Math 265 Exam 3 Solutions

Transcription

1 C Roettger, Fall 16 Math 265 Exam 3 Solutions Problem 1 Let D be the region inside the circle r 5 sin θ but outside the cardioid r 2 + sin θ. Find the area of D. Note that r and θ denote polar coordinates. Solution. Looking at the inequalities r 5 sin θ for the circle and r 2 + sin θ for the cardioid, the region D is defined by the inequalities 2 + sin θ r 5 sin θ. We can see the lower and upper limits for r here. For θ, we see that the values are limited by the points where where circle and cardioid intersect. From 5 sin θ 2 + sin θ, we get 4 sin θ 2, so sin θ 1/2 and θ 5. We are ready to set

2 up the integral (use symmetry with regard to the y-axis in the 3rd step) A sin θ 2+sin θ 5 π/2 π/2 π/2 r dr dθ 25 (2 + sin θ) 2 dθ 25 sin 2 θ (2 + sin θ) 2 dθ 24 sin 2 θ 4 4 sin θ dθ 12(1 cos 2θ) 4 4 sin θ dθ [8θ 6 sin 2θ + 4 cos θ] π/2 8π Problem 2 a) Sketch the region of integration for the following integral. Label each bounding curve with its equation. Then rewrite the integral with the reverse order of integration dx dy. 1 x+3 3 x+3 f(x, y) dy dx. b) If f(x, y) 5y 2 is the density of a thin plate based on the region of part a), find the center of mass of that thin plate. Solution. a) The region is determined by y 2 3 x 1. Combining these two inequalities, we get y 2 3 1, so y 2. Look how these rewritten inequalities describe the same region as the given integral! It can therefore be rewritten as 2 y 2 3 f(x, y) dx dy. b) With the density 5y 2, the mass is m 2 y 2 3 5y 2 dx dy

3 The moments are M x M y y y 3 dx dy 5y 2 (1 y 2 + 3) dy, y , ȳ, x y 2 x dx dy 5y 2 [1 (y 2 3) 2 ] dx dy 5y 2 ( y 2 + 6y 8)) dx dy even integrand So the center of mass is ( 1/7, ). You could also argue that ȳ because of symmetry with regard to the x-axis. And you could also compute the mass and moments using the original form of the integral, as dy dx-integral.

4 Problem 3 Let S be the tetrahedron with vertices (4,, ), ( 4,, ), (, 4, ), (,, 4) (meaning there are four faces, each of which is a triangle having three out of the four given vertices). a) Find the inequalities defining this solid. b) Set up the integral for the volume of S in order dx dy dz. c) Set up the integral for the volume of S in order dz dy dx. You will have to use a minimum expression. d) If you had to compute the center of mass, which moment would be zero? (there is only one) DO NOT ACTUALLY COMPUTE any of these integrals. Solution. a) The faces of this tetrahedron are given by four planes. They must contain three out of four of the given vertices. We find the planes z, y, x + y + z 4, x + y + z 4. Looking at the fourth vertex for each plane, we get the inequalities. Eg (,, 4) is omitted for the first plane, z. That vertex satisfies z. So this must be true for the entire solid (because all of it is on the same side of that plane). Similarly, y, x + y + z 4, x + y + z 4. b) The shadow of the solid in the yz-plane is a triangle with vertices (y, z) (, ), (4, ), and (, 4). This tells me the limits for z and y. Alternatively, you could look at the inequalities (set x to get the extreme values for z, y). Vol(S) 4 4 z 4 y z y+z 4 1 dx dy dz. c) The shadow of the solid in the xy-plane is a triangle with vertices (y, z) (4, ), ( 4, ), and (, 4). This tells me the limits for x and y. Alternatively, you could look at the inequalities (set z to get the extreme values for x, y). Vol(S) 4 min{4 x,4+x} min{4 x y,4+x y} 4 1 dz dy dx. d) M yz, because of symmetry with regard to the yz-plane.

5 Problem 4 Consider the solid S above the cone z x 2 + y 2 and below the sphere x 2 + y 2 + z 2 9. a) Set up the integral for the volume of S in cylindrical coordinates. b) Compute the mass of the solid S if the density is δ 5. Solution. a) Limits for z are x2 + y 2 z 9 (x 2 + y 2 ), and this gives points in space above a disk about the origin in the xy-plane. To actually have any points, we do need x2 + y 2 9 (x 2 + y 2 ) which we can solve by squaring both sides: x 2 + y That disk in polar coordinates is described by r 3/ 2. So Vol(S) 3/ 2 2π 9 r 2 r r dz dθ dr.

6 b) Using the same limits as in part a), m 3/ 2 2π 9 r 2 1π 3/ 2 1π [ (9 r2 ) 3/2 3 r 5r dz dθ dr r 9 r 2 r 2 dr ] 3/ 2 r3 3 1π(27 27/ 8 27/ 8) 3 1π(9 9/ 2) 45π(2 2). Problem 5 Find the work done by the force field F (2x + 3y)i + (3x 3y 2 )j moving a particle along the semicircle given by x 2 + y 2 4, y. Solution. Solution 1: We could just use a parametrization of the semicircle, eg r(t) (2 cos t, 2 sin t) with t π. Then the work is π W F dr (4 cos t + 6 sin t)( sin t) + (6 cos t 12 sin 2 t)(2 cos t) dt C π 8 cos t sin t 12 sin 2 t + 12 cos 2 t 24 sin 2 t cos t dt [ 4 cos 2 t + 6 sin 2t 8 sin 3 t ] π. Solution 2: We could try to find a potential function. It turns out that one exists, namely f x 2 + 3xy y 3. So this force field is conservative, and the work equals W f(, ) f(2, ). Solution 3: if we just see that F is conservative, but don t want/can t find a potential, we still know that this line integral is independent of path. So we could decide to just go along the diameter of the semicircle instead of the arc, making y (with r(t) (2 t, )) and an easier integral. Still gives.