MAT 21D – Lecture 5 – Triple Integrals in Rectangular Coordinates PDF

Preview

Summary

Shiqian Ma...

Description

MAT 21D – Lecture 5 – Triple Integrals in Rectangular Coordinates 

The double integral in rectangular or Cartesian coordinates over a region R,

∬ f ( x , y ) dxdy can be converted into polar form over a region G, ∬ f ( rcosθ , rsinθ) rdrdθ where x=rcosθ and y=rsinθ .  

R and G represent the same region. Example: Convert

1

√ 1−x 2

0

0

∫ ∫ ( x 2 + y 2 ) dydx

into polar form and evaluate it.

The region R: 0 ≤ x ≤1, 0 ≤ y ≤ √ 1−x 2 . Based on the sketch of the region R, our π new region G: 0 ≤θ ≤ , 0≤ r ≤1 ⟹ 2 1 4 π 2

π

()

π 1 θ= 1 π 1 4 r=1 = (¿r ¿ r=0 )dθ=∫ dθ= θ ¿θ=02 = 4 2 8 . 4 4 0 π 2

1

π 2 1

π 2

0

0

0 0

0

∫ ∫ ( r 2 ) r drdθ=∫ ∫r 3 drdθ=∫ ¿



Example 2: Compute the volume bounded above by z=9−x 2− y 2 below by the unit disk in the x-y plane. The unit disk refers to the unit circle of radius 1. Therefore, the region, G: 0 ≤θ ≤ 2 π , 0 ≤ r≤ 1 .

2π 1

2π 1

2π

2π

17 17 θ =2 π 17 9 2 1 4 r =1 V =∫ ∫( 9−r 2 ) rdrdθ=∫ ∫ (9 r −r 3 ) drdθ=∫( r − r )¿r =0 dθ=∫ dθ= θ ¿θ =0 = ( 2 π ) = 4 4 2 4 4 0 0 0 0 0 0 .

  

Suppose we have a function, f (x , y , z ) . Then

∭ f ( x , y, z ) dxdydz

. Triple

integrals are studied because not all regions lie on the x-y plane. Fubini’s Theorem applies since integration can change either, for example, from dxdzdy or dydzdx or dydydx . D is a bounded region in the xyz space. The volume over the region D, vol ( D )=∭ dV =∭ dxdydz . To compute the volume, use the following steps: o 1) Sketch the region, D. o 2) Find z limits where z=f 1 (x , y ) , z=f 2 (x , y ) . o 3) Find the x and y limits. To find those limits, only examine the region R and use a double integral. b y= g 2(x) z=f 2(x , y)



Then the triple integral becomes

∫ ∫

∫

f ( x , y , z ) dzdydx .

a y= g 1(x) z=f 1(x , y)



Example: Find the volume of a region, D enclosed by two surfaces: z=x 2+3 y 2 and z=8−x 2− y 2 . From the sketch of region, D we can easily find the z limits. The lower bound is just equal to z=x 2+ 3 y 2 and the upper bound is equal to z=8−x 2− y 2 . To find the region R, begin by setting z=x 2+ 3 y 2 and z=8−x 2− y 2 equal to

each other such that x 2+3 y 2= 8− x 2− y 2 ⟺ 2 x 2 +4 y 2=8 ⟺ x2 +2 y 2=4 .

√

√

2 2 Based on the sketch for region R, the y limits are − 4−x ≤ y ≤ 4−x 2 2 the x limits are −2≤ x ≤ 2 . Thus, the volume

2

V =∫ −2

√

4 −x 2

2 2

8− x − y

∫

∫

√

− 4− x 2

2

2

x +3 y

2

2

dzdydx=∫ −2

2

√

4 −x 2

2

2

∫

√

− 4− x 2

8−x − y −(x +3 y )dydx=∫ 2

2

2

2

−2 −

2

√

4−x 2

and

2

2

∫

√

4 −x 2

(8−2 x −4 y )dydx=∫ 2

2

−

2

. Applying trigonometric substitution, let x=2 sinθ ⟹ dx =2 cosθdθ . When x = 2, then θ= π 2

V =∫ −π 2

(

π 2

and when x = -2, θ= π 2

π 2

) 2 cosθdθ=∫ 163 (2 cos θ) 2 cosθdθ= ∫ 163 (2 √2 cos θ )2 cosθdθ= 163 4 √

2 3 2

16 4−(2 sinθ) 3 2

−π . Then, 2 2

−π 2

3 2

3

−π 2

.



The average value of f (x , y , z ) over D, A v =

1 f ( x , y , z ) dV . vol of D ∭...