The line integral $\int $C $x y d s ,$ where C is the curve $x = 4 t , y = t ^ { 2 } , 0 \leq t \leq 2$ is

A) $- \frac { 512 ( 1 + \sqrt { 2 } ) } { 15 }$
B) $\frac { 512 ( 1 + \sqrt { 2 } ) } { 5 }$
C) $\frac { 512 ( 1 - \sqrt { 2 } ) } { 5 }$
D) $\frac { 512 ( 1 + \sqrt { 2 } ) } { 15 }$
E) $\frac { 512 ( 1 - \sqrt { 2 } ) } { 15 }$

Question

The line integral $\int $C $x y d s ,$ where C is the curve $x = 4 t , y = t ^ { 2 } , 0 \leq t \leq 2$ is

A) $- \frac { 512 ( 1 + \sqrt { 2 } ) } { 15 }$
B) $\frac { 512 ( 1 + \sqrt { 2 } ) } { 5 }$
C) $\frac { 512 ( 1 - \sqrt { 2 } ) } { 5 }$
D) $\frac { 512 ( 1 + \sqrt { 2 } ) } { 15 }$
E) $\frac { 512 ( 1 - \sqrt { 2 } ) } { 15 }$

Quizplus · Accepted Answer

To solve the line integral $\int_C xy \, ds$, we first express $x$, $y$, and $ds$ in terms of the parameter $t$. Given $x = 4t$ and $y = t^2$, we find $dx = 4 \, dt$ and $dy = 2t \, dt$. The differential arc length $ds$ is given by $ds = \sqrt{(dx/dt)^2 + (dy/dt)^2} \, dt = \sqrt{16 + 4t^2} \, dt$. Substituting $x$, $y$, and $ds$ into the integral gives $\int_0^2 4t \cdot t^2 \cdot \sqrt{16 + 4t^2} \, dt$, which simplifies to $\int_0^2 4t^3 \sqrt{16 + 4t^2} \, dt$. Solving this integral yields $\frac{512(1 + \sqrt{2})}{15}$, making option D the correct answer.

The Line Integral ∫\int ∫ C xyds,x y d s ,xyds, Where C Is the Curve

The Line Integral $\int$ C $x y d s ,$ Where C Is the Curve