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Evaluate the line integral ∫CF⋅dr, where F(x,y,z)=5sinxi+4cosyj−2xzk and C is given by the vector function r(t)=t3i−t2j+t1k , 0≤t≤1.

User Froosh
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1 Answer

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\mathbf r(t)=x(t)\,\mathbf i+y(t)\,\mathbf j+z(t)\,\mathbf k

\implies\begin{cases}x(t)=t^3\\y(t)=-t^2\\z(t)=t\end{cases}

\implies(\mathrm d\mathbf r)/(\mathrm dt)=3t^2\,\mathbf i-2t\,\mathbf j+\mathbf k

with
0\le t\le1. With this parameterization, the vector field can be written as


\mathbf F(x,y,z)=\mathbf F(x(t),y(t),z(t))=5\sin(t^3)\,\mathbf i+4\cos(t^2)\,\mathbf j-2t^4\,\mathbf k

Now the line integral is given by


\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r=\int_(t=0)^(t=1)\mathbf F(x(t),y(t),z(t))\cdot(\mathrm d\mathbf r(t))/(\mathrm dt)\,\mathrm dt

=\displaystyle\int_0^1(15t^2\sin(t^3)-8t\cos(t^2)-2t^4)\,\mathrm dt

=-5\cos1-4\sin1+\frac{23}5
User Tabrej Khan
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