Q:

Let F(x, y, z) = xi + yj + zk. Evaluate the integral of F along each of the following paths. (a) c(t) = (t, t, t), 0 ≤ t ≤ 5 Incorrect: Your answer is incorrect. (b) c(t) = (2 cos(t), 2 sin(t), 0), 0 ≤ t ≤ 8π Correct: Your answer is correct. (c) c(t) = (2 sin(t), 0, 2 cos(t)), 0 ≤ t ≤ 8π (d) c(t) = (2t2, 3t, t3), −2 ≤ t ≤ 3

Accepted Solution

A:
a.[tex]c(t)=(t,t,t)\implies c'(t)=(1,1,1)[/tex][tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec c=\int_0^5t(\vec\imath+\vec\jmath+\vec k)\cdot(\vec\imath+\vec\jmath+\vec k)\,\mathrm dt=3\int_0^5t\,\mathrm dt=\boxed{\frac{75}2}[/tex]b.[tex]c(t)=(2\cos t,2\sin t,0)\implies c'(t)=(-2\sin t,2\cos t,0)[/tex][tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec c=\int_0^{8\pi}(2\cos t\,\vec\imath+2\sin t\,\vec\jmath)\cdot(-2\sin t\,\vec\imath+2\cos t\,\vec\jmath)\,\mathrm dt=\int_0^{8\pi}0\,\mathrm dt=\boxed0[/tex]c. Same as (b), [tex](x,y,z)[/tex] is essentially replaced with [tex](y,z,x)[/tex].d.[tex]c(t)=(2t^2,3t,t^3)\implies c'(t)=(4t,3,3t^2)[/tex][tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec c=\int_{-2}^3(2t^2\,\vec\imath+3t\,\vec\jmath+t^3\,\vec k)\cdot(4t\,\vec\imath+3\,\vec\jmath+3t^2\,\vec k)=\int_{-2}^3(8t^3+9t+3t^5)\,\mathrm dt=\boxed{485}[/tex]