Evaluate the line integral over the given curve $C$.
$$\int _ { C } \left( 6 x + 2 y ^ { 2 } \right) d s ; C : \mathbf { r } ( t ) = ( t - 6 ) \mathbf { i } + t \mathbf { j } , 0 \leq t \leq 1$$
A) 28
B) $- 26$
C) $40 \sqrt { 2 }$
D) $- \frac { 97 } { 3 } \sqrt { 2 }$

Question

Quizplus · Accepted Answer

To evaluate the line integral $\int _ { C } \left( 6 x + 2 y ^ { 2 } \right) d s$, we first express $x$ and $y$ in terms of $t$ using the given curve $C : \mathbf { r } ( t ) = ( t - 6 ) \mathbf { i } + t \mathbf { j }$, where $0 \leq t \leq 1$. Thus, $x = t - 6$ and $y = t$.Next, we find $ds$, the differential arc length, which is given by $ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$. From the given curve, $\frac{dx}{dt} = 1$ and $\frac{dy}{dt} = 1$, so $ds = \sqrt{1^2 + 1^2} dt = \sqrt{2} dt$.Substituting $x$, $y$, and $ds$ into the integral gives:$$\int_{0}^{1} (6(t-6) + 2t^2) \sqrt{2} dt = \sqrt{2} \int_{0}^{1} (6t - 36 + 2t^2) dt$$Evaluating the integral:$$\sqrt{2} \left[\int (6t - 36) dt + \int 2t^2 dt\right] = \sqrt{2} \left[3t^2 - 36t + \frac{2}{3}t^3\right]_0^1$$$$= \sqrt{2} \left[\left(3(1)^2 - 36(1) + \frac{2}{3}(1)^3\right) - \left(0\right)\right]$$$$= \sqrt{2} \left[3 - 36 + \frac{2}{3}\right] = \sqrt{2} \left[-33 + \frac{2}{3}\right] = \sqrt{2} \left[-\frac{97}{3}\right]$$$$= -\frac{97}{3} \sqrt{2}$$Therefore, the correct answer is $-\frac{97}{3} \sqrt{2}$, which corresponds to choice D.

Evaluate the Line Integral Over the Given Curve CCC ∫C(6x+2y2)ds;C:r(t)=(t−6)i+tj,0≤t≤1\int _ { C } \left( 6 x + 2 y ^ { 2 } \right) d s ; C : \mathbf { r } ( t ) = ( t - 6 ) \mathbf { i } + t \mathbf { j } , 0 \leq t \leq 1∫C​(6x+2y2)ds;C:r(t)=(t−6)i+tj,0≤t≤1

Evaluate the Line Integral Over the Given Curve $C$ $\int _ { C } \left( 6 x + 2 y ^ { 2 } \right) d s ; C : \mathbf { r } ( t ) = ( t - 6 ) \mathbf { i } + t \mathbf { j } , 0 \leq t \leq 1$