Evaluate the line integral along the curve C.
-$\int _ { C } ( x y + y z ) d s , C$ is the path from $( 1,1,0 )$ to $\left( e ^ { 2 } , e ^ { 2 } , 1 \right)$ given by:
$C _ { 1 } : \mathbf { r } ( \mathrm { t } ) = \mathrm { e } ^ { 2 } \mathrm { t } _ { \mathbf { i } } + \mathrm { e } ^ { 2 } \mathrm { t } \mathbf { j } , 0 \leq \mathrm { t } \leq 1$
$C _ { 2 } : \mathbf { r } ( \mathrm { t } ) = \mathrm { e } ^ { 2 } \mathbf { i } + \mathrm { e } ^ { 2 } \mathbf { j } + 9 \mathrm { t } \mathbf { k } , 0 \leq \mathrm { t } \leq 1$
A) $\frac { \sqrt { 2 } } { 3 } \left( e ^ { 6 } - 1 \right)$
B) $\frac { \sqrt { 2 } } { 3 } \left( e ^ { 6 } - 1 \right) + 9 e ^ { 4 } + 81 e ^ { 2 }$
C) $\frac { \sqrt { 2 } } { 3 } \left( e ^ { 6 } - 1 \right) + 9 e ^ { 4 } + \frac { 81 } { 2 } e ^ { 2 }$
D) $\frac { \sqrt { 2 } } { 3 } \left( e ^ { 6 } - 1 \right) + e ^ { 4 } + \frac { 1 } { 2 } e ^ { 2 }$

Question

Quizplus · Accepted Answer

The integral is split into two parts: along $C_1$ and $C_2$. For $C_1$, the integral evaluates to $\frac{\sqrt{2}}{3}(e^6 - 1)$. For $C_2$, the integral evaluates to $9e^4 + \frac{81}{2}e^2$. Adding these results gives the total integral value.

Evaluate the Line Integral Along the Curve C ∫C(xy+yz)ds,C\int _ { C } ( x y + y z ) d s , C∫C​(xy+yz)ds,C

Evaluate the Line Integral Along the Curve C $\int _ { C } ( x y + y z ) d s , C$