Hasil dari [tex]\displaystyle{\int\limits_C {(x^3+y)} \, dS,~x=3t,~y=t^3,~0\leq t\leq 1 }[/tex] adalah 14(2√2 - 1).
Hasil dari [tex]\displaystyle{\int\limits_C {(2x+9z)} \, dS,~x=t,~y=t^2,~z=t^3~0\leq t\leq 1 }[/tex] adalah [tex]\displaystyle{\boldsymbol{\frac{1}{6}\left ( 14\sqrt{14}-1 \right )} }[/tex].
PEMBAHASAN
Untuk menghitung suatu integral garis pada suatu persamaan parameter kurva mulus C dengan batas a ≤ t ≤ b adalah :
Pada bidang xy, dimana x = x(t) dan y = y(t) :[tex]\displaystyle{\int\limits_C {f(x,y)} \, dS=\int\limits^b_a {f[x(t),y(t)]\sqrt{[x'(t)]^2+[y'(t)]^2} } \, dt }[/tex]
Pada ruang xyz, dimana x = x(t), y = y(t), dan z = z(t) : [tex]\displaystyle{\int\limits_C {f(x,y,z)} \, dS=\int\limits^b_a {f[x(t),y(t),z(t)]\sqrt{[x'(t)]^2+[y'(t)]^2+[z'(t)]^2} } \, dt }[/tex]
Verified answer
PEMBAHASAN
Untuk menghitung suatu integral garis pada suatu persamaan parameter kurva mulus C dengan batas a ≤ t ≤ b adalah :
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DIKETAHUI
[tex]\displaystyle{1.~\int\limits_C {(x^3+y)} \, dS,~x=3t,~y=t^3,~0\leq t\leq 1 }[/tex]
[tex]\displaystyle{2.~\int\limits_C {(2x+9z)} \, dS,~x=t,~y=t^2,~z=t^3~0\leq t\leq 1 }[/tex]
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DITANYA
Tentukan hasil integral garisnya.
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PENYELESAIAN
> Soal 1.
[tex]x(t)=3t~\to~x'(t)=3[/tex]
[tex]y(t)=t^3~\to~y'(t)=3t^2[/tex]
a = 0
b = 1
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Maka :
[tex]\displaystyle{\int\limits_C {(x^3+y)} \, dS}[/tex]
[tex]\displaystyle{=\int\limits^b_a {f[x(t),y(t)]\sqrt{[x'(t)]^2+[y'(t)]^2} } \, dt }[/tex]
[tex]\displaystyle{=\int\limits^1_0 {[(3t)^3+t^3]\sqrt{(3)^2+(3t^2)^2} } \, dt}[/tex]
[tex]\displaystyle{=\int\limits^1_0 {(27t^3+t^3)\sqrt{9+9t^4} } \, dt}[/tex]
[tex]\displaystyle{=\int\limits^1_0 {28t^3\sqrt{9(1+t^4)} } \, dt}[/tex]
[tex]-----------[/tex]
Misal :
[tex]u=1+t^4~\to~du=4t^3dt[/tex]
[tex]untuk~t=0~\to~u=1+0^3=1[/tex]
[tex]untuk~t=1~\to~u=1+1^3=2[/tex]
[tex]-----------[/tex]
[tex]\displaystyle{=\int\limits^2_1 {28t^3\sqrt{9u} } \, \left ( \frac{du}{4t^3} \right )}[/tex]
[tex]\displaystyle{=\int\limits^2_1 {7\sqrt{9u} } \, du}[/tex]
[tex]\displaystyle{=\int\limits^2_1 {21u^{\frac{1}{2}} } \, du}[/tex]
[tex]\displaystyle{=\frac{21}{\frac{1}{2}+1}u^{\frac{1}{2}+1}\Bigr|^2_1 }[/tex]
[tex]\displaystyle{=\frac{21}{\frac{3}{2}}u^{\frac{3}{2}}\Bigr|^2_1 }[/tex]
[tex]\displaystyle{=14u\sqrt{u}\Bigr|^2_1 }[/tex]
[tex]\displaystyle{=14\left ( 2\sqrt{2}-1\sqrt{1} \right ) }[/tex]
[tex]\displaystyle{=14\left ( 2\sqrt{2}-1 \right ) }[/tex]
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> Soal 2.
[tex]x(t)=t~\to~x'(t)=1[/tex]
[tex]y(t)=t^2~\to~y'(t)=2t[/tex]
[tex]z(t)=t^3~\to~z'(t)=3t^2[/tex]
a = 0
b = 1
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Maka :
[tex]\displaystyle{\int\limits_C {(2x+9z)} \, dS }[/tex]
[tex]\displaystyle{=\int\limits^b_a {f[x(t),y(t),z(t)]\sqrt{[x'(t)]^2+[y'(t)]^2+[z'(t)]^2} } \, dt }[/tex]
[tex]\displaystyle{=\int\limits^1_0 {[2(t)+9(t^3)]\sqrt{(1)^2+(2t)^2+(3t^2)^2} } \, dt }[/tex]
[tex]\displaystyle{=\int\limits^1_0 {(2t+9t^3)\sqrt{1+4t^2+9t^4} } \, dt }[/tex]
[tex]-----------[/tex]
Misal :
[tex]u=1+4t^2+9t^4~\to~du=(8t+36t^3)dt[/tex]
[tex]untuk~t=0~\to~u=1+4(0)^2+9(0)^4=1[/tex]
[tex]untuk~t=1~\to~u=1+4(1)^2+9(1)^4=14[/tex]
[tex]-----------[/tex]
[tex]\displaystyle{=\int\limits^{14}_1 {(2t+9t^3)\sqrt{u} } \, \left ( \frac{du}{8t+36t^3} \right ) }[/tex]
[tex]\displaystyle{=\int\limits^{14}_1 {(2t+9t^3)\sqrt{u} } \, \left ( \frac{du}{4(2t+9t^3)} \right ) }[/tex]
[tex]\displaystyle{=\frac{1}{4}\int\limits^{14}_1 {u^{\frac{1}{2}} } \, du }[/tex]
[tex]\displaystyle{=\frac{1}{4}\left ( \frac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1} \right )\Bigr|^{14}_1 }[/tex]
[tex]\displaystyle{=\frac{1}{4}\left ( \frac{2}{3}u^{\frac{3}{2}} \right )\Bigr|^{14}_1 }[/tex]
[tex]\displaystyle{=\frac{1}{6}u\sqrt{u}\Bigr|^{14}_1 }[/tex]
[tex]\displaystyle{=\frac{1}{6}\left ( 14\sqrt{14}-1\sqrt{1} \right ) }[/tex]
[tex]\displaystyle{=\frac{1}{6}\left ( 14\sqrt{14}-1 \right ) }[/tex]
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KESIMPULAN
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PELAJARI LEBIH LANJUT
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DETAIL JAWABAN
Kelas : -
Mapel: Matematika
Bab : Integral garis
Kode Kategorisasi: x.x.x