (1 pts.) Let X be the number of minutes that your math professor spends telling jokes during a 5 minute lecture, and let Y be the number of minutes he spends erasing the chalkboard. Suppose that the joint probability density function for X and Y is given by { CXY if X 5, Y 5 f(x, Y ) = otherwise, where C is a constant. (a 5 pts) Describe the process you would use to find the value of the constant C. (A sentence or two will suffice). We know that if f(x, Y ) is a joint probability density function then f(x, Y ) dx dy = 1. In our case, f(x, Y ) = unless X, Y 5, so we have 1 = f(x, Y ) dx dy = 5 5 CXY dx dy = C We can then solve this equation for the constant C, i.e., 1 C = 5 5 XY dx dy. 5 5 XY dx dy. (b 5 pts) Set up an integral which computes the probability that your professor spends more than 1 minutes telling jokes, and less than minutes erasing the board. We want to know the probability that 1 X 5 and Y. This is computed by the integral 5 1 CXY dydx. 1

5 Problem. ( pts.) Do any three of the following (and denote clearly which three those are): { (a 1 pts.) Let D be the lamina bounded by y = x x + if x and y = and suppose that the density at the point x + otherwise. (x, y) is given by the function ρ(x, y) = Set up integrals to compute the center of mass of the lamina. x Note that density does not depend on the x and y coordinates, and thus the mass of the region described by { {x, y} x, x y x + } is exactly 1/ of the mass of the entire lamina. So mass = x+ x dy dx. Because of the symmetry of the lamina and the fact that the density is not affected by the x-coordinate, it is clear that x =. To compute y, we calculate y = mass Note that we have multiplied by because by symmetry. x+ x+ x y dy dx = x y dy dx. D y da 1

6 6 (b 1 pts.) Let f(x, y, z) = x +y +z y ex and let E be the region inside the sphere x + y + z =, above the xy-plane, and x + y below the cone z =. Set up the integral f(x, y, z) dv using spherical coordinates. Spherical coordinates fit this particular region fairly well. The cone and the sphere intersect when ( ) = x + y + z = x + y x + y + = (x + y ), or x + y =. We can use this information to find the angle that the cone makes with the z-axis (and thus find a lower bound for φ). In particular, we can form a right triangle using the z-axis, the side of the cone, and a chord perpendicular to the z-axis connecting to the circle constituting the intersection of the cone and the sphere. Our calculation above tells us that this chord has length, while the side of the cone (which gives the hypotenuse of our triangle) has length ) (because it intersects the sphere). The angle the side of the cone ( makes with the z-axis is thus sin 1 The integral we are interested in is thus { {ρ, θ, φ} ρ, θ π, π π = π π π π E = π. So we can describe the given region as π φ π }. ρ cos(θ) sin(π) ρ sin(θ) sin(φ) eρ ρ sin(φ) dρdθdφ cot(θ)e ρ ρ sin(φ) dρdθdφ. 1

7 (c 1 pts.) Consider the integral 1 1 x 1 x (i.e., find the appropriate limits of integration). (x + y + 1) dz dy dx. Rewrite this integral as We can describe E as E = { {x, y, z} x 1, y 1 x, z 1 x }. In order to integrate in the given order, we must then re-describe E as So the integral in question is E = { {x, y, z} y 1 x, x 1 z, z 1 }. 1 1 z 1 x (x + y + 1) dy dx dz. E 7 (x + y + 1) dy dx dz (d 1 pts.) Let E be the region consisting of the intersection of the sphere of radius (centered at the origin) with the cylinder (x 1) + y = 1. Set up an integral to compute the volume of E. The key to this problem is describing the region using cylindrical coordinates. The circle of radius 1 with center (1, ) can be written parametrically as { {r, θ} r cos(θ), π θ π }. We get this by considering the equation (x 1) + y = 1 (the equation of the circle we are interested in) and changing to polar coordinates. That is, (r cos(θ) 1) + (r sin(θ)) = 1, so simplifying r cos (θ) r cos(θ) r sin (θ) = 1, r (cos (θ) + sin (θ)) = r cos(θ), and r = cos(θ). Then the region E can be described as E = {{r, θ, z} r cos(θ), π θ π, 1 x y z } 1 x y = {{r, θ, z} r cos(θ), π θ π, 1 r z } 1 r. The integral in question is then π π cos(θ) r r r dzdrdθ. 1 1

8 8 (e 1 pts.) Let f(x, y, z) = x + 1 y + 1 and let E be the region in the first octant bounded by the planes z = x and y = 1, and the curve x = y. Set up an integral to compute f(x, y, z) dv. We set up this integral using rectangular coordinates. The region E can be described as E = { {x, y, z} x 1, x y 1, z x }. E So the integral in question is 1 1 x x The other possible correct answers are: 1 y x x + 1 y + 1 dzdydx. x + 1 y + 1 dzdxdy, 1 1 y x + 1 y + 1 dxdydz, 1 y y x + 1 y + 1 dxdzdy, z z 1 x 1 x x + 1 y + 1 dydzdx, and z z x x + 1 y + 1 dydxdz. 1

9 (f 1 pts.) Suppose that E is the region bounded by the ellipsoid x 9 + y +z = 1. Set up an integral to evaluate using the transformation x = u, y = v, and z = c. E x dv 9 Replacing x, y, and z with u, v, and c respectively, the ellipsoid becomes (u) + (v) + c = 1, or 9 u + v + c = 1, the sphere of radius 1 in the uvc-plane (we will refer to this sphere as S). The Jacobian of this transformation is x x x u v c y y y u v c = z z z 1 = 6. u v c So the integral we want is (x, y, z) f(u, v, c) S (u, v, c) dv, which in spherical coordinates becomes (with u acting like x, v like y, and c like z) π π 1 ρ sin(φ) cos(θ)6ρ sin(φ) dρ dθ dφ. (Note that the integrand x became u, which we replaced with ρ sin(φ) cos(θ)). 1

10 1 Problem 5. (1 pts.) Match the following functions with the plots of their gradient vector fields and their level curves. Function Plot of Gradient Vector Field Level Curve (a) f(x, y) = x y + y V D (b) f(x, y) = sin(x) + y I C (c) f(x, y) = ln(x + y + 1) II B (d) f(x, y) = sin(y) + x IV A (e) f(x, y) = x + y III E (I) (II) (III) (IV) (V) (A) (B) (C) (D) (E) - - 1