2( 2 r 2 2r) rdrdθ. 4. Your result fits the correct answer: get 2 pts, if you make a slight mistake, get 1 pt. 0 r 1

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1 Page 1 of 微甲 7-11 班期末考解答和評分標準 1. (1%) Find the volume of the solid bounded below by the cone z 2 4(x 2 + y 2 ) and above by the ellipsoid 4(x 2 + y 2 ) + z 2 8. Method 1 Use cylindrical coordinates: Note that the region is bounded below by the cone and above by the ellipsoid, so it only consist of the part above the xy-plane! V ( 8 4(x 2 + y 2 ) 4(x 2 + y 2 )) dxdy x 2 +y 2 1 ˆ 2π ˆ 1 8π 3 ( 2 1) Scoring to parts of this method: 1. Integral domain x 2 + y 2 1: 2 pts 2( 2 r 2 2r) rdrdθ 2. Upper bound of z, i.e. 8 4(x 2 + y 2 ): get 2 pts 3. Lower bound of z, i.e. 4(x 2 + y 2 ): get 2 pts 4. Jacobian of polar coordinates: get 2 pts 5. Your result fits the correct answer: get 2 pts, if you make a slight mistake, get 1 pt. Method 2 Use spherical coordinates: Let (x, y, z) ( 2r sin φ cos θ, 2r sin φ sin θ, 2 2r cos φ). The Jacobian is (x, y, z) J(u, v) (r, θ, φ) 2 sin φ cos θ 2r sin φ sin θ 2 2 cos φ 2r sin φ sin θ 2r sin φ cos θ 2r cos φ cos θ 2r cos φ sin θ 2 2r sin φ 4 2r 2 sin φ Then the corresponding domain is r 1 φ π 4 θ 2π Scoring to parts of this method: V 1. Integral domain for r: 2 pts 2. Integral domain for φ: get 2 pts ˆ π ˆ 4 2π ˆ 1 4 2r 2 sin φdrdφdθ 3. Jacobian of spherical coordinate: get 4 pts, and if you miss the multiple before r 2 sin φ, get 2 pts. 4. Your result fits the correct answer: get 2 pts, if you make a slight mistake, get 1 pt.

2 Page 2 of 1 2. (12%) Evaluate e 4x2 9y 2 dxdy, where is the region satisfying 2x 3y and x. Let x r 2 cos θ y r 3 sin θ (3%) then J r 6 (3%) Therefore ˆ ˆ ˆ π/2 ˆ π/4 π (3%) 48 e 4x2 9y 2 dxdy e r2 J drdθ (3%)

3 Page 3 of 1 3. (12%) Evaluate the surface integral lies inside the cylinder x 2 + y 2 4. S (x 2 + y 2 )zdσ, where S is the part of the plane z 4 + x + y that To parametrize this surface: r (x, y, 4 + x + y), where x 2 + y 2 4. ( ) z 2 ( ) z 2 So dσ dxdy 3dxdy x y (x 2 + y 2 )zdσ S (x 2 + y 2 )(4 + x + y) 3dxdy x 2 +y 2 4 ˆ 2π ˆ 2 ˆ 2π ˆ 2 ˆ 2π ˆ 2 ˆ 2π 32 3π (r 2 )(4 + r cos θ + r sin θ) 3rdrdθ (r 2 )(4) 3rdrdθ (by symmetry) 4 3r 3 drdθ 16 3dθ 評分標準如下 : 將曲面參數化 : 2 分算出曲面的 Jacobian: 2 分將面積分換成可處理的積分式, 並且寫出完整的積分區域 : 4 分計算積分 : 4 分 其餘錯誤酌量扣分 將此積分視為三重積分者一律不給分!!!!

4 Page 4 of 1 4. (12%) Find the line integral ˆ (2x sin(πy) e z ) dx + (πx 2 cos(πy) 3e z ) dy xe z dz C along the curve C (x, y, z) z ln } 1 + x 2, y x, x 1. Since F (2x sin(πy) e z )i + (πx 2 cos(πy) 3e z )j xe z k has F + 3e z j (x 2 sin(πy) xe z ) (4 points), so ˆ C F dr ˆ 1 ˆ 1 3e z dy (3 points) 1 + x 2 dx (3 points) ( ln( 2 + 1))/2 (2 points)

5 Page 5 of 1 ffi 5. (1%) Evaluate (x 2 y + y)dx (xy 2 x)dy with the curve oriented counterclockwise. r1 cos θ By Green s theorem, r1 cos θ (x 2 y + y)dx (xy 2 x)dy (x 2 + y 2 )da ˆ 2π ˆ 1 cos θ 1 4 ˆ 2π π. Green s thm : 3pts, region : 3pts,Jacobian : 2pts, computation : 2pts. r 2 rdrdθ (1 cos θ) 4 dθ

6 Page 6 of 1 6. (12%) Let V (2x y)i+(2y +z)j+x 2 y 2 z 2 k and let S be the upper half of the ellipsoid x2 4 + y2 9 + z Find the flux of curlv in the direction of the upper unit normal n (pointing away from the origin.). Solution 1 By using Stokes Theorem, ( V) n ds V(r)dr S C (4 points) C : r(θ) 2 cos θi + 3 sin θj, θ [, 2π] ˆ 2π V(r)dr (1 sin θ cos θ + 6 sin 2 θ) dθ (4 points) C [ 5 2 ( cos 2θ) + 3(θ 1 sin 2θ)] 2π 2 6π. (4 points) Solution 2 V (2yx 2 z 2 1)i + ( 2xy 2 z 2 )j + k. ( V) By using divergence theorem, ( V) n ds ( V)dv (3points) (2points) (3points) In the suface of the buttom, n k. (1points) We can find that the solution is [ (2 3π)] 6π. (3points) ps. Using Stokes Theorem twice is permitted. ps2. If students observe the unit normal of the bottom surface, and they only calculate the k-component of curl V. They do NOT pay for without calculating other components.

7 Page 7 of 1 7. (12%) Evaluate the flux of V(x, y, z) (z 2 x + y 2 z)i + ( ) 1 3 y3 + z tan x j + (x 2 z + 2y 2 + 1)k across S: the upper half sphere x 2 + y 2 + z 2 1, z with normal pointing away from the origin. x 2 + y 2 + z 2 1 S : upward. z z S 1 : x 2 + y 2 downward. 1 x 2 + y 2 + z 2 1 : z Let Then : S : ρ 2 1 x ρ sin φ cos θ y ρ sin φ sin θ z ρ cos φ ρ cos φ ρ 2 1 ρ cos φ ρ 1 φ π/2 θ 2π ρ 1 φ π/2 θ 2π.. Let x r cos θ y r sin θ z z. Then S 1 : z r 2 1 z r 1 θ 2π d area k r dr dθ By divergent theorem, V d volumn V d area S+S 1 S V d area + V d area S 1 (2 pts). V d volumn z 2 + y 2 + x 2 d volumn ˆ 2π ˆ π/2 ˆ 1 ρ5 5 θ φ ρ 1 ( cos φ) π/2 ρ 4 sin φ dρ dφ dθ ρ 2 ρ 2 sin φ dρ dφ dθ (3 pts) 2π 2 π (1 pt) 5

8 Page 8 of 1 V d area (x 2 z + y 2 + 1) r dr dθ S 1 S 1 ˆ 2π ˆ 1 θ 2 r4 4 r ( (r sin θ) ) r dr dθ (3 pts) 1 π π 3π 2 (1 pt) S V d area 19π 1 V d volumn V d area S 1 (2 pts)

9 Page 9 of 1 8. (1%) Find stationary points of f 3xy x 3 y Determine which are local maximum, local minimum or a saddle point. f (3y 3x 2, 3x 3y 2 ) (, ) y x 2 and x y 2 (x, y) (, ) and (1, 1) (stationary points) D f xx f yy f 2 xy 36xy 9 D(, ) 9 < (,) is a saddle point. D(1, 1) 27 > and f xx (1, 1) 6 < (1,1) is a local maximum. 2 pts if stationary points are all wrong. 5 pts if D is wrong or you just get one stationary point. 6 8 pts if you have some mistakes.

10 Page 1 of 1 9. (1%) Use Lagrange multiplier to find the maximum and the minimum of f(x, y) 3x 2 2y 2 for x, y on the curve 2x 2 2xy + y 2 1. (You don t have to give the locations of these extrema.) by Lagrange multiplier, we have 6x λ(4x 2y) and 4y λ( 2x + 2y) and 2x 2 2xy + y 2 1 if λ then x y it is not satisfies 2x 2 2xy + y 2 1 so, λ we get 6x λ 4x 2y and 4y λ 2y 2x and 2x2 2xy + y 2 1 ( 6 λ 4) 2 λ 2 or λ 2 if λ 2 then x 2y and y x 2 from 2x 2 2xy + y 2 1 we get 8y 2 4y 2 + y 2 1 y and 2x 2 x 2 + x2 4 1 x2 4 5 so, 3x 2 2y 2 2 (maximum) if λ 3 then 3x y and x y 3 from 2x 2 2xy + y 2 1 we get 2x 2 6x 2 + 9x 2 1 x and 2 9 y2 2 3 y2 + y 2 1 y so, 3x 2 2y 2 3 (minimum) no point if you don t use Lagrange multiplier. 2 3 pts if you don t get any extreme values. 5 pts if both extreme values are wrong. 7 8 pts if one of the extreme values is wrong.