Calculus Final Exam

Calculus Final Exam

1. Evaluate the integral.

(a) (8 points) Z ₁

0

Z ₂

2y

sin x x dxdy

Solution:

Z ₁

0

Z ₂

2y

sin x

x dxdy = Z ₂

0

Z ^x

2 0

sin x

x dydx = −cos x

2 |²₀=1 − cos 2 2 (b) (8 points)

Z _∞

0

e^−x2dx.

Solution: Let I = Z _∞

0

e^−x2dx. Then I²=¡^{Z ∞}

0

e^−x2dx¢¡^{Z ∞}

0

e^−y2dy¢

= Z _∞

0

Z _∞

0

e^−(x2+y2)dxdy = Z _π/2

0

Z _∞

0

e^−r2rdrdθ ^=¡−e−r

2

2 |^∞₀¢¡π 2

¢= π

4. Therefore, I = Z _∞

0

e^−x2dx =

√π 2 .

(c) (8 points) Z ₂

0

Z √

4−y² 0

e

√

x²+y²

dxdy

Solution:

Z ₂

0

Z √

4−y² 0

e

√

x²+y²

dxdy = Z _π/2

0

Z ₂

0

e^rrdrdθ ^{=¡rer− er|2}0

¢¡π 2

¢= (e²+ 1)π 2 (d) (8 points)

ZZZ

T

1

(x²+ y²+ z²)dxdydz,

where T = {(x, y, z) ∈ R³| 0 ≤ x ≤ 1, 0 ≤ y ≤p

1 − x², 0 ≤ z ≤p

1 − x²− y²}.

Solution: Let (x, y, z) = (ρsinφcosθ,ρsinφsinθ,ρcosφ), where 0 ≤ρ≤ 1, 0 ≤φ≤π 2, and 0 ≤θ ≤ π

2. Then dxdydz =ρ²sinφdρdφdθ, and ZZZ

T

1

(x²+ y²+ z²)dxdydz

= Z _π/2

0

Z _π/2

0

Z ₁

0

1

ρ²ρ^2sinφ^dρ^dφ^dθ ^{=¡− cos}φ^|π₀^/2¢¡π 2

¢= π 2

2. Let T be the ellipsoid

x² a²+y²

b²+z² c² ≤ 1 parametrized by

x = aρ^sinφ^cosθ^{, y = b}ρ^sinφ^sinθ^{, z = c}ρ^cosφ^, where 0 ≤φ ^≤π^{, and 0 ≤}θ ^{≤ 2}π^.

(a) (6 points) Find the Jacobian J = J(ρ^,φ^,θ^).

Solution: J = J(ρ^,φ^,θ^{) = abc}ρ^2sinφ (b) (4 points) Calculate the volume of T.

Calculus Final Exam

Solution: V (T ) = Z _2π

0

Z _π

0

Z ₁

0

abcρ^2sinφ^dρ^dφ^dθ ^{= 2abc}π 3

¡− cosφ^|π₀^{¢= 4abc}π 3 (c) (8 points) Locate the centroid of the upper half of T.

Solution: Let T⁼ denote the upper half of T. Then V (T⁺) = 1

2V (T ) = 2abcπ 3 , and zV (T⁺) =

ZZZ

T⁺

zdxdydz = Z _2π

0

Z _π/2

0

Z ₁

0

abc²ρ^3sinφ^cosφ^dρ^dφ^dθ

=abc²π 4

¡sin²φ^|π/2₀ ^¢=abc2π

4 . Therefore, the centroid is at (x, y, z) = (0, 0,3c 8).

3. Let S be the unit sphere x²+ y²+ z²= 1 parametrized by

r = r(φ^,θ) = (x, y, z) = (sinφ^cosθ^{, sin}φ^sinθ^{, cos}φ^),

where 0 ≤φ ^≤π^{, and 0 ≤}θ ^{≤ 2}π^. (a) (4 points) Find N = r_φ× r_θ.

Solution: Since r_φ = (cosφ^cosθ^{, cos}φ^sinθ^{, − sin}φ^),

and r_θ = (− sinφ^sinθ^{, sin}φ^cosθ, 0), N = (sin²φ^cosθ^{, sin2}φ^sinθ^{, sin}φ^cosφ^).

(b) (6 points) Calculate the flux of v = (x, y, 0) across S in the outward direction.

Solution: Flux = ZZ

S

v · ndσ, where n = (sinφ^cosθ^{, sin}φ^sinθ^{, cos}φ^{), and d}σ ^{= kNkd}φ^dθ^. Hence, Flux =

Z _2π

0

Z _π

0

sin²φ^sinφ^dφ^dθ

= Z _2π

0

Z _π

0

(1 − cos²φ^{) sin}φ^dφ^dθ ^{=¡− cos}φ^+cos3φ 3 |^π₀¢

2π^{= 8}π 3 .

4. (8 points) Evaluate the surface integral ZZ

S

dσ, where S is the graph of z = f (x, y) = xy over the unit discΩ: 0 ≤ x²+ y²≤ 1.

Solution: r(x, y) = (x, y, f ), for 0 ≤ x²+y²≤ 1, is a parametrization of S, and dσ^{= kr}x×r_ykdxdy = k(1, 0, f_x) × (0, 1, f_y)kdxdy = k(1, 0, y) × (0, 1, x)kdxdy =p

1 + x²+ y²dxdy.

Hence, ZZ

S

dσ⁼ ZZ

Ω

p

1 + x²+ y²dxdy = Z _2π

0

Z ₁

0

p

1 + r²rdrdθ ⁼²π

3 (1 + r²)^3/2|¹₀= 2π 3

¡2^3/2− 1¢

5. (8 points) Let T be a solid with volume

V = ZZZ

T

dxdydz = Z ₂

0

Z _9−x2

0

Z _2−x

0

dzdydx.

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Calculus Final Exam

Find the limits of integration for the formula of V in ZZZ

T

dzdxdy = Z ₅

0

Z _¤

¤

Z _¤

¤ dzdxdy + Z ₉

5

Z _¤

¤

Z _¤

¤ dzdxdy.

Solution:

ZZZ

T

dzdxdy = Z ₅

0

Z ₂

0

Z _2−x

0

dzdxdy + Z ₉

5

Z ^√_9−y

0

Z _2−x

0

dzdxdy.

6. (6 points) Let S be the upper half of the unit sphere x²+ y²+ z²= 1 and take n as the upper unit normal.

Let v = v(x, y, z) = (y, −x, z), find _ZZ

S[(∇× v) · n]dσ by Stokes’s theorem.

Solution: Since ZZ

S[(∇× v) · n]dσ ⁼ I

C

v(r)dr, where C : r(u) = (cos u, sin u, 0) for u ∈ [0, 2π^].

Hence, I

C

v(r)dr = Z _2π

0

(sin u, − cos u, 0) · (− sin u, cos u, 0)du = −2π^.

7. (6 points) Use the divergence theorem to find the total flux of v(x, y, z) = (2x, y, z²) out of the cylinder T = {(x, y, z) | 0 ≤ x²+ y²≤ 9, 0 ≤ z ≤ 1}.

Solution: Flux = ZZ

∂Tv · ndσ, where n is the unit outward normal to∂T. The divergence theorem implies that =

ZZ

∂Tv · ndσ⁼ ZZZ

T∇· vdzdydx = Z _2π

0

Z ₃

0

Z ₁

0

(3 + 2z)dzrdrdθ ^{= 36}π

8. (a) (6 points) Find a function f = f (x, y) such that∇f = h(x, y) = (xy²+ 2x, yx²+ 2y).

Solution: Set h = (P, Q). Since ∂^Q

∂x = 2xy = ∂^P

∂y, there exists a function f satisfying that

∇f = h(x, y) = (xy²+2x, yx²+2y), and f = Z

Pdx =1

2x²y²+x²+g(y), and from the equation f_y= Q, we get g⁰(y) = 2y, and g(y) = y²+ C, where C is an arbitrary constant. Hence, f = 1

2x²y²+ x²+ y²+C.

(b) (6 points) Let C₁: r₁(u), u ∈ [a, b], and C₂: r₂(v), v ∈ [c, d] be any two smooth paths from p to q in the xy−plane. Show that

Z

C1

h(r₁) · dr₁= Z

C2

h(r₂) · dr₂.

Solution: The fundamental theorem of line integral implies that Z

C1

h(r₁) · dr₁= Z

C1

∇f (r₁) · dr₁= f (q) − f (p) =

Z

C₂∇f (r₂) · dr₂= Z

C₂

h(r₂) · dr₂.

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