Question #220118

find workdone integral of F



d

~

r

where

~

F

(

x

,

y

,

z

)

=

e

2

x

~

i

+

z

(

y

+

1

)

~

j

+

z

3

~

k

along the curve

C

~

r

(

t

)

=

t

3

~

i

+


(

1

3

t

)

~

j

+

e

t

~

k

for 0

t

2



1
Expert's answer
2021-07-26T09:45:45-0400
F(x,y,z)=e2xi+z(y+1)j+z3k\vec F(x, y, z)=e^{2x}\vec i+z(y+1)\vec j+z^3\vec k

C:r(t)=t3i+(1t)j+etk,0t2C:\vec r(t)=t^3\vec i+(1-t)\vec j+e^t\vec k, 0\leq t\leq 2

r(t)=3t2ij+etk\vec {r'}(t)=3t^2\vec i-\vec j+e^t\vec k

F(r(t))=e2t3i+et(1t+1)j+e3tk\vec F(\vec r(t))=e^{2t^3}\vec i+e^t(1-t+1)\vec j+e^{3t}\vec k

F(r(t))r(t)=3t2e2t3+et(t2)+e4t\vec F(\vec r(t))\cdot\vec {r'}(t)=3t^2e^{2t^3}+e^t(t-2)+e^{4t}


W=CFdr=02(3t2e2t3+et(t2)+e4t)dtW=\int_C\vec F\cdot d\vec r=\displaystyle\int_{0}^{2}(3t^2e^{2t^3}+e^t(t-2)+e^{4t})dt

=[12e2t3+(t3)et+14e4t]20=\bigg[\dfrac{1}{2}e^{2t^3}+(t-3)e^t+\dfrac{1}{4}e^{4t}\bigg]\begin{matrix} 2 \\ 0 \end{matrix}

=12e16e2+14e8(123+14)=\dfrac{1}{2}e^{16}-e^2+\dfrac{1}{4}e^{8}-(\dfrac{1}{2}-3+\dfrac{1}{4})

=12e16e2+14e8+94=\dfrac{1}{2}e^{16}-e^2+\dfrac{1}{4}e^{8}+\dfrac{9}{4}


W=12e16e2+14e8+94W=\dfrac{1}{2}e^{16}-e^2+\dfrac{1}{4}e^{8}+\dfrac{9}{4}



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