Math 52 Homework 2 Solutions

Transcription

1 Math 52 Homework 2 Solutions October 3, 28 Problem. If is a verticall simple region then we know that the double integral da computes the area of. On the other hand, one can also compute the area of as a single integral, b describing the region in terms of the graph of functions of a single variable. B direct manipulation of the double integral da, show that these two was of computing area are actuall the same. Solution. Since the region is verticall simple, we can describe it as the region between two functions () and 2 () and between two values, 2 of, as shown in the diagram. 2 () () 2 2 From single-variable integration, we know that the area of the region is given b ( 2 () ()) d. Using double integration, we have Area da 2 2 () () Hence the two methods are the same. dd 2 [ ] 2() () d 2 ( 2 () ()) d.

2 Problem 2. B direct computation with double integrals, show that the volume of a sphere of radius r is given b 4 3 πr3. Solution. Let be the disk of radius r in the -plane centered at the origin. The volume of the upper hemisphere is then r2 2 2 da. This integral is quite difficult, so we simplif it using polar co-ordinates (ρ, θ), where ρ is the distance from the origin and θ tan (/) is the angle with the positive -ais. In the region, the radius ρ varies from to r, and the angle θ varies from to 2π. We also know that da ρdρdθ. In these polar co-ordinates, the integral therefore becomes [ 3 ] ρr (r2 ρ 2 ) 3/2 2π r ρ r 2 ρ 2 dρdθ 2π 2π [ 3 r3 θ 2 3 πr3. 3 r3 dθ ] 2π dθ ρ Hence the volume of the whole sphere is 4 3 πr3. (Since the sphere has so man smmetries, there are man other was one could compute this.) Problem 3. Let be the region { π, sin() + }. Suppose we fill with a material of densit δ(, ). B approimating b rectangles as in class, estimate the center of mass. Solution. Approimating the region b 6 rectangles and numbering them from to 6 gives the following picture sin + 2

3 We then find the area of each rectangle and then multipl it b the values of,, and + + at the bottom left corner. These calculations are summarised in the following table, + given to 5 significant figures. So we have the approimations δ(, )da 3.389, Dividing gives us the approimation ectangle Area Area Area Sum δ(, )da , centroid (.3828,.45456). δ(, )da In fact, the centroid is (.3646,.674) to five significant figures, so we aren t too far off. Problem 4. Sketch the solid in the first octant:,, and z below the surface given b the equation z 3, and label the corner vertices in our sketch. Then calculate the volume of the solid using a triple integral. Solution. z 3 3

4 To compute the volume, we will need the limits of integration. First, we have z /3. Net, in the -plane, we have the boundaries,, + 3 3, so we have /3. Finall, we have 3. Integrating to find the volume, we have Volume dv solid 3 /3 /3 3 /3 3 / dzdd [ z ] z /3 z dd ( /3 )dd [ 3 2 ] /3 2 d ( 3 3 ( 3 ) ( 3 ) ) 2 2 d ( ) 8 2 d [ ] 3 Problem 5. Find the volume of the region bounded b the surface 2 + 2z 2 and the surface 2 2. z Solution. We start b finding limits of integration. The surfaces give us eas limits for : we have 2 + 2z Thinking about limits for and z: if we re looking at the 4

5 z-plane, the limits for and z will be given b the (, z)-co-ordinates of the intersection of the two surfaces, so we solve 2 + 2z z z 2. So the (, z)-region is the unit disk centered at in the z-plane, which we call. Then the volume is 2 2 [ ] 2 2 dddz ddz 2 +2z z 2 ( z 2 )ddz. Now this is a double integral over a disk, so we can make things easier b using polar coordinates. Let r cos θ, z r sin θ. We can describe the unit disk centered at zero b the limits r, θ 2π. We have z 2 2 2( 2 + z 2 ) 2 2r 2, and ddz rdrdθ. Hence we have ( z 2 )ddz 2π 2π 2π 2π π. (2 2r 2 )rdrdθ (2r 2r 3 )drdθ [r 2 2 r4 ] r 2 dθ dθ r Problem 6. Sketch the solid bounded b the graphs of the equations z, 2, 4, z. Calculate the volume of the solid using a triple integral. Then calculate the centroid, i.e. the center of mass with uniform densit δ(,, z), of the solid. Solution. First we need to find integration limits for this solid. If we start with z, we have z. Net, we have the region in the -plane bounded b 2 and 4. This 5

6 gives us limits 2 4 and 2. Then Volume dzdd [ ] z z dd 2 z dd [ 2 ] d (8 2 4 ) d [ 8 5 ] Now we need to find the co-ordinates of the centroid. The solid is smmetric about, so that saves us an integral: we know that the -co-ordinate will be. For the other co-ordinates, we compute dzdd [ ] z z dd 2 z 2 dd [ 2 ] ( [ 64 3 ] ) d 6

7 So the -co-ordinate of the centroid is 52/7 2. For the z-co-ordinate, we have 28/ [ ] z zdzddz 2 2 z2 dd 2 z ] 4 2 [ dd 2 d ( 32 3 ) 6 6 d [ So the z-co-ordinate of the centroid is 256/7. Hence the centroid is (, 2/7, /7). 28/5 7 ] 2 7