SMAM 345 Assignment 3 Name____________ Before attemping this assignment please review double integrals and double integrals in polar coordinates. The material should be available in Stewart or any standard Calculus Textbook. Do these problems by hand and show your work. Put your final solutions in the space provided. 1. Find the constant k where k (2x + 3y)dydx = 1 0 2 ∫ 1 3 ∫ (2x + 3y)dydx = 2xy + 3 2 y 2 1 3 ∫ 0 2 ∫ 1 3 ∫ 0 2 dx = (4x + 6 1 3 ∫ )dx = 2x 2 + 6x 1 3 = 28 28k = 1 k = 1 / 28 2. A.Evaluate 4xydxdy R ∫∫ where R = {(x,y): 0<x<y<1} [Draw the region before you set up the integral.]
Transcript
SMAM 345 Assignment 3 Name____________ Before attemping this assignment please review double integrals and double integrals in polar coordinates. The material should be available in Stewart or any standard Calculus Textbook. Do these problems by hand and show your work. Put your final solutions in the space provided. 1. Find the constant k where
k (2x + 3y)dydx = 10
2
∫1
3
∫(2x + 3y)dydx = 2xy + 3
2 y2
1
3
∫0
2
∫1
3
∫ 0
2
dx = (4x + 61
3
∫ )dx = 2x2 + 6x1
3= 28
28k = 1k = 1/ 28
2. A.Evaluate
4xydxdy
R∫∫ where R ={(x,y) : 0<x<y<1} [Draw the region before you
set up the integral.]
4xydxdy = 2x2y
0
1
∫0
y
∫0
1
∫ 0
y
dy = 2y3 dy =120
1
∫
B. Set up and evaluate the integral in A with the order of integration reversed.
4xydydx
x
1
∫0
1
∫ = 2xy2
0
1
∫ x
1
dx = (2x − 2x3
0
1
∫ )dx = x2 − 12 x4
0
1=
12
3.A. Evaluate
23
xydydxR∫∫ R ={(x,y) :0 ≤ x ≤ 2,0 ≤ y ≤ 3,0 < 3x + 2y ≤ 6}
230
3− 32
x
∫0
2
∫ xydydx = 130
2
∫ xy2
0
3− 3x2 dx = 1
30
2
∫ x(3− 32 x)2dx = (3x − 3x2
0
2
∫ + 34 x3)dx
= 32 x2 − x3 + 3
16 x4
0
2= 1
B. Set up and evaluate the integral in part A with the order of integration reversed.
230
2− 23
y
∫0
3
∫ xydxdy = 130
3
∫ x2y0
2− 23
ydy = 1
30
3
∫ (2 − 23 y)2 ydy = 4
3 (y − 230
3
∫ y2 + 19 y3)dy
= 43 ( y2
2 − 19 y3 + 1
36 y4 )0
3
= 1
4. A.Evaluate
e−(x+2y) dydx
R∫∫ R ={(x,y) :0 ≤ 2x + y ≤1}
e−(x+2y) dydx0
1−2x
∫0
12∫ = − 1
2 e−(x+2y)
0
12∫ 0
1−2x
dx = (− 120
12∫ e−(2−3x) + 1
2 e−x )dx
= − 16 e−(2−3x) − 1
2 e−x
0
12 = 1
2 +16 e−2 − 2
3 e− 12 = .118
B. Set up and evaluate the integral in A with the order of integration reversed.
e−(x+2y) dx dy = −e−(x+2y)
0
1
∫0
1−y2∫0
1
∫ 0
1−y2
dy = (−e( 3y+12
)
0
1
∫ + e−2y )dy
= 23 e−( 3y+1
2) − 1
2 e−2y
0
1
= 16 e−2 − 2
3 e− 12 + 1
2 = .118
5. Evaluate
e− 1
2(x2+y2 )
0
∞
∫0
∞
∫ dxdy by using polar coordinates.
e− 12(x2+y2 )
0
∞
∫0
∞
∫ dxdy = e− 12r2
0
π2∫0
∞
∫ rdθdr =π2 a→∞lim r
0
a
∫ e− 12r2
dr =π2 a→∞lim(−e− r2
2 )0
a
=π2
Note that if
I = e−x2 / 2 dx0
∞
∫ , I2 = e−x2 / 2 dx0
∞
∫ e−y2 / 2 dy0
∞
∫ = e− x2+y2
2
0
∞
∫0
∞
∫ dxdy =π2
I =π2
2I = e−x2 / 2
−∞
∞
∫ dx = 2π2= 2π
so 12π
e−x2 / 2
−∞
∞
∫ dx = 1
and
letting z=x − µσ
,so dz=dxσ
12πσ−∞
∞
∫ exp −(x − µ)2
2σ2
⎡
⎣⎢
⎤
⎦⎥dx =
12π
e−z2 / 2
−∞
∞
∫ dx = 1
,
Thus, the normal probability density function is indeed a bonafide pdf.
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