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Tentukan lintasan yang menghasilkan usaha dari medan gaya $\\mathbf{F}=x\\,\\mathbf{i}+y\\,\\mathbf{j}$ terbesar untuk memindahkan partikel dari titik $(0,0)$ ke $(1,1)$.

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Perubahan pengurutan integral yang benar dari integral $\\displaystyle\\int_{-1}^1\\int_{x^2}^1\\int_0^{1-y} dz\\,dy\\,dx$ adalah $\\ldots$.

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Misalkan daerah segi empat $R$ dibatasi oleh $x-2y=0$, $x-2y=\\var{a}$, $3x-y=1$, dan $3x-y=\\var{b}$.

Akan diperoleh $\\displaystyle\\iint_R \\dfrac{x-2y}{3x-y}\\, dA=P\\ln(\\var{b})$. Maka nilai $P=$[[0]].

Misalkan turunan berarah fungsi $f(x,y)=xy+y^2$ pada titik $P(3,2)$ bernilai $0$ pada arah vektor satuan $\\mathbf{u}=\\langle a,b \\rangle$.

Maka $ab=\\ldots$.

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Misalkan lintasan $C$ merupakan batas dari daerah dalam kordinat polar yang memenuhi pertidaksamaan $1\\leq r \\leq \\var{r}$ dan $0\\leq \\theta \\leq \\pi$. Jika $\\mathbf{F}=\\left( \\tan^{-1}\\left(\\dfrac{y}{x}\\right) \\right)\\,\\mathbf{i}+\\ln(x^2+y^2)\\,\\mathbf{j}$,

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maka $\\displaystyle\\oint_C \\mathbf{F} \\cdot \\mathbf{n} \\, ds=$ [[0]] dan $\\displaystyle\\oint_C \\mathbf{F} \\cdot \\mathbf{T} \\, ds=$ [[1]].

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Volume terbesar dari balok yang berada pada oktan pertama dengan tiga sisi berada pada bidang-$xy$, bidang-$xz$, dan bidang-$yz$ serta satu titik sudut balok berada pada bidang $\\dfrac{x}{\\var{a}}+\\dfrac{y}{\\var{b}}+\\dfrac{z}{\\var{c}}=1$ adalah $\\ldots$.

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Diberikan medan vektor $\\mathbf{F}(x,y,z)=xe^y\\,\\mathbf{i}+(z-e^y)\\,\\mathbf{j}-x^2y^2\\,\\mathbf{k}$ dan $S$ merupakan daerah yang dibatasi oleh permukaan setengah bola $z=\\sqrt{{\\var{rkuadrat}}-x^2-y^2}$ untuk $z\\geq 0$ dan bidang $z=0$.

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Tentukan fluks yang keluar dari medan vektor $\\mathbf{F}$ dari seluruh permukaan $S$.

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Fluks yang keluar dari medan vektor $\\mathbf{F}$ dari alas bidang $z=0$ daerah $S$ adalah $A\\pi$, dengan $A=$ [[0]].

Fluks yang keluar dari medan vektor $\\mathbf{F}$ dari permukaan $z=\\sqrt{{\\var{rkuadrat}}-x^2-y^2}$ daerah $S$ adalah $B\\pi$, dengan $B=$ [[0]].

jika $\\mathbf{n}_1$ menyatakan vektor normal yang mengarah ke atas, maka $\\displaystyle\\iint_S \\text{curl}(\\mathbf{F})\\cdot \\mathbf{n}_1\\, dS=$[[0]]$\\pi$.
jika $\\mathbf{n}_2$ menyatakan vektor normal yang mengarah ke bawah, maka $\\displaystyle\\iint_S \\text{curl}(\\mathbf{F})\\cdot \\mathbf{n}_2\\, dS=$[[1]]$\\pi$.

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Waktunya 5 menit lagi ya.

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Selamat datang!

Silakan gunakan set soal ini dengan bijak. Kalian bisa mencoba set soal ini terus-menerus.

Set soal ini dapat digunakan oleh siapapun secara gratis.

Petunjuk pengerjaan soal:

Jawaban dapat dituliskan dalam bentuk desimal maupun pecahan.
Jika jawaban dituliskan dalam bentuk desimal, pisahkan dengan tanda titik dan bulatkan hingga dua angka di belakang titik. Contoh: ketik 0.67 untuk menjawab \$\\frac{2}{3}\$.
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Jika ada kunci jawaban yang salah, kalian dapat memberitahu kami di Instagram @meongmeongproject atau OA Line @eog7710d.

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