Let F(x, y, z) = (5ex sin(y))i + (5ex cos(y))j + 7z2k. Evaluate the integral C F · ds, where c(t) = 8 t , t3, exp( t ) , 0 ≤ t ≤ 1. (Note that exp(u) = eu.)

Respuesta :

I'm going to assume this reads

[tex]\vec F(x,y,z)=5e^x\sin y\,\vec\imath+5e^x\cos y\,\vec\jmath+7z^2\,\vec k[/tex]

and the path [tex]C[/tex] is parameterized by

[tex]\vec c(t)=8t\,\vec\imath+t^3\,\vec\jmath+e^t\,\vec k[/tex]

with [tex]0\le t\le1[/tex]. Under this parameterization,

[tex]\vec F(x,y,z)=\vec F(x(t),y(t),z(t))=5e^{8t}\sin(t^3)\,\vec\imath+5e^{8t}\cos(t^3)\,\vec\jmath+7e^{2t}\,\vec k[/tex]

and

[tex]\mathrm d\vec c=\dfrac{\mathrm d\vec c}{\mathrm dt}\,\mathrm dt=(8\,\vec\imath+3t^2\,\vec\jmath+e^t\,\vec k)\,\mathrm dt[/tex]

Then in the integral,

[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec s=\int_0^1\vec F(x(t),y(t),z(t))\cdot\frac{\mathrm d\vec c}{\mathrm dt}\,\mathrm dt[/tex]

[tex]=\displaystyle\int_0^1(7e^{3t}+15e^{8t}t^2\cos(t^3)+40e^{8t}\sin(t^3))\,\mathrm dt\approx\boxed{12586.5}[/tex]

(It's unlikely that an exact answer can be found in terms of elementary functions)

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