Example triple integral
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The document provides 4 examples of calculating triple integrals over different regions. Example 1 calculates a triple integral over a region bounded by a paraboloid and plane in rectangular coordinates, then evaluates it in polar coordinates. Example 2 calculates a triple integral over a hemisphere in spherical coordinates. Example 3 finds the volume under a paraboloid and above a rectangle using a double integral. Example 4 calculates a triple integral over a tetrahedron bounded by 4 planes.
![1 1 1 π
y 2 z 2 dV = ( θ − sin(4θ))|2π =
0 .
E 86 2 8 86
Example 2.Calculate the value of multiple integral H
z 3 x2 + y 2 + z 2 dV
where H is the solid hemisphere with the center at the origin, radius 1, that
lies above xy-plane.
Solution.
Using the spherical coordinates we can describe the hemisphere H as
π
H = {(x, y, z)|0 ≤ ρ ≤ 1, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ }.
2
Then we have
π
2 2π 1
3
z x2 + y 2 + z 2 dV = ρ3 cos3 φρ3 sinφdρdθdφ
H 0 0 0
π π
2π 1 2π
2
6 3
2 ρ7
= ρ cos φsinφdρdθdφ = cos3 φsinφ|1 dθdφ
0
0 0 0 0 0 7
π π
2π
1 2 2π 2 2π π
= cos3 φsinφdθdφ = cos3 φsinφdφ = − cos4 φ|02
7 0 0 7 0 28
2π
= .
28
Example 3. Find the volume of the given solid. Under the paraboloid
z = x2 + 4y 2 and above the rectangle R = [0, 2] × [1, 4].
Solution.
2 4 2
4
volume = (x2 +4y 2 )dA = (x2 +4y 2 )dydx = (x2 y+ y 3 )|4 dx =
1
R 0 1 0 3
2
(3x2 + 84)dx = (x3 + 84x)|2 = 8 + 168 = 176.
0
0
Example 4. Calculate the multiple integral T
ydV , where T is the
tetrahedron bounded by the planes x = 0, y = 0, z = 0 and 2x + y + z = 2.
Solution.
2](https://image.slidesharecdn.com/exampletripleintegral-121110065749-phpapp02/85/Example-triple-integral-2-320.jpg)
Example triple integral
- 1. TRIPLE INTEGRALS Example 1.Calculate the value of multiple integral E y 2 z 2 dV where 2 2 E is bounded by the paraboloid x = 1 − y − z and plane x = 0. Solution. In the rectangular coordinate system we can describe our region E as E = {(x, y, z)|(y, z) ∈ D, 0 ≤ x ≤ 1 − y 2 − z 2 } where D = {(x, y)|y 2 + z 2 ≤ 1}. Therefore 2 −z 2 y 2 z 2 dV = y 2 z 2 x|1−y 0 dA = y 2 z 2 (1 − y 2 − z 2 )dA. E D D In order to evaluate the double integral over D it is better to use the polar coordinate system. We set z = rcosθ, y = rsinθ. In the polar coordinate system we can describe the region D as D = {(r, θ)|0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}. So we have 2π 1 2π 1 2 2 2 2 2 2 2 sin2 (2θ) y z dV = r cos θr sin θ(1−r )rdrdθ = (r5 −r7 ) drdθ = E 0 0 0 0 4 2π 2π r6 r8 sin2 (2θ) 1 1 ( − ) |0 dθ = sin2 (2θ)dθ. 0 6 8 4 86 0 Note that 1 1 sin2 (2θ) = θ − sin(4θ). 2 8 Typeset by AMS-TEX 1
- 2. 1 1 1 π y 2 z 2 dV = ( θ − sin(4θ))|2π = 0 . E 86 2 8 86 Example 2.Calculate the value of multiple integral H z 3 x2 + y 2 + z 2 dV where H is the solid hemisphere with the center at the origin, radius 1, that lies above xy-plane. Solution. Using the spherical coordinates we can describe the hemisphere H as π H = {(x, y, z)|0 ≤ ρ ≤ 1, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ }. 2 Then we have π 2 2π 1 3 z x2 + y 2 + z 2 dV = ρ3 cos3 φρ3 sinφdρdθdφ H 0 0 0 π π 2π 1 2π 2 6 3 2 ρ7 = ρ cos φsinφdρdθdφ = cos3 φsinφ|1 dθdφ 0 0 0 0 0 0 7 π π 2π 1 2 2π 2 2π π = cos3 φsinφdθdφ = cos3 φsinφdφ = − cos4 φ|02 7 0 0 7 0 28 2π = . 28 Example 3. Find the volume of the given solid. Under the paraboloid z = x2 + 4y 2 and above the rectangle R = [0, 2] × [1, 4]. Solution. 2 4 2 4 volume = (x2 +4y 2 )dA = (x2 +4y 2 )dydx = (x2 y+ y 3 )|4 dx = 1 R 0 1 0 3 2 (3x2 + 84)dx = (x3 + 84x)|2 = 8 + 168 = 176. 0 0 Example 4. Calculate the multiple integral T ydV , where T is the tetrahedron bounded by the planes x = 0, y = 0, z = 0 and 2x + y + z = 2. Solution. 2
- 3. 1 2−2x 2−2x−y 1 2−2x 2−2x−y ydV = ydzdydx = zy|0 dydx = T 0 0 0 0 0 1 2−2x (2y − 2xy − y 2 )dydx = 0 0 1 3 1 y 2−2x (2 − 2x)3 (y 2 − xy 2 − )| dx = ((2 − 2x)2 − x(2 − 2x)2 − dx = 0 3 0 0 3 1 4 1 1 (1 − x)3 dx = − (1 − x)4 |1 = . 0 3 0 3 3 3