The level curve has equation 2 = (1 + cos(4θ))/r. Solving for r gives the polar form:

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1 19 Nov 4 MATH 63 UB ID: Page of 5 pages 5] 1. An antenna at the origin emits a signal whose strength at the point with polar coordinates r, θ] is f(r, θ) 1+cos(4θ), r >, π r 4 <θ<π 4. (a) Write the level curve f(r, θ) in polar function form r r(θ), π 4 <θ<π 4. (c) ketch the region in the xy plane consisting of all points whose polar coordinates obey the equation r r(θ) of part (a). Indicate the region where f(r, θ). Find the area of the region described in part. (a) The level curve has equation (1 + cos(4θ))/r. olving for r gives the polar form: r r(θ) 1+cos(4θ), π 4 <θ<π 4. The curve r r(θ) encloses a single lobe along the x-axis. The rightmost point of the lobe is at (x, y) (1, ). One has f(r, θ) at points on and inside the closed curve just mentioned. y θπ/4 r.5*(1+cos(4θ)) f(r,θ)> 1 x θ (c) all the region. Its area is 1 da (1+cos(4θ)) rdrdθ r r 1 (1+cos(4θ)) 1 + cos(4θ)) 8 (1 dθ 1 (1 + cos(4θ)+cos (4θ))dθ def J. There are several ways to find J. One is to let u 4θ, du 4dθ: da 1 π + cos u +cos 3 π(1 u)du 1 (u +sinu + u ) uπ sin u u π ( 3 )) π 1 ( 3 3 π 3 3 π Or, one could use basic geometry to make three simple observations: dθ π, cos(4θ) dθ, r umming these values gives J 3π/4, so A J/8 3π/3, as before. dθ cos (4θ) dθ π 4. ontinued on page 3

2 19 Nov 4 MATH 63 UB ID: Page 3 of 5 pages 5]. Let denote the solid defined by the system of inequalities x, y, z, z 1 x, x+ y + z. (a) Express the volume of as an iterated triple integral. ompute the volume of. (a) Looking at the figure below from the side (standing far out on the y axis) z x + y + z z 1 x y x we see a base region in the xz plane consisting of x 1, z 1 x. The corresponding triple integral is 1 1 x x z V dx dz dy. The volume is V x dx dz ( x z) dx ( x)(1 x ) 1 (1 x ) ] dx 3 x x + x 3 1 x4] ontinued on page 4

3 19 Nov 4 MATH 63 UB ID: Page 4 of 5 pages 5] 3. Let be the curve from P (1,, ) to Q (,π/,π/) along the intersection of these surfaces: x cos(y), y z. hoose specific numbers A and B (state your choices clearly!) and then use them to evaluate both I 1 (ye x Ax cos(z)) dx +(e x + By 4 z ) dy +(y 5 z x 3 sin(z)) dz and I ye x Ax cos(z)+3sin (y), e x + By 4 z, y 5 z x 3 sin(z) dr. Hint: You can replace A and B with any values you like. Efficient choices would be best; taking A and B isnot efficient at all. Both I 1 and I are line integrals of vector fields: I 1 F dr and I I 1 + G dr, where F(x, y, z) ye x Ax cos(z), e x + By 4 z, y 5 z x 3 sin(z), G(x, y, z) 3sin (y),,. Line integrals are easy to evaluate when they represent work done by a conservative vector field. ould F be conservative? Only when it passes the screening test, i.e., when F 1 z F 3 x, i.e., Ax sin(z) 3x sin(z), i.e., A 3, and F z F 3 y, i.e., By4 z 1y 4 z i.e., B 5. With these choices, F, and it is not hard to see that F φ for the function φ(x, y, z) ye x + y 5 z + x 3 cos(z). onsequently I 1 F dr φ dr φ(q) φ(p ) π + ( ) 7 π +] ++1] ( ) 7 π + ( ) π 1. With the same choices for A and B, I I 1 + 3sin (y) dx. A simple parametrization for is given by x cos(t), y t, z t, t π/; note dx sin(t) dt, dy dt, dz dt. Hence π/ ] π/ ( ) 7 ( ) π π I I 1 +3 sin (t)( sin(t) dt) I 1 cos 3 (t) 3cos(t) I t t The integral of sin 3 (t) is given on the formula sheet. One may also write sin 3 (t) 1 cos (t)] sin(t) and then substitute u cos(t).] ontinued on page 5

4 19 Nov 4 MATH 63 UB ID: Page 5 of 5 pages 5] 4. Let be the piece of the paraboloid z 1 x y where 1 z 6. ompute 4x +4y +1d. Method 1: ectangular oordinates (then switch to polar). We parametrize by r(x, y) x, y, f(x, y), wheref(x, y) 1 x y. Then we know that ( ) ( ) d (f x ) +(f y ) +1 dx dy 4x +4y +1 dx dy. Also note that if z 6thenr 4andifz 1thenr 9, so we are integrating over an annulus with inner radius and outer radius 3, which we will denote by. Hence ( 4x +4y +1) d (4x +4y +1)dx dy π 3 π π 3 (4r +1)rdrdθ (4r 3 + r) dr r 4 + r / ] 3 π( /) ( 4 + /)] π(81 + 9/ 16 ) π( ) 135π. Method : ylindrical oordinates. We parametrize in terms of (r, θ) bys(r, θ) r cos θ, r sin θ, g(r, θ), whereg(r, θ) 1 r. Then we know that d ((g θ ) +(rg r ) + r ) 1/ dr dθ (4r 4 + r ) 1/ dr dθ. Also note that if z 6thenr 4andifz 1thenr 9, so we are integrating over an annulus with inner radius and outer radius 3, which we will denote by. Hence ( 4x +4y +1) d (4r +1) 1/ (4r 4 + r ) 1/ dr dθ π 3 π π 3 r(4r +1)dr dθ (4r 3 + r) dr r 4 + r / ] 3 π( /) ( 4 + /)] 135π. The End

5 This examination has 5 pages including this cover The University of British olumbia Midterm Examination 19 Nov 4 Mathematics 63 Multivariable and Vector alculus losed book examination Time: 5 minutes Name ignature tudent Number pecial Instructions: To receive full credit, all answers must be supported with clear and correct derivations. No calculators, notes, or other aids are allowed. A formula sheet is provided with the test. ules governing examinations 1. All candidates should be prepared to produce their library/am cards upon request.. ead and observe the following rules: No candidate shall be permitted to enter the examination room after the expiration of one half hour, or to leave during the first half hour of the examination. andidates are not permitted to ask questions of the invigilators, except in cases of supposed errors or ambiguities in examination questions. AUTION - andidates guilty of any of the following or similar practices shall be immediately dismissed from the examination and shall be liable to disciplinary action. (a) Making use of any books, papers or memoranda, other than those authorized by the examiners. peaking or communicating with other candidates. (c) Purposely exposing written papers to the view of other candidates. The plea of accident or forgetfulness shall not be received. 3. moking is not permitted during examinations Total 1