Answer to Question #220118 in Calculus for hari

Question

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

Answer 1

$$\vec F(x, y, z)=e^{2x}\vec i+z(y+1)\vec j+z^3\vec k$$

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

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

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

$$\vec F(\vec r(t))\cdot\vec {r'}(t)=3t^2e^{2t^3}+e^t(t-2)+e^{4t}$$

$$W=\int_C\vec F\cdot d\vec r=\displaystyle\int_{0}^{2}(3t^2e^{2t^3}+e^t(t-2)+e^{4t})dt$$

$$=\bigg[\dfrac{1}{2}e^{2t^3}+(t-3)e^t+\dfrac{1}{4}e^{4t}\bigg]\begin{matrix} 2 \\ 0 \end{matrix}$$

$$=\dfrac{1}{2}e^{16}-e^2+\dfrac{1}{4}e^{8}-(\dfrac{1}{2}-3+\dfrac{1}{4})$$

$$=\dfrac{1}{2}e^{16}-e^2+\dfrac{1}{4}e^{8}+\dfrac{9}{4}$$

$$W=\dfrac{1}{2}e^{16}-e^2+\dfrac{1}{4}e^{8}+\dfrac{9}{4}$$