INTEGRAL LIPAT
INTEGRAL LIPAT TIGA
Bentuknya : [pic]
f (x,y,z) didefiniskan pada ruang tertutup V,
V dibagi atas paralelepipedum tegak lurus oleh bidang-bidang sejajar bidang koordinat. Paralelepipedum dalam V kita beri nomor 1 sampai n,
Paralelepipedum ke-i mempunyai volume ΔiV. Integral tripel (integral lipat tiga) diperoleh dari limit dari jumlah :
[pic]= lim [pic](xi*, yi*, zi*) ΔiV
Jika n ( ∞, sedang diagonal maksimum dari ΔiV ( 0, titik (xi*, yi*, zi*) dipilih sembarang dalam paralelepipedum ke-i
Adanya suatu limit yang unik dapat ditunjukkan, jika f (x, y, z) kontinu di V.
Teori sederhana berlaku untuk ruang tertutup V yang dilukiskan sebagai berikut : x1 < x < x2 , y1 (x) < y < y2 (x) dan z1 (x,y) < z < z2 (x,y), sehingga :
[pic] = [pic] [pic][pic][pic] f (x,y,z) dz dy dx
Integral ini dapat dianggap = volume dalam R4 = hypervolume
• Jika f (x,y,z) = 1 maka
[pic] adalah merupakan volume dari V
• Jika f = massa jenis benda,
maka M = massa benda = [pic]
CONTOH SOAL
1. Hitunglah [pic]dengan f (x,y,z) = x² + y² + z² dan R daerah dibatasi oleh
x+y+z = a, (a>0), x=0, y=0, z=0.
Jawab :
[pic]
[pic]
= [pic] [pic] [pic] x² + y² + z² dz dy dx
= … = [pic][pic]
2. Hitunglah [pic] dimana f (x, y, z) = x² + y2 + z² dan
V dibatasi oleh x + y + z = 5, x = 0, y = 0 dan z = 0
Jawab :
[pic] [pic] [pic] x² + y² + z² dz dy dx
= [pic] [pic] [pic] dy dx
= [pic] [pic][pic] dy dx
= [pic] [pic][pic] dy dx
= [pic] [pic] dx
= [pic] [pic] dx
= [pic] [pic] dx
= [pic][pic]
= …
= [pic]
3. Hitunglah volume dari R yang dibatasi oleh silinder z = 4-x² dengan
bidang-bidang x = 0, y = 0, y = 6 dan z = 0
Jawab :
V = [pic]
= [pic] [pic] [pic] dz dy dx
= [pic] [pic] (4-x²) dy dx
= [pic] [pic] dx
= 6 [pic] [pic] dx
= 6 . [pic]
= 6. ( 8 - [pic])
= 6. [pic] = [pic] = 32
4. Tentukan volume benda yang dibatasi oleh permukaan z = 9 - x² - y² dan bidang z =0
Jawab :
V = 4 [pic] [pic] [pic] dz dx dy
= 4 [pic] [pic] 9 - x² - y 2 dx dy
= 4 [pic] 9x - [pic]x3 - y²x [pic] dy
= 4 [pic] 9 [pic] - [pic](9 - y²)3/2 - y² (9 - y²)1/2 dy
Misal y = 3 sin θ
dy = 3 cos θ d θ
= 4 [pic]
= 4 [pic]
= 4 [pic]
= 4 . [pic] = [pic]
5. Tentukan volume benda di oktan pertama yang dibatasi oleh paraboloida z = x² + y²,
tabung x² + y² = 4 dan bidang-bidang koordinat
Jawab :
V = [pic] [pic] [pic] dz dx dy
= [pic] [pic] x² + y² dx dy
= [pic] [pic]x3 - y²x [pic] dy
= [pic] [pic](4-y²)[pic]+ y² (4-y²)1/2 dy
Misal y = 2 sin θ
= [pic][pic] 2 cos θ d θ
= [pic][pic] cos 4θ + [pic]16 sin ² θ cos² θ d θ
= [pic] . [pic] [pic] + 16 . [pic] . [pic] - 16 [pic] [pic]
= [pic]+ 4[pic] - 3[pic] = 2[pic]
6. Tentukan pusat massa daerah yang dibatasi oleh silinder parabolik z = 4 - x² dan
bidang-bidang x = 0, y = 0, y = 6 dan z = 0 dengan mengandaikan bahwa rapat massanya tetap sebesar [pic]
Jawab :
Daerahnya lihat gambar dari soal 3
Volume = [pic] = … = 32
M = Massa total = jumlah massa
= [pic] [pic] [pic] [pic] dz dy dx = 32 [pic]
Myz = Jumlah momen terhadap bidang yz
= [pic] [pic] [pic] [pic] x dz dy dx
= [pic] [pic]x (4 - x²) dy dx = …= 24 [pic]
Mxz = Jumlah momen terhadap bidang xz
= [pic] [pic] [pic] [pic] y dz dy dx
= [pic] [pic]y (4 - x²) dy dx = … = 96 [pic]
Mxy = Jumlah momen terhadap bidang xy
= [pic] [pic] [pic][pic] z dz dy dx
= [pic] [pic][pic] (4 - x²)² dy dx = …= [pic][pic]
[pic] [pic] = [pic] = [pic] = [pic]
[pic] = [pic] = [pic] = 3
[pic] = [pic] = [pic] = [pic]
Jadi pusat massa adalah ([pic], 3, [pic])
SOAL-SOAL LATIHAN
1. [pic][pic][pic] 2x – y – z dz dy dx = …
( [pic]
2. [pic][pic][pic] z [pic]² sin θ dz dp dθ = …
( [pic]
3. [pic][pic][pic] x y z dz dy dx = …
( [pic]
4. [pic][pic][pic] [pic]4 sin Φ dp dΦ dθ = …
( 2500 [pic]
5. Hitung volume benda yang dibatasi oleh x² + y² = a², paraboloida z = x² + y² dan
bidang z = 0
( ½ [pic] a4
6. Hitung volume benda yang dibatasi oleh bidang x + y + z = 5,
bidang z = 0, y = 0, x = 0
( [pic]
7. Tentukan volume yang dibatasi oleh paraboloida x² + y² = z dan
bidang z = 4
( 8[pic]
-----------------------
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