$\text { Evaluate } \int _ { C _ { 1 } } y ^ { 3 } d x + \left( 27 x - x ^ { 3 } \right) d y \text {, where } C _ { 1 } \text { is the unit circle given by }$
$\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } , 0 \leq t \leq 2 \pi$
A) $\frac { 51 \pi } { 2 }$
B) $\frac { 49 \pi } { 2 }$
C) $\frac { 57 \pi } { 2 }$
D) $\frac { 27 \pi } { 2 }$
E) $\frac { 59 \pi } { 2 }$

Question

Quizplus · Accepted Answer

The integral is a line integral over a closed curve, and the vector field is conservative. The integral of a conservative field over a closed path is zero. However, if we calculate directly, the integral evaluates to $\frac{51 \pi}{2}$.

r(t)=cos⁡ti+sin⁡tj,0≤t≤2π\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } , 0 \leq t \leq 2 \pir(t)=costi+sintj,0≤t≤2π

$\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } , 0 \leq t \leq 2 \pi$